3  Topology

So far we have worked in \(\mathbb{R}^n\), where for example we have the notions of open set, continuous function and compact set. Topology is what allows us to extend these notions to arbitrary sets.

Definition 1: Topological space

Let \(X\) be a set and \(\mathcal{T}\) a collection of subsets of \(X\). We say that \(\mathcal{T}\) is a topology on \(X\) if the following 3 properties hold:

• (A1) The sets \(\emptyset, X\) belong to \(\mathcal{T}\),

• (A2) If \(\{A_i\}_{i \in I}\) is an arbitrary family of elements of \(\mathcal{T}\), then \[ \bigcup_{i \in I} A_i \in \mathcal{T}\,. \]

• (A3) If \(A,B \in \mathcal{T}\) then \(A \cap B \in \mathcal{T}\).

Further, we say:

• The pair \((X,\mathcal{T})\) is a topological space.
• The elements of \(X\) are called points.
• The sets in the topology \(\mathcal{T}\) are called open sets.

Remark 2

The intersection property of \(\mathcal{T}\), Property (A3) in Definition 1, is equivalent to the following:

• (A3’) If \(A_1, \ldots, A_M \in \mathcal{T}\) for some \(M \in \mathbb{N}\), then \[ \bigcap_{n=1}^M A_n \in \mathcal{T}\,. \]

The equivalence between (A3) and (A3’) can be immediately obtained by induction.

Warning

Notice:

• The union property (A2) of \(\mathcal{T}\) holds for an arbitrary number of sets, even uncountable!
• The intersection property (A3’) of \(\mathcal{T}\) holds only for a finite number of sets.

There are two main examples of topologies that one should always keep in mind. These are:

• Trivial topology: The topology with the smallest possible number of sets.
• Discrete topology: The topology with the highest possible number of sets.

Definition 3: Trivial topology

Let \(X\) be a set. The trivial topology on \(X\) is the collection of sets \[ \mathcal{T}_{\textrm{trivial}}:= \{ \emptyset , X \} \,. \]

Proof: \(\mathcal{T}_{\textrm{trivial}}\) is a topology on \(X\)

To prove \(\mathcal{T}_{\textrm{trivial}}\) is a topology on \(X\), we need to check the axioms:

• (A1) By definition of \(\mathcal{T}_{\textrm{trivial}}\), we have \(\emptyset , X \in \mathcal{T}_{\textrm{trivial}}\).

• (A2) Assume \(\{A_i\}_{i \in I}\) is an arbitrary family of elements of \(\mathcal{T}_{\textrm{trivial}}\). There are two possibilities

  • If all the sets \(A_i\) are empty, then \(\bigcup_i A_i = \emptyset \in \mathcal{T}_{\textrm{trivial}}\).
  • If \(A_i = X\) for at least one index \(i\), then \(\bigcup_i A_i = X \in \mathcal{T}_{\textrm{trivial}}\).

  In both cases, \(\bigcup_i A_i \in \mathcal{T}_{\textrm{trivial}}\), so that (A2) holds.

• (A3) Assume \(A,B \in \mathcal{T}_{\textrm{trivial}}\). We have 3 cases:

  • \(A=B= \emptyset\). Then \(A \cap B = \emptyset \in \mathcal{T}_{\textrm{trivial}}\).
  • \(A = X\) and \(B = \emptyset\). Then \(A \cap B = \emptyset \in \mathcal{T}_{\textrm{trivial}}\).
  • \(A = B=X\). Then \(A \cap B = X \in \mathcal{T}_{\textrm{trivial}}\).

  In all the 3 cases we have \(A \cap B \in \mathcal{T}_{\textrm{trivial}}\), so that (A3) holds.

Therefore \(\mathcal{T}_{\textrm{trivial}}\) is a topology on \(X\).

Definition 4: Discrete topology

Let \(X\) be a set. The discrete topology on \(X\) is the collection of all subsets of \(X\) \[ \mathcal{T}_{\textrm{discrete}}:= \{ A \, \colon \,A \subseteq X \} \,. \]

Proof: \(\mathcal{T}_{\textrm{discrete}}\) is a topology on \(X\)

To prove \(\mathcal{T}_{\textrm{discrete}}\) is a topology on \(X\), we need to check the axioms:

• (A1) We have \(\emptyset , X \in \mathcal{T}_{\textrm{discrete}}\), since \(\emptyset\) and \(X\) are subsets of \(X\).

• (A2) The arbitrary union of subsets of \(X\) is still a subset of \(X\). Therefore \(\bigcup A_i \in \mathcal{T}_{\textrm{discrete}}\) whenever \(A_i \in \mathcal{T}_{\textrm{discrete}}\) for all \(i \in I\).

• (A3) The intersection of two subsets of \(X\) is still a subset of \(X\). Therefore \(A \cap B \in \mathcal{T}_{\textrm{discrete}}\) whenever \(A, B \in \mathcal{T}_{\textrm{discrete}}\).

Therefore \(\mathcal{T}_{\textrm{discrete}}\) is a topology on \(X\).

We anticipated that topology is the extension of familiar concepts of open set, continuity, etc. that we have in \(\mathbb{R}^n\). Let us see how the usual definition of open set of \(\mathbb{R}^n\) can fit in our new abstract framework of topology.

Definition 5: Open set of \(\mathbb{R}^n\)

Let \(A \subseteq \mathbb{R}^n\). We say that the set \(A\) is open if it holds: \[ \forall \, \mathbf{x}\in A \,, \,\, \exists \, r > 0 \, \text{ s.t. } \, B_r(\mathbf{x}) \subseteq A \,, \tag{3.1}\] where \(B_r(\mathbf{x})\) is the ball of radius \(r>0\) centered at \(\mathbf{x}\) \[ B_r(\mathbf{x}) := \{ \mathbf{y}\in \mathbb{R}^n \, \colon \,\left\| \mathbf{y}- \mathbf{x} \right\| < r \} \,, \] and the Euclidean norm of \(\mathbf{x}\in \mathbb{R}^n\) is defined by \[ \| \mathbf{x}\| := \sqrt{ \sum_{i=1}^n x_i^2 } \,. \]

See Figure 3.1 for a schematic picture of an open set.

Figure 3.1: The set \(A \subseteq \mathbb{R}^n\) is open if for every \(\mathbf{x}\in A\) there exists \(r>0\) such that \(B_r(\mathbf{x}) \subseteq A\).

Definition 6: Euclidean topology of \(\mathbb{R}^n\)

The Euclidean topology on \(\mathbb{R}^n\) is the collection of sets \[ \mathcal{T}_{\textrm{euclid}}:= \{ A \, \colon \,A \subseteq \mathbb{R}^n \,, \,\, A \, \mbox{ is open} \} \,. \]

Proof: \(\mathcal{T}_{\textrm{euclid}}\) is a topology on \(\mathbb{R}^n\)

To prove \(\mathcal{T}_{\textrm{euclid}}\) is a topology on \(\mathbb{R}^n\), we need to check the axioms:

• (A1) We have \(\emptyset , \mathbb{R}^n \in \mathcal{T}_{\textrm{euclid}}\): Indeed \(\emptyset\) is open because there is no point \(\mathbf{x}\) for which (3.1) needs to be checked. Moreover, \(\mathbb{R}^n\) is open because (3.1) holds with any radius \(r>0\).

• (A2) Let \(A_i \in \mathcal{T}_{\textrm{euclid}}\) for all \(i \in I\). Define the union \(A =\bigcup_i A_i\). We need to check that \(A\) is open. Let \(\mathbf{x}\in A\). By definition of union, there exists an index \(i_0 \in I\) such that \(\mathbf{x}\in A_{i_0}\). Since \(A_{i_0}\) is open, by (3.1) there exists \(r>0\) such that \(B_r(\mathbf{x}) \subseteq A_{i_0}\). As \(A_{i_0} \subseteq A\), we conclude that \(B_r(\mathbf{x}) \subseteq A\), so that \(A \in \mathcal{T}_{\textrm{euclid}}\).

• (A3) Let \(A, B \in \mathcal{T}_{\textrm{euclid}}\). We need to check that \(A \cap B\) is open. Let \(\mathbf{x}\in A \cap B\). Therefore \(\mathbf{x}\in A\) and \(\mathbf{x}\in B\). Since \(A\) and \(B\) are open, by (3.1) there exist \(r_1,r_2>0\) such that \(B_{r_1}(\mathbf{x}) \subseteq A\) and \(B_{r_2}(\mathbf{x}) \subseteq B\). Set \(r := \min\{ r_1,r_2\}\). Then \[ B_r(\mathbf{x}) \subseteq B_{r_1}(\mathbf{x}) \subseteq A \,, \quad B_r(\mathbf{x}) \subseteq B_{r_2}(\mathbf{x}) \subseteq B \,, \] Hence \(B_r(\mathbf{x}) \subseteq A \cap B\), showing that \(A \cap B \in \mathcal{T}_{\textrm{euclid}}\).

This proves that \(\mathcal{T}_{\textrm{euclid}}\) is a topology on \(\mathbb{R}^n\).

Let us make a basic, but useful, observation: balls in \(\mathbb{R}^n\) are open for the Euclidean topology.

Proposition 7: \(B_r(\mathbf{x})\) is an open set of \(\mathcal{T}_{\textrm{euclid}}\)

Let \(\mathbb{R}^n\) be equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\). Let \(r>0\) and \(\mathbf{x}\in \mathbb{R}^n\). Then \(B_r(\mathbf{x}) \in \mathcal{T}_{\textrm{euclid}}\).

Proof

To prove that \(B_r(\mathbf{x}) \in \mathcal{T}_{\textrm{euclid}}\), we need to show that \(B_r(\mathbf{x})\) satisfies (3.1). Therefore, let \(\mathbf{y}\in B_r(\mathbf{x})\). In particular \[ \left\| \mathbf{x}- \mathbf{y} \right\|<r \,. \tag{3.2}\] Define \[ \varepsilon:= r - \left\| \mathbf{x}- \mathbf{y} \right\| \,. \] Note that \(\varepsilon>0\) by (3.2). We claim that \[ B_{\varepsilon} (\mathbf{y}) \subseteq B_r (\mathbf{x}) \,, \tag{3.3}\] see Figure 3.2. Indeed, let \(\mathbf{z}\in B_{\varepsilon}(\mathbf{y})\). By triangle inequality we have \[ \| \mathbf{z}- \mathbf{x}\| \leq \left\| \mathbf{x}- \mathbf{y} \right\| + \left\| \mathbf{y}- \mathbf{z} \right\| < \left\| \mathbf{x}- \mathbf{y} \right\| + \varepsilon= r \,, \] where we used that \(\left\| \mathbf{y}- \mathbf{z} \right\| < \varepsilon\) and the definition of \(\varepsilon\). Hence \(\mathbf{z}\in B_r(\mathbf{x})\), proving (3.3). Thus, \(B_r(\mathbf{x})\) satisfies (3.1), ending the proof.

Figure 3.2: The ball \(B_{\varepsilon}(\mathbf{y})\) is contained in \(B_r(\mathbf{x})\) if \(\varepsilon:= r - \left\| \mathbf{x}-\mathbf{y} \right\|\).

3.1 Closed sets

The opposite of open sets are closed sets.

Definition 8: Closed set

Let \((X,\mathcal{T})\) be a topological space. A set \(C \subseteq X\) is closed if \[ C^c \in \mathcal{T}\,, \] where \(C^c:= X \smallsetminus C\) is the complement of \(C\) in \(X\).

In words, a set is closed if its complement is open.

Warning

There are sets which are neither open nor closed. For example consider \(\mathbb{R}\) equipped with Euclidean topology. Then the interval \[ A:=[0,1) \] is neither open nor closed.

Note: For the moment we do not have the tools to prove this. We will have them shortly.

We could have defined a topology starting from closed sets. We would have had to replace the properties (A1)-(A2)-(A3) with suitable properties for closed sets, as detailed in the following proposition.

Proposition 9

Let \((X,\mathcal{T})\) be a topological space. Properties (A1)-(A2)-(A3) of \(\mathcal{T}\) are equivalent to (C1)-(C2)-(C3), where

• (C1) \(\emptyset, X\) are closed.
• (C2) If \(C_i\) is closed for all \(i \in I\), then \(\bigcap_{i \in I} \, C_i\) is closed.
• (C3) If \(C_1,C_2\) are closed then \(C_1 \cup C_2\) is closed.

Proof

We have 3 points to check:

1. The equivalence between (A1) and (C1) is clear, since \[ \emptyset^c = X \,, \quad X^c = \emptyset \,. \]

2. Suppose \(C_i\) is closed for all \(i \in I\). Therefore \(C_i^c\) is open for all \(i \in I\). By De Morgan’s laws we have that \[ \left( \bigcap_{i \in I} \, C_i \right)^c = \bigcup_{i \in I} \, C_i^c \] showing that \[ \bigcap_{i \in I} \, C_i \, \mbox{ is closed} \quad \iff \quad \bigcup_{i \in I} \, C_i^c \, \mbox{ is open} \,. \] Therefore (A2) and (C2) are equivalent.

3. Suppose \(C_1,C_2\) are closed. Therefore \(C_1^c, C_2^c\) are open. By De Morgan’s laws we have that \[ \left( C_1 \cup C_2 \right)^c = C_1^c \cap C_2^c \] showing that \[ C_1 \cup C_2 \, \mbox{ is closed} \quad \iff \quad C_1^c \cap C_2^c \, \mbox{ is open} \,. \] Therefore (A3) and (C3) are equivalent.

As a consequence of the above proposition, we can define a topology by declaring what the closed sets are. We then need to verify that (C1)-(C2)-(C3) are satisfied by such topology. Let us make an example.

Example 10: The Zariski topology

Let \((\mathbb{K}, + , \cdot)\) be a field. Define \[ X:=\mathbb{K}^n := \{ (a_1, \ldots, a_n) \, \colon \,a_i \in \mathbb{K} \} \,. \] Consider the ring of polynomials with coefficients in the field \[ \mathbb{K}[x_1,\ldots,x_n] \,. \] Therefore \(f \in \mathbb{K}[x_1,\ldots,x_n]\) has the form \[ f(x_1,\ldots,x_n) = \lambda_1 x_1^{k_1} + \ldots + \lambda_n x_n^{k_n} \,, \] where \(\lambda_1, \ldots, \lambda_n\) are given elements of \(\mathbb{K}\) and \(k_1, \ldots, k_n \in \mathbb{N}\). For a collection of polynomials \(I \subset \mathbb{K}[x_1,\ldots,x_n]\) define \[ V(I):= \{ (a_1,\ldots,a_n) \in \mathbb{K}^n \, \colon \, f(a_1,\ldots,a_n) = 0 \,, \,\, \forall \, f \in I \} \,. \] The set \(V(I)\) is called an algebraic set. Define \[ \mathcal{C} := \{ V(I) \, \colon \,I \subset \mathbb{K}[x_1,\ldots,x_n] \} \,. \] The collection \(\mathcal{C}\) is known as the Zariski topology on the space \(\mathbb{K}^n\). This topology provides a natural framework for studying Affine Varieties – generalized surfaces obtained by gluing together algebraic sets of the form \(V(I)\). The area of mathematics studying these objects is known as Algebraic Geometry. For more information, see this Wikipedia page and this paper.

An example of affine variety is the Quadrifolium, which is the curve defined by the polar coordinates equation \(r = \sin(2\theta)\), see Figure 3.3. It can be easily seen that the Quadrifolium is an affine variety in \(\mathbb{R}^2\), which can be described by using just one algebraic set, namely \(V( (x^2+y^2)^3 - 4x^2 y^2)\).

Question. Prove that \(\mathcal{C}\) satisfies (C1), (C2) and (C3).

Solution. Easy check, left by exercise.

Figure 3.3: The Quadrifolium is an affine variety with algebraic set \(V( (x^2+y^2)^3 - 4x^2 y^2)\).

3.2 Comparing topologies

Consider the situation where you have two topologies \(\mathcal{T}_1\) and \(\mathcal{T}_2\) on the same set \(X\). We would like to have some notions of comparison between \(\mathcal{T}_1\) and \(\mathcal{T}_2\).

Definition 11: Comparing topologies

Let \(X\) be a set and let \(\mathcal{T}_1\), \(\mathcal{T}_2\) be topologies on \(X\).

1. \(\mathcal{T}_1\) is finer than \(\mathcal{T}_2\) if \(\mathcal{T}_2 \subseteq \mathcal{T}_1\).

2. \(\mathcal{T}_1\) is strictly finer than \(\mathcal{T}_2\) if \(\mathcal{T}_2 \subsetneq \mathcal{T}_1\).

3. \(\mathcal{T}_1\) and \(\mathcal{T}_2\) are the same topology if \(\mathcal{T}_1 = \mathcal{T}_2\).

Example 12: Comparing \(\mathcal{T}_{\textrm{trivial}}\) and \(\mathcal{T}_{\textrm{discrete}}\)

Let \(X\) be a set. Then \(\mathcal{T}_{\textrm{trivial}} \subseteq \mathcal{T}_{\textrm{discrete}}\).

Another interesting example is given by the cofinite topology on \(\mathbb{R}\). The sets in this topology are open if they are either empty, or coincide with \(\mathbb{R}\) with a finite number of points removed.

Example 13: Cofinite topology on \(\mathbb{R}\)

Question. The cofinite topology on \(\mathbb{R}\) is the collection of sets \[ \mathcal{T}_{\textrm{cofinite}}:= \{ U \subseteq \mathbb{R}\, \colon \,U^c \, \mbox{ is finite, } \mbox{or } U^c = \mathbb{R}\}\,. \]

1. Prove that \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is a topological space.
2. Prove that \(\mathcal{T}_{\textrm{cofinite}}\subseteq \mathcal{T}_{\textrm{euclid}}\).
3. Prove that \(\mathcal{T}_{\textrm{cofinite}}\neq \mathcal{T}_{\textrm{euclid}}\).

Solution. Part 1. Show that the topology properties are satisfied:

(A1) We have \(\emptyset \in \mathcal{T}_{\textrm{cofinite}}\), since \(\emptyset^c = \mathbb{R}\). We have \(\mathbb{R}\in \mathcal{T}_{\textrm{cofinite}}\) because \({\mathbb{R}}^c = \emptyset\) is finite.

(A2) Let \(U_i \in \mathcal{T}_{\textrm{cofinite}}\) for all \(i \in I\), and define \(U := \bigcup_{i \in I} \, U_i\). By the De Morgan’s laws we have \[ U^c = \left( \cup_{i \in I} \, U_i \right)^c = \cap_{i \in I} \, U_i^c \,. \] We have two cases:

1. There exists \(i_0 \in I\) such that \(U_{i_0}^c\) is finite. Then \[ U^c = \cap_{i \in I} U_i^c \subset U_{i_0}^c \,, \] and therefore \(U^c\) is finite, showing that \(U \in \mathcal{T}_{\textrm{cofinite}}\).

2. None of the sets \(U_i^c\) is finite. Therefore \(U_i^c = \mathbb{R}\) for all \(i \in I\), from which we deduce \[ U^c = \cap_{i \in I} U_i^c = \mathbb{R}\quad \implies \quad U \in \mathcal{T}_{\textrm{cofinite}}\,. \]

In both cases, we have \(U \in \mathcal{T}_{\textrm{cofinite}}\), so that (A2) holds.

(A3) Let \(U,V \in \mathcal{T}_{\textrm{cofinite}}\). Set \(A=U \cap V\). Then \[ A^c = U^c \cup V^c \,. \] We have \(2\) possibilities:

1. \(U^c, V^c\) finite: Then \(A^c\) is finite, and \(A \in \mathcal{T}_{\textrm{cofinite}}\).

2. \(U^c = \mathbb{R}\) or \(V^c = \mathbb{R}\): Then \(A^c = \mathbb{R}\), and \(A \in \mathcal{T}_{\textrm{cofinite}}\).

In all cases, we have shown that \(A \in \mathcal{T}_{\textrm{cofinite}}\), so that (A3) holds.

Part 2. Let \(U \in \mathcal{T}_{\textrm{cofinite}}\). We have two cases:

• \(U^c\) is finite. Then \(U^c = \{x_1, \ldots,x_n\}\) for some points \(x_i \in \mathbb{R}\). Up to relabeling the points, we can assume that \(x_i < x_j\) when \(i < j\). Therefore, \[ U = \{x_1, \ldots,x_n\}^c = \bigcup_{i=0}^{n} (x_i,x_{i+1}) \,, \quad x_0 := -\infty,\quad x_{n+1} := \infty \,. \] The sets \((x_i,x_{i+1})\) are open in \(\mathcal{T}_{\textrm{euclid}}\), and therefore \(U \in \mathcal{T}_{\textrm{euclid}}\).

• \(U^c = \mathbb{R}\). Then \(U = \emptyset\), which belongs to \(\mathcal{T}_{\textrm{euclid}}\) by (A1).

In both cases, \(U \in \mathcal{T}_{\textrm{euclid}}\). Therefore \(\mathcal{T}_{\textrm{cofinite}}\subseteq \mathcal{T}_{\textrm{euclid}}\).

Part 3. consider the interval \(U=(0,1)\). Then \(U \in \mathcal{T}_{\textrm{euclid}}\). However \(U^c\) is neither \(\mathbb{R}\), nor finite. Thus \(U \notin \mathcal{T}_{\textrm{cofinite}}\).

3.3 Convergence

We have generalized the notion of open set to arbitrary sets. Next we generalize the notion of convergence of sequences.

Definition 14: Convergent sequence

Let \((X,\mathcal{T})\) be a topological space. Consider a sequence \(\{x_n\} \subseteq X\) and a point \(x \in X\). We say that \(x_n\) converges to \(x_0\) in the topology \(\mathcal{T}\), if the following property holds: \[ \begin{aligned} & \forall \, U \in \mathcal{T}\, \text{ s.t. } \, x_0 \in U\,, \,\, \exists \, N = N(U) \in \mathbb{N}\, \text{ s.t. } \, \\ & x_n \in U \,, \, \forall \, n \geq N \,. \end{aligned} \tag{3.4}\] The convergence of \(x_n\) to \(x_0\) is denoted by \(x_n \to x_0\).

Let us analyze the definition of convergence in the topologies we have encountered so far. We will have that:

• Trivial topology: Every sequence converges to every point.
• Discrete topology: A sequence converges if and only if it is eventually constant.
• Euclidean topology: Topological convergence coincides with classical notion of convergence.

We now precisely state and prove the above claims.

Proposition 15: Convergent sequences in \(\mathcal{T}_{\textrm{trivial}}\)

Let \(X\) be equipped with \(\mathcal{T}_{\textrm{trivial}}\). Let \(\{x_n\} \subseteq X\), \(x_0 \in X\). Then \(x_n \to x_0\).

Proof

To show that \(x_n \to x_0\) we need to check that (3.4) holds. Let \(U \in \mathcal{T}_{\textrm{trivial}}\) with \(x_0 \in U\). We have two cases:

• \(U = \emptyset\): There is nothing to prove, since \(x_0\) cannot be in \(U\).

• \(U = X\): Take \(N=1\). Since \(U = X\), we have \(x_n \in U\) for all \(n \geq 1\).

Thus (3.4) holds for all the sets \(U \in \mathcal{T}_{\textrm{trivial}}\), showing that \(x_n \to x_0\).

Warning

Proposition 15 shows the topological limit may not be unique!

Proposition 16: Convergent sequences in \(\mathcal{T}_{\textrm{discrete}}\)

Let \(X\) be equipped with \(\mathcal{T}_{\textrm{discrete}}\). Let \(\{x_n\} \subseteq X\), \(x_0 \in X\). They are equivalent:

1. \(x_n \to x_0\) in the topology \(\mathcal{T}_{\textrm{discrete}}\).
2. \(\{x_n\}\) is eventually constant: \(\exists \, N \in \mathbb{N}\, \, \text{ s.t. } \, \, x_n = x_0 , \, \forall \, n \geq N\)

Proof

Part 1. Assume that \(x_n \to x_0\). Let \(U = \{x_0\}\). Then \(U \in \mathcal{T}_{\textrm{discrete}}\). Since \(x_n \to x_0\), by (3.4) there exists \(N \in \mathbb{N}\) such that \[ x_n \in U \,, \quad \forall \, n \geq N \,. \] As \(U = \{x_0\}\), we infer \(x_n = x_0\) for all \(n \geq N\). Hence \(x_n\) is eventually constant.

Part 2. Assume that \(x_n\) is eventually equal to \(x_0\), that is, there exists \(N \in \mathbb{N}\) such that \[ x_n = x_0 \,, \quad \forall \, n \geq N \,. \tag{3.5}\] Let \(U \in \mathcal{T}\) be an open set such that \(x_0 \in U\). By (3.5) we have that \[ x_n \in U \,, \quad \forall \, n \geq N \,. \] Since \(U\) was arbitrary, we conclude that \(x_n \to x_0\).

Before proceeding to examining convergence in the Euclidean topology, let us recall the classical definition of convergence in \(\mathbb{R}^n\).

Definition 17: Classical convergence in \(\mathbb{R}^n\)

Let \(\{\mathbf{x}_n\} \subseteq \mathbb{R}^n\) and \(\mathbf{x}_0 \in \mathbb{R}^n\). We say that \(\mathbf{x}_n\) converges \(\mathbf{x}_0\) in the classical sense if \(\left\| \mathbf{x}_n - \mathbf{x}_0 \right\| \to 0\), that is, \[ \forall \, \varepsilon>0 , \, \exists \, N \in \mathbb{N}, \, \, \text{ s.t. } \, \left\| \mathbf{x}_n - \mathbf{x}_0 \right\| < \varepsilon\,, \, \forall \, n \geq N \,. \]

Proposition 18: Convergent sequences in \(\mathcal{T}_{\textrm{euclid}}\)

Let \(\mathbb{R}^n\) be equipped with \(\mathcal{T}_{\textrm{euclid}}\). Let \(\{x_n\} \subseteq \mathbb{R}^n\), \(x_0 \in \mathbb{R}^n\). They are equivalent:

1. \(\mathbf{x}_n \to \mathbf{x}_0\) in the topology \(\mathcal{T}_{\textrm{euclid}}\).
2. \(\mathbf{x}_n \to \mathbf{x}_0\) in the classical sense.

Proof

Part 1. Assume \(\mathbf{x}_n \to \mathbf{x}_0\) with respect to \(\mathcal{T}_{\textrm{euclid}}\). Fix \(\varepsilon>0\) and define \(U := B_{\varepsilon}(\mathbf{x}_0)\). By Proposition 7, we have \(U \in \mathcal{T}\). Moreover \(\mathbf{x}_0 \in U\). As \(\mathbf{x}_n \to \mathbf{x}_0\) with respect to \(\mathcal{T}_{\textrm{euclid}}\), there exists \(N \in \mathbb{N}\) such that \[ \mathbf{x}_n \in U \,, \quad \forall \, n \geq N \,. \] As \(U = B_{\varepsilon}(\mathbf{x}_0)\), the above reads \[ \left\| \mathbf{x}_n - \mathbf{x}_0 \right\| < \varepsilon\,, \quad \forall \, n \geq N \,, \] showing that \(\mathbf{x}_n \to \mathbf{x}_0\) in the classical sense.

Part 2. Assume \(\mathbf{x}_n \to \mathbf{x}_0\) in the classical sense. Let \(U \in \mathcal{T}_{\textrm{euclid}}\) be an arbitrary set such that \(\mathbf{x}_0 \in U\). By definition of Euclidean topology, this means that there exists \(r>0\) such that \(B_r(\mathbf{x}_0) \subseteq U\). As \(\mathbf{x}_n \to \mathbf{x}_0\) in the classical sense, there exists \(N \in \mathbb{N}\) such that \[ \left\| \mathbf{x}_n - \mathbf{x}_0 \right\| < r \,, \quad \forall \, n \geq N \,. \] By definition of \(B_r(\mathbf{x}_0)\), and since \(B_r(\mathbf{x}_0) \subseteq U\), the above is equivalent to \[ \mathbf{x}_n \in B_{r}(\mathbf{x}_0) \subseteq U \,, \quad \forall \, n \geq N \,. \] As \(U\) is arbitrary, we infer \(\mathbf{x}_n \to \mathbf{x}_0\) with respect to \(\mathcal{T}_{\textrm{euclid}}\).

Notation

Since classical convergence in \(\mathbb{R}^n\) agrees with topological convergence with respect to \(\mathcal{T}_{\textrm{euclid}}\), we will just say that \(\mathbf{x}_n \to \mathbf{x}_0\) in \(\mathbb{R}^n\) without ambiguity.

We conclude with a useful proposition which relates convergences when multiple topologies are present.

Proposition 19

Let \(X\) be a set and \(\mathcal{T}_1,\mathcal{T}_2\) be topologies on \(X\). Suppose that \(\mathcal{T}_2 \subseteq \mathcal{T}_1\). Let \(\{x_n\} \subset X\) and \(x_0 \in X\). We have \[ x_n \to x_0 \,\, \mbox{ in } \,\, \mathcal{T}_1 \quad \implies \quad x_n \to x_0 \,\, \mbox{ in } \,\, \mathcal{T}_2 \,. \]

Proof

Assume \(x_n \to x_0\) in \(\mathcal{T}_1\). We need to prove that \(x_n \to x_0\) in \(\mathcal{T}_2\). Therefore, let \(U \in \mathcal{T}_2\) be such that \(x_0 \in U\). Since \(\mathcal{T}_2 \subseteq \mathcal{T}_1\), we have that \(U \in \mathcal{T}_1\). As \(x_n \to x_0\) in \(\mathcal{T}_1\), there exists \(N \in \mathbb{N}\) such that \[ x_n \in U \,, \quad \forall \, n \geq N \,. \tag{3.6}\] Since \(U \in \mathcal{T}_2\), condition (3.6) shows that \(x_n \to x_0\) in \(\mathcal{T}_2\).

3.4 Metric spaces

We will now define a class of topological spaces known as metric spaces.

Definition 20: Distance and Metric space

Let \(X\) be a set. A distance on \(X\) is a function \(d \colon X \times X \to \mathbb{R}\) such that, for all \(x,y,z \in X\) they hold:

• (M1) Positivity: \(d(x,y) \geq 0\) and \(d(x,y) = 0 \, \iff \, x=y\)

• (M2) Symmetry: \(d(x,y) = d(y,x)\)

• (M3) Triangle Inequality: \(d(x,z) \leq d(x,y) + d(y,z)\)

The pair \((X,d)\) is called a metric space.

Definition 21: Euclidean distance on \(\mathbb{R}^n\)

The Euclidean distance over \(\mathbb{R}^n\) is defined by \[ d(\mathbf{x},\mathbf{y}) := \left\| \mathbf{x}- \mathbf{y} \right\| = \left( \sum_{i=1}^n |x_i - y_i|^2 \right)^{1/2} \,, \quad \forall \mathbf{x},\mathbf{y}\in \mathbb{R}^n \,. \]

Proposition 22

Let \(d\) be the Euclidean distance on \(\mathbb{R}^n\). Then \((\mathbb{R}^n,d)\) is a metric space.

Proof

It is trivial that \(d\) satisfies (M1) and (M2). To show (M3), recall the triangle inequality in \(\mathbb{R}^n\): \[ \| \mathbf{x}+ \mathbf{y}\| \leq \| \mathbf{x}\| + \| \mathbf{y}\| \,, \quad \forall \, \mathbf{x}, \mathbf{y}\in \mathbb{R}^n \,. \] Using the above, we obtain \[\begin{align*} d(\mathbf{x},\mathbf{y}) & = \| \mathbf{x}- \mathbf{y}\| \\ & = \| (\mathbf{x}- \mathbf{z}) + (\mathbf{z}- \mathbf{y}) \| \\ & \leq \| \mathbf{x}- \mathbf{z}\| + \| \mathbf{z}- \mathbf{y}\| \\ & = d (\mathbf{x},\mathbf{z}) + d (\mathbf{z},\mathbf{y}) \,. \end{align*}\] Thus, \(d\) satisfies (M3) and \((\mathbb{R}^n,d)\) is a metric space.

Definition 23: \(p\)-distance on \(\mathbb{R}^n\)

Let \(p \in [1,\infty)\). The \(p\)-distance over \(\mathbb{R}^n\) is \[ d_p ( \mathbf{x}, \mathbf{y}) := \left( \sum_{i=1}^n |x_i - y_i|^p \right)^{\frac1p} \,. \] Note that \(d_2\) coincides with the Euclidean distance. For \(p = \infty\) we set \[ d_{\infty} ( \mathbf{x}, \mathbf{y}) := \max_{i = 1 \ldots, n} |x_i - y_i| \,. \]

Proposition 24

Let \(d_p\) be the \(p\)-distance over \(\mathbb{R}^n\), with \(p \in [1,\infty]\). Then \((\mathbb{R}^n,d_p)\) is a metric space.

Proof

Properties (M1)-(M2) hold trivially. The triangle inequality is also trivially satisfied by \(d_{\infty}\). We are left with checking the triangle inequality for \(d_p\) with \(p \in [1,\infty)\). To this end, define \[ \| \mathbf{x}\|_p := \left( \sum_{i=1}^n |x_i|^p \right)^{\frac1p} \,. \] Minkowski’s inequality, see Wikipedia page, states that \[ \| \mathbf{x}+ \mathbf{y}\|_p \leq \| \mathbf{x}\|_p + \| \mathbf{y}\|_p \,, \quad \, \forall \mathbf{x},\mathbf{y}\in \mathbb{R}^n\,. \] Therefore \[\begin{align*} d_p(\mathbf{x},\mathbf{y}) & = \| \mathbf{x}- \mathbf{y}\|_p \\ & = \| (\mathbf{x}- \mathbf{z}) + (\mathbf{z}- \mathbf{y}) \|_p \\ & \leq \| \mathbf{x}- \mathbf{z}\|_p + \| \mathbf{z}- \mathbf{y}\|_p \\ & = d_p (\mathbf{x},\mathbf{z}) + d_p (\mathbf{z},\mathbf{y}) \,, \end{align*}\] proving that \(d_p\) satisfies (M3). Hence \((\mathbb{R}^n,d_p)\) is a metric space.

A metric \(d\) on a set \(X\) naturally induces a topology which is compatible with the metric.

Definition 25: Topology induced by the metric

Let \((X,d)\) be a metric space. The set \(A \subseteq X\) is open if it holds \[ \forall \, x \in U \,, \, \exists \, r \in \mathbb{R}, \, r > 0 \, \, \text{ s.t. } \, \, B_r(x) \subseteq U \,, \] where \(B_r(x)\) is the ball centered at \(x\) of radius \(r\), defined by \[ B_r(x) = \{ y \in X \, \colon \, d(x,y)<r \} \,. \] The topology induced by the metric \(d\) is the collection of sets \[ \mathcal{T}_d = \{ U\, \colon \, U \subseteq X, \, U \text{ open} \}\,. \]

The proof that \(\mathcal{T}_d\) is a topology on \(X\) follows, line by line, the proof that the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\) is indeed a topology, see proof immediately below Definition 6. This is left as an exercise.

Remark 26: Topology induced by Euclidean distance

Consider the metric space \((\mathbb{R}^n,d)\) with \(d\) the Euclidean distance. Then \[ \mathcal{T}_d = \mathcal{T}_{\textrm{euclid}}\,, \] where \(\mathcal{T}_{\textrm{euclid}}\) is the Euclidean topology on \(\mathbb{R}^n\).

The proof is trivial, since the metric \(d\) is the Euclidean distance.

Example 27: Discrete distance

Question. Let \(X\) be a set. The discrete distance is the function \(d \colon X \times X \to \mathbb{R}\) defined by \[ d(x,y) := \begin{cases} 0 & \mbox{ if } \, x = y \\ 1 & \mbox{ if } \, x \neq y \end{cases} \]

1. Prove that \((X,d)\) is a metric space.
2. Prove that \(\mathcal{T}_d = \mathcal{T}_{\textrm{discrete}}\).

Solution. See Question 3 in Homework 3.

The following proposition tells us that balls in a metric space \(X\) are open sets. Moreover balls are the building blocks of all open sets in \(X\). The proof is left as an exercise.

Proposition 28

Let \((X,d)\) be a metric space, \(\mathcal{T}_d\) the topology induced by \(d\). Then:

• For all \(x \in X\), \(r>0\) we have \(B_r(x) \in \mathcal{T}_d\).
• \(U \in \mathcal{T}_d\) if and only if \(\exists \, I\) family of indices s.t. \[ U = \bigcup_{ i \in I } B_{r_i}(x_i) \,, \quad x_i \in X , \,\, r_i > 0 \,. \]

We now define the concept of equivalent metrics.

Definition 29: Equivalent metrics

Let \(X\) be a set and \(d_1,d_2\) be metrics on \(X\). We say that \(d_1\) and \(d_2\) are equivalent if \[ \mathcal{T}_{d_1} = \mathcal{T}_{d_2} \,. \]

The following proposition gives a sufficent condition for the equivalence of two metrics.

Proposition 30

Let \(X\) be a set and \(d_1,d_2\) be metrics on \(X\). Suppose that there exists a constant \(\alpha > 0\) such that \[ \frac{1}{\alpha} \, d_2 (x,y) \leq d_1 (x,y) \leq \alpha \, d_2(x,y) \,, \quad \forall \, x,y \in X \,. \] Then \(d_1\) and \(d_2\) are equivalent metrics.

The proof of Proposition 30 is trivial, and is left as an exercise.

Example 31

Let \(p > 1\). The metrics \(d_p\) and \(d_{\infty}\) on \(\mathbb{R}^n\) are equivalent.

Proof Follows from Proposition 30 and the estimate \[ d_{\infty} (\mathbf{x},\mathbf{y}) \leq d_p (\mathbf{x},\mathbf{y}) \leq n \, d_{\infty} (\mathbf{x},\mathbf{y}) \,, \quad \forall \mathbf{x}\,, \, \mathbf{y}\in \mathbb{R}^n \,. \]

Warning

If two metrics are equivalent, that does not mean they have the same balls. For example the balls of the metrics \(d_1\), \(d_2\) and \(d_{\infty}\) on \(\mathbb{R}^n\) look very different, see Figure 3.4.

Figure 3.4: Balls \(B_r(0)\) for the metrics \(d_2, d_\infty, d_1\) in \(\mathbb{R}^2\).

We can characterize the convergence of sequences in metric spaces.

Proposition 32: Convergence in metric space

Suppose \((X,d)\) is a metric space and \(\mathcal{T}_d\) the topology induced by \(d\). Let \(\{x_n\} \subseteq X\) and \(x_0 \in X\). They are equivalent:

1. \(x_n \to x_0\) with respect to the topology \(\mathcal{T}_d\).
2. \(d(x_n,x_0) \to 0\) in \(\mathbb{R}\).
3. For all \(\varepsilon>0\) there exists \(N \in \mathbb{N}\) such that \[ x_n \in B_r(x_0) \, , \,\, \forall \, n \geq \mathbb{N}\,. \]

The proof is similar to the one of Proposition 18, and it is left as an exercise.

3.5 Interior, closure and boundary

We now define interior, closure and boundary of a set \(A\) contained in a topological space.

Definition 33: Interior of a set

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). The interior of \(A\) is \[ \mathop{\mathrm{Int}}{A} := \bigcup_{ \substack{U \subseteq A \\ U \in \mathcal{T}} } \, U \,. \]

Remark 34

The definition of \(\mathop{\mathrm{Int}}{A}\) is well-posed, since \(\emptyset \subseteq A\) and \(\emptyset \in \mathcal{T}\). Therefore the union is taken over a non-empty family.

Proposition 35: \(\mathop{\mathrm{Int}}{A}\) is the largest open set contained in \(A\)

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). Then \(\mathop{\mathrm{Int}}{A}\) is the largest open set contained in \(A\), that is:

1. \(\mathop{\mathrm{Int}}{A}\) is open.
2. \(\mathop{\mathrm{Int}}{A} \subseteq A\).
3. If \(V \in \mathcal{T}\) and \(V \subseteq A\), then \(V \subseteq \mathop{\mathrm{Int}}{A}\).
4. \(A\) is open if and only if \(A = \mathop{\mathrm{Int}}{A}\).

Proof

1. \(\mathop{\mathrm{Int}}{A}\) is open, since it is union of open sets, see property (A2).

2. \(\mathop{\mathrm{Int}}{A} \subseteq A\), since \(\mathop{\mathrm{Int}}{A}\) is union of sets contained in \(A\).

3. Suppose \(V \in \mathcal{T}\) and \(V \subseteq A\). Therefore \[ V \subseteq \bigcup_{ \substack{U \subseteq A \\ U \in \mathcal{T}} } \, U = \mathop{\mathrm{Int}}{A} \,. \]

4. Suppose that \(A\) is open. Then \[ A \subseteq \bigcup_{ \substack{U \subseteq A \\ U \in \mathcal{T}} } \, U = \mathop{\mathrm{Int}}{A} \,. \] As we already know that \(\mathop{\mathrm{Int}}{A} \subseteq A\), we conclude that \(A = \mathop{\mathrm{Int}}{A}\).
   Conversely, suppose that \(A = \mathop{\mathrm{Int}}{A}\). Since \(\mathop{\mathrm{Int}}{A}\) is open, then also \(A\) is open.

Definition 36: Closure of a set

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). The closure of \(A\) is \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} := \bigcap_{ \substack{A \subseteq C \\ C \, \text{closed} } } \, C \,. \]

Remark 37

The definition of \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is well-posed, since \(A \subseteq X\), and \(X\) is closed. Therefore the intersection is taken over a non-empty family.

Proposition 38: \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is the smallest closed set containing \(A\)

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). Then \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is the smallest closed set containing \(A\), that is:

1. \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is closed.
2. \(A \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\).
3. If \(V\) is closed \(A \subseteq V\), then \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \subseteq V\).
4. \(A\) is closed if and only if \(A = {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\).

Proof

1. \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is closed, since it is intersection of closed sets, see property (C2).

2. \(A \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\), since \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is intersection of sets which contain \(A\).

3. Suppose \(V\) is closed and \(A \subseteq V\). Therefore \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = \bigcap_{ \substack{A \subseteq C \\ C \, \text{closed} } } \, C \subseteq V \,. \]

4. Suppose that \(A\) is closed. Then \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = \bigcap_{ \substack{A \subseteq C \\ C \, \text{closed}} } \, C \subseteq A \,, \] showing that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \subseteq A\). As we already know that \(A \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\), we conclude that \(A = {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\).
   Conversely, suppose that \(A = {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\). Since \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is closed, then also \(A\) is closed.

Lemma 39

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). They are equivalent:

1. \(x_0 \in {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\).
2. For every \(U \in \mathcal{T}\) such that \(x_0 \in U\), it holds \[ U \cap A \neq \emptyset \,. \]

Proof

We prove the contronominal statement: \[ x_0 \notin {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \quad \iff \quad \exists \,\, U \in \mathcal{T}\, \text{ s.t. } \, x_0 \in U \,, \,\, \, U \cap A = \emptyset \,. \]

Let us check the two implications hold:

• Suppose \(x_0 \notin {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\). Then \(x_0 \in U := ({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {})^c\). Note that \(U\) is open, since \(U^c = {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is closed. Since \(A \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\), we have \[ A \cap U = A \cap (\overline{A})^c = \emptyset \,. \]

• Assume there exists \(U \in \mathcal{T}\) such that \(x_0 \in U\) and \(U \cap A = \emptyset\). Therefore \(A \subseteq U^c\). Since \(U\) is open, \(U^c\) is closed. Then \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = \bigcap_{ \substack{ A \subseteq C \\ C \, \text{closed} } } \, C \subseteq U^c \,. \] Since \(x_0 \notin U^c\), we conclude that \(x_0 \notin {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\).

Definition 40: Boundary of a set

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). The boundary of \(A\) is \[ \partial A := {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \smallsetminus \mathop{\mathrm{Int}}{A} \,. \]

Proposition 41

Let \((X,\mathcal{T})\) be a topological space, and \(A \subseteq X\). Then \(\partial A\) is closed.

Proof

We can write \[ \partial A = {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \smallsetminus \mathop{\mathrm{Int}}{A} = {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \cap (\mathop{\mathrm{Int}}{A})^c \,. \] Note that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is closed and \((\mathop{\mathrm{Int}}{A})^c\) is closed, since \(\mathop{\mathrm{Int}}{A}\) is open. Then \(\partial A\) is intersection of two closed sets, and in hence closed by (C2).

We can characterize \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) as the set of limit points of sequences in \(A\).

Definition 42: Set of limit points

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). The set of limit points of \(A\) is defined as \[ L(A):=\{ x \in X \, \colon \,\exists \, \{x_n \} \subseteq A \, \text{ s.t. } \, x_n \to x \} \,. \]

Proposition 43

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\) a set. Let \(\{x_n\} \subseteq A\) and \(x_0 \in X\) be such that \(x_n \to x_0\). Then \(x_0 \in {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\). In particular, \[ L(A) \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \,. \]

Proof

Suppose by contradiction \(x_0 \notin {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\), so that \[ x_0 \in ({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {})^c \,. \] Since \(({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {})^c\) is open and \(x_n \to x_0\), there exists \(N \in \mathbb{N}\) such that \[ x_n \in ({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {})^c \,, \quad \forall \, n \geq N \,. \] This is a contradiction, since we were assuming that \(\{x_n\} \subseteq A\). This shows \(x_0 \in {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) and therefore \(L(A) \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\).

Warning

1. The converse of Proposition 43 is false in general, that is, \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \not\subset L(A) \,. \] We show a counterexample in Example 44.

2. The relation
   \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = L(A) \,. \] holds in the so-called first countable topological spaces, such as metric spaces, see Proposition 45 below.

Example 44: Co-countable topology on \(\mathbb{R}\)

Question. The co-countable topology on \(\mathbb{R}\) is the collection of sets \[ \mathcal{T}_{\textrm{cc}}:= \{ A \subseteq \mathbb{R}\, \colon \,A^c = \mathbb{R}\,\, \mbox{ or } \,\, A^c \, \mbox{ countable }\} \,. \]

1. Prove that \(\mathcal{T}_{\textrm{cc}}\) is a topology on \(\mathbb{R}\).

2. Prove that a sequence \(\{x_n\}\) is convergent in \(\mathcal{T}_{\textrm{cc}}\) if and only if it is eventually constant.

3. Define the set \(A = (-\infty,0]\). Prove that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = \mathbb{R}\).

4. Conclude that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \not\subset L(A)\).

Solution.

1. See Question 2 in Homework 3.

2. See Question 2 in Homework 3.

3. Assume \(C\) is a closed set such that \(A \subseteq C\). Since \(C\) is closed, it follows that \(C^c \in \mathcal{T}_{\textrm{cc}}\). Therefore \((C^c)^c = C\) is either countable, or equal to \(\mathbb{R}\). As \(A \subseteq C\), we have that \(C\) is uncountable. Therefore, \(C = \mathbb{R}\). As \(C\) is an arbitrary closed set containing \(A\), we conclude that \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} =\bigcap_{ \substack{ A \subseteq C \\ C \, \text{closed} } } \, C = \bigcap_{ \substack{ A \subseteq C \\ C \, \text{closed} } } \, \mathbb{R}= \mathbb{R}\,. \]

4. By Point 2, convergent sequences are eventually constant. Therefore, if \(\{x_n\} \subseteq A\) converges to \(x_0\), we conclude that \(x_0 \in A\). This shows
   \[ L(A) = A = [-\infty , 0) \,. \] By Point 3, we have \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = \mathbb{R}\). We conclude that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \not\subset L(A)\).

In metric spaces we can characterize the interior of a set and the closure in the following way.

Proposition 45: Characterization of \(\mathop{\mathrm{Int}}{A}\) and \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) in metric space

Let \((X,d)\) be a metric space. Denote by \(\mathcal{T}_d\) the topology induced by \(d\). For any \(A \subseteq X\), we have

1. \(\mathop{\mathrm{Int}}{A} = \{ x \in A \, \colon \,\exists \,\, r>0 \, \text{ s.t. } \, B_r(x) \subseteq A \}\) ,

2. \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = L(A) = \{ x \in X \, \text{ s.t. } \, \exists \,\, \{ x_n \} \subseteq A \, \text{ s.t. } \, x_n \to x \}\).

Proof

1. See Question 4 in Homework 3.

2. The inclusion \(L(A) \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) holds by Proposition 43. We are left to show that \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \subseteq L(A) \,. \] To this end, let \(x_0 \in {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\). For \(n \in \mathbb{N}\), consider the ball \(B_{1/n}(x_0)\). Since \(B_{1/n}(x_0) \in \mathcal{T}_d\) and \(x_0 \in B_{\varepsilon}(x_0)\), we can apply Lemma 39 and deduce that \[ B_{1/n}(x_0) \cap A \neq \emptyset \,. \] Let \(x_n \in B_{1/n}(x_0) \cap A\). Since \(n\) was arbitrary, we have constructed a sequence \(\{x_n\} \subseteq A\) such that \[ x_n \in B_{1/n}(x_0) \,, \quad \forall \, n \in \mathbb{N}\,. \] In particular, we have that \[ 0 \leq d(x_n,x_0) < \frac{1}{n} \to 0 \] as \(n \to \infty\). Thus \(x_n \to x_0\), showing that \(x_0 \in L(A)\).

Example 46

Question. Consider \(\mathbb{R}\) equipped with the Euclidean topology. Let \(A =[0,1)\). Prove that: \[ \mathop{\mathrm{Int}}{A} = (0,1) \,, \quad {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = [0,1] \,, \quad \partial A = \{0,1\} \,. \]

Note: In particular, this shows \[ \mathop{\mathrm{Int}}{A} \neq A \,, \quad {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \neq A \,, \] so that \(A\) is neither open, nor closed.

Solution. See Question 5 in Homework 3.

3.6 Density

Definition 47: Density

Let \((X,\mathcal{T})\) be a topological space. We say that a subset \(A \subseteq X\) is dense in \(X\), if \[ A \cap U \neq \emptyset \,, \quad \, \forall \,\, U \in \mathcal{T}\,, \,\, U \neq \emptyset \,. \]

Density can be characterized in terms of closure.

Proposition 48: Characterization of density

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq X\). They are equivalent:

1. \(A\) is dense in \(X\).
2. It holds \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = X\).

Proof

Part 1. Let \(A\) be dense in \(X\). Suppose by contradiction that \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \neq X \,. \] This means \(({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {})^c \neq \emptyset\). Note that \(({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {})^c\) is open, being \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) closed. By density of \(A\) in \(X\) we have \[ A \cap ({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {})^c \neq \emptyset \,. \] Since \(A \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\), the above is a contradiction.

Part 2. Suppose that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = X\). Let \(U \in \mathcal{T}\) with \(U \neq \emptyset\). By contradiction, assume that \[ A \cap U = \emptyset \,. \] Therefore \(A \subseteq U^c\). As \(U^c\) is closed, we have \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \subseteq U^c\,, \] because \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) is the smallest closed set containing \(A\). Recalling that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = X\), we conclude that \(U^c = X\). Therefore \(U = \emptyset\), which is a contradiction.

Example 49: \(\mathbb{Q}\) is dense in \((\mathbb{R},\mathcal{T}_{\textrm{euclid}})\)

Question. Consider \(\mathbb{R}\) equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\).

1. Prove that \(\mathbb{Q}\) is dense in \(\mathbb{R}\), that is, \({}\mkern 2mu \overline{\mkern-2mu \mathbb{Q} \mkern-2mu}\mkern 2mu {} = \mathbb{R}\).

2. Prove that \(\mathop{\mathrm{Int}}{Q} = \emptyset\).

Note: This shows \(\mathbb{Q}\) is neither open, nor closed, since \[ \mathop{\mathrm{Int}}{\mathbb{Q}} \neq \mathbb{Q}\,, \quad {}\mkern 2mu \overline{\mkern-2mu \mathbb{Q} \mkern-2mu}\mkern 2mu {} \neq \mathbb{Q}\,. \]

Solution. To solve the exercise we will need the following well-known analysis result:

Density Thoerem. Let \(x \in \mathbb{R}\) and \(\varepsilon>0\). There exists \(q \in \mathbb{Q}\) such that \[ |x-q| < \varepsilon\,. \]

1. We want to prove that \(\mathbb{Q}\) is dense in \(\mathbb{R}\) according to the topological definition. Therefore, let \(U\) be a non-empty open set in \(\mathbb{R}\). Let \(x \in U\). Since \(U\) is open, there exists \(r>0\) such that \[ (x-r,x+r) \subseteq U \,. \] By the Density Theorem, there exists \(q \in \mathbb{R}\) such that \[ |x-q| < r \,. \] In particular, this shows \(q \in (x-r,x+r)\), so that \(q \in U\). Therefore \(\mathbb{Q}\cap U \neq \emptyset\), proving that \(\mathbb{Q}\) is dense in \(\mathbb{R}\). In particular, by Proposition 49, we conclude that \({}\mkern 2mu \overline{\mkern-2mu \mathbb{Q} \mkern-2mu}\mkern 2mu {} = \mathbb{R}\).

2. Recall that \((\mathbb{R},\mathcal{T}_{\textrm{euclid}})\) is a metric space. Therefore, we can apply Proposition 45, and infer that \[ \mathop{\mathrm{Int}}{\mathbb{Q}} = \{ q \in \mathbb{Q}\, \colon \,\exists \,\, r>0 \, \text{ s.t. } \, (q-r,q+r) \subseteq \mathbb{Q}\} \tag{3.7}\] Assume by contradiction that \(\mathop{\mathrm{Int}}{\mathbb{Q}} \neq \emptyset\). Let \(q \in \mathop{\mathrm{Int}}{\mathbb{Q}}\). By (3.7), there exists \(r > 0\) such that \[ (q-r,q+r) \subseteq \mathbb{Q}\,. \] But \((q-r,q+r)\) is an interval of \(\mathbb{R}\), and therefore it will contain an irrational number \(x \in \mathbb{R}\smallsetminus \mathbb{Q}\). Contradiction. Hence, \(\mathop{\mathrm{Int}}{\mathbb{Q}} = \emptyset\).

Example 50: \(\mathbb{Z}\) is not dense in \((\mathbb{R},\mathcal{T}_{\textrm{euclid}})\)

Question. Consider \(\mathbb{R}\) equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\). Prove that the set of integers \(\mathbb{Z}\) is not dense in \(\mathbb{R}\), with \[ {}\mkern 2mu \overline{\mkern-2mu \mathbb{Z} \mkern-2mu}\mkern 2mu {} = \mathbb{Z}\,. \]

Solution. The set of integers \(\mathbb{Z}\) satisfies \[ \mathbb{Z}^c = \bigcup_{ z \in \mathbb{Z}} \, (z , z + 1 ) \,. \] Since \((z,z+1)\) is open in \(\mathbb{R}\), by (A2) we conclude that \(\mathbb{Z}^c\) is open, so that \(\mathbb{Z}\) is closed. Therefore \[ {}\mkern 2mu \overline{\mkern-2mu \mathbb{Z} \mkern-2mu}\mkern 2mu {} = \mathbb{Z}\,. \] As \({}\mkern 2mu \overline{\mkern-2mu \mathbb{Z} \mkern-2mu}\mkern 2mu {} \neq \mathbb{R}\), by Proposition 49, we have that \(\mathbb{Z}\) is not dense in \(\mathbb{R}\).

If we change topologies, the closure might change.

Example 51: \(\mathbb{Z}\) is dense in \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\)

Question. Consider \(\mathbb{R}\) equipped with the cofinite topology \[ \mathcal{T}_{\textrm{cofinite}}= \{ U \subset \mathbb{R}\, \colon \,U^c \, \mbox{ is finite, } \mbox{ or } U^c = \mathbb{R}\}\,. \] Prove that \(\mathbb{Z}\) is dense in \(\mathbb{R}\). In particular, \[ {}\mkern 2mu \overline{\mkern-2mu \mathbb{Z} \mkern-2mu}\mkern 2mu {} = \mathbb{R}\,, \]

Solution. Suppose \(C\) is a closed set such that \(\mathbb{Z}\subseteq C\). By definition of \(\mathcal{T}_{\textrm{cofinite}}\), we have that \((C^c)^c = C\) is either finite, or it coincides with \(\mathbb{R}\). Since \(\mathbb{Z}\subseteq C\), and \(\mathbb{Z}\) is not finite, we conclude \(C = \mathbb{R}\). As \(C\) is an arbitrary closed set containing \(\mathbb{Z}\), we conclude that \[ {}\mkern 2mu \overline{\mkern-2mu \mathbb{Z} \mkern-2mu}\mkern 2mu {} =\bigcap_{ \substack{ \mathbb{Z}\subseteq C \\ C \, \text{closed} } } \, C = \bigcap_{ \substack{ \mathbb{Z}\subseteq C \\ C \, \text{closed} } } \, \mathbb{R}= \mathbb{R}\,. \] In particular, by Proposition 49, \(\mathbb{Z}\) is dense in \(\mathbb{R}\).

3.7 Hausdorff spaces

Hausdorff space are topological spaces in which points can be separated by means of disjoint open sets.

Definition 52: Hausdorff space

We say that a topological space \((X,\mathcal{T})\) is Hausdorff if for every \(x,y \in X\) with \(x \neq y\), there exist \(U,V \in \mathcal{T}\) such that \[ x \in U \,, \quad y \in V \,, \quad U \cap V = \emptyset \,. \]

The main example of Hausdorff spaces are metrizable spaces.

Proposition 53

Let \((X,d)\) be a metric space, \(\mathcal{T}_d\) the topology induced by \(d\). Then \((X, \mathcal{T}_d)\) is a Hausdorff space.

Proof

Let \(x,y \in X\) with \(x \neq y\). Define \[ U := B_{\varepsilon}(x)\,, \quad V := B_{\varepsilon}(y) \,, \quad \varepsilon:= \frac12 \, d(x,y) \,. \] By Proposition 28, we know that \(U, V \in \mathcal{T}_d\). Moreover \(x \in U\), \(y \in V\). We are left to show that \(U \cap V = \emptyset\). Suppose by contradiction that \(U \cap V \neq \emptyset\) and let \(z \in U \cap V\). Therefore \[ d(x,z) < \varepsilon\,, \quad d(y,z) < \varepsilon\,. \] By triangle inequality we have \[ d(x,y) \leq d(x,z) + d(y,z) < \varepsilon+ \varepsilon= d(x,y) \,, \] where in the last inequality we used the definition of \(\varepsilon\). This is a contradiction. Therefore \(U \cap V = \emptyset\) and \((X,\mathcal{T}_d)\) is Hausdorff.

In general, every metrizable space is Hausdorff.

Definition 54: Metrizable space

Let \((X,\mathcal{T})\) be a topological space. We say that the topology \(\mathcal{T}\) is metrizable if there exists a metric \(d\) on \(X\) such that \[ \mathcal{T}= \mathcal{T}_d \,, \] with \(\mathcal{T}_d\) the topology induced by \(d\).

Corollary 55

Let \((X,\mathcal{T})\) be a metrizable space. Then \(X\) is Hausforff.

Proof

Since \((X,\mathcal{T})\) is metrizable, there exists a metric \(d\) on \(X\) such that \[ \mathcal{T}= \mathcal{T}_d \,. \] By Proposition 53 we know that \((X,\mathcal{T}_d)\) is Hausdorff. Hence \((X,\mathcal{T})\) is Hausdorff.

As a conseuqence of Corollary 55 we have that spaces which are not metrizable are not Hausdorff. Let us make a few examples.

Example 56: \((X,\mathcal{T}_{\textrm{trivial}})\) is not Hausdorff

Question. Let \(X\) be equipped with the trivial topology \(\mathcal{T}_{\textrm{trivial}}\). Then \(X\) is not Hausdorff.

Solution. Assume by contradiction \((X,\mathcal{T}_{\textrm{trivial}})\) is Hausdorff and let \(x,y \in X\) with \(x \neq y\). Then, there exist \(U,V \in \mathcal{T}_{\textrm{trivial}}\) such that \[ x \in U \,, \quad y \in V \,, \quad U \cap V = \emptyset \,. \] In particular \(U \neq \emptyset\) and \(V \neq \emptyset\). Since \(\mathcal{T}= \{ \emptyset, X \}\), we conclude that \[ U = V = X \quad \implies \quad U \cap V = X \neq \emptyset \,. \] This is a contradiction, and thus \((X,\mathcal{T}_{\textrm{trivial}})\) is not Hausdorff.

Example 57: \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is not Hausdorff

Question. Consider the cofinite topology on \(\mathbb{R}\) \[ \mathcal{T}_{\textrm{cofinite}}= \{ U \subseteq \mathbb{R}\, \colon \,U^c \, \mbox{ is finite, } \mbox{or } U^c = \mathbb{R}\} \,. \] Prove that \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is not Hausdorff.

Solution. Assume by contradiction \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is Hausdorff and let \(x, y \in \mathbb{R}\) with \(x \neq y\). Then, there exist \(U,V \in \mathcal{T}_{\textrm{cofinite}}\) such that \[ x \in U \,, \quad y \in V \,, \quad U \cap V = \emptyset \,. \] Taking the complement of \(U \cap V = \emptyset\), we infer \[ \mathbb{R}= (U \cap V)^c = U^c \cup V^c \,. \tag{3.8}\] There are two possibilities:

1. \(U^c\) and \(V^c\) are finite. Then \(U^c \cup V^c\) is finite, so that (3.8) is a contradiction.

2. Either \(U^c = \mathbb{R}\) or \(U^c = \mathbb{R}\). If \(U^c = \mathbb{R}\), then \(U = \emptyset\). This is a contradiction, since \(x \in U\). If \(V^c = \mathbb{R}\), then \(V = \emptyset\). This is a contradiction, since \(y \in V\).

Hence \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is not Hausdorff.

Example 58: Lower-limit topology on \(\mathbb{R}\) is not Hausdorff

Question. The lower-limit topology on \(\mathbb{R}\) is the collection of sets \[ \mathcal{T}_{\textrm{LL}}=\{\emptyset, \mathbb{R}\} \cup \{ (a,+\infty) \, \colon \, a \in \mathbb{R}\} \,. \]

1. Prove that \((\mathbb{R},\mathcal{T}_{\textrm{LL}})\) is a topological space.
2. Prove that \((\mathbb{R},\mathcal{T}_{\textrm{LL}})\) is not Hausdorff.

Solution. Part 1. We show that \((\mathbb{R}, \mathcal{T}_{\textrm{LL}})\) is a topological space by verifying the axioms:

(A1) By definition \(\emptyset, \mathbb{R}\in \mathcal{T}_{\textrm{LL}}\).

(A2) Let \(A_i \in \mathcal{T}_{\textrm{LL}}\) for all \(i \in I\). We have 2 cases:

• If \(A_i = \emptyset\) for all \(i\), then \(\cup_i A_i = \emptyset \in \mathcal{T}_{\textrm{LL}}\).

• At least one of the sets \(A_i\) is non-empty. As empty-sets do not contribute to the union, we can discard them. Therefore, \(A_i = (-\infty, a_i)\) with \(a_i \in \mathbb{R}\cup \{\infty\}\). Define: \[ a := \sup_{i \in I} a_i, \quad A := (-\infty, a). \] Then \(A \in \mathcal{T}\) and: \[ A = \cup_{i \in I} A_i. \] To prove this, let \(x \in A\). Then \(x < a\), so there exists \(i_0 \in I\) such that \(x < a_{i_0}\). Thus, \(x \in A_{i_0}\), showing \(A \subseteq \cup_{i \in I} A_i\). Conversely, if \(x \in \cup_{i \in I} A_i\), then \(x \in A_{i_0}\) for some \(i_0 \in I\), implying \(x < a_{i_0} \leq a\). Thus, \(x \in A\), proving \(\cup_{i \in I} A_i \subseteq A\).

(A3) Let \(A, B \in \mathcal{T}_{\textrm{LL}}\). We have 3 cases:

• \(A = \emptyset\) or \(B = \emptyset\). Then \(A \cap B = \emptyset \in \mathcal{T}_{\textrm{LL}}\).

• \(A \neq \emptyset\) and \(B \neq \emptyset\). Therefore, \(A = (-\infty, a)\) and \(B = (-\infty, b)\) with \(a, b \in \mathbb{R}\cup \{\infty\}\). Define \[ U := A \cap B, \quad z := \min\{a, b\}. \] Then \(U = (-\infty, z) \in \mathcal{T}_{\textrm{LL}}\).

Thus, \((\mathbb{R}, \mathcal{T}_{\textrm{LL}})\) is a topological space.

Part 2. To show \((\mathbb{R}, \mathcal{T}_{\textrm{LL}})\) is not Hausdorff, assume otherwise. Let \(x, y \in \mathbb{R}\) with \(x \neq y\). Then there exist \(U, V \in \mathcal{T}_{\textrm{LL}}\) such that: \[ x \in U, \quad y \in V, \quad U \cap V = \emptyset. \] As \(U,V\) are non-empty, by definition of \(\mathcal{T}_{\textrm{LL}}\), there exist \(a, b \in \mathbb{R}\cup \{\infty\}\) such that \(U = (-\infty, a)\) and \(V = (-\infty, b)\). Define: \[ z := \min\{a, b\}, \quad Z := U \cap V = (-\infty, z). \] Hence \(Z \neq \emptyset\), contradicting \(U \cap V = \emptyset\). Thus, \((\mathbb{R}, \mathcal{T}_{\textrm{LL}})\) is not Hausdorff.

In Hausdorff spaces the limit of sequences is unique.

Proposition 59: Uniqueness of limit in Hausdorff spaces

Let \((X,\mathcal{T})\) be a Hausdorff space. If a sequence \(\{x_n\} \subseteq X\) converges, then the limit is unique.

Proof

Let \(\{x_n\} \subseteq X\) be convergent, and suppose by contradiction that \[ x_n \to x_0 \,, \quad x_n \to y_0 \,, \quad x_0 \neq y_0 \,. \] Since \(X\) is Hausdorff, there exist \(U,V \in \mathcal{T}\) such that \[ x_0 \in U \,, \quad y_0 \in V \,, \quad U \cap V = \emptyset \,. \] As \(x_n \to x_0\) and \(U \in \mathcal{T}\) with \(x_0 \in U\), there exists \(N_1 \in \mathbb{N}\) such that \[ x_n \in U \,, \quad \forall \, n \geq N_1 \,. \] Similarly, since \(x_n \to y_0\) and \(V \in \mathcal{T}\) with \(y_0 \in U\), there exists \(N_2 \in \mathbb{N}\) such that \[ x_n \in V \,, \quad \forall \, n \geq N_2 \,. \] Take \(N:=\max\{N_1,N_2\}\). Then \[ x_n \in U \cap V \,, \quad \forall \, n \geq N \,. \] Since \(U \cap V = \emptyset\), the above is a contradiction. Therefore the limit is unique.

3.8 Continuity

We extend the notion of continuity to topological spaces. To this end, we need the concept of pre-image of a set under a function.

Definition 60: Images and Pre-images

Let \(X, Y\) be sets and \(f \colon X \to Y\) be a function.

1. Let \(U \subseteq X\). The image of \(U\) under \(f\) is the subset of \(Y\) defined by \[ f (U) := \{ y \in Y \, \colon \,\exists \, x \in X \, \text{ s.t. } \, y = f(x) \} = \{ f(x) \, \colon \,x \in X \} \,. \]

2. Let \(V \subseteq Y\). The pre-image of \(V\) under \(f\) is the subset of \(X\) defined by \[ f^{-1} (V) := \{ x \in X \, \colon \,f(x) \in V \} \,. \]

Warning

The notation \(f^{-1}(V)\) does not mean that we are inverting \(f\). In fact, the pre-image is defined for all functions.

Let us gather useful properties of images and pre-images.

Proposition 61

Let \(X,Y\) be sets and \(f \colon X \to Y\). We denote with the letter \(A\) sets in \(X\) and with the letter \(B\) sets in \(Y\). We have

• \(A \subseteq f^{-1}(f(A))\)
• \(A = f^{-1}(f(A))\) if \(f\) is injective
• \(f(f^{-1}(B)) \subseteq B\)
• \(f(f^{-1}(B)) = B\) if \(f\) is surjective
• If \(A_1 \subseteq A_2\) then \(f(A_1) \subseteq f(A_2)\)
• If \(B_1 \subseteq B_2\) then \(f^{-1}(B_1) \subseteq f^{-1}(B_2)\)
• If \(A_i \subseteq X\) for \(i \in I\) we have \[\begin{gather*} f \left( \bigcup_{i \in I} A_i \right) = \bigcup_{i \in I} f(A_i) \\ f \left( \bigcap_{i \in I} A_i \right) \subseteq \bigcap_{i \in I} f(A_i) \end{gather*}\]
• If \(B_i \subseteq Y\) for \(i \in I\) we have \[\begin{gather*} f^{-1} \left( \bigcup_{i \in I} B_i \right) = \bigcup_{i \in I} f^{-1}(B_i) \\ f^{-1} \left( \bigcap_{i \in I} B_i \right) = \bigcap_{i \in I} f^{-1}(B_i) \end{gather*}\]

Suppose \(Z\) is another set and \(g \colon Y \to Z\). Let \(C \subseteq Z\). Then \[\begin{align*} & (g \circ f)(A) = g(f(A)) \\ & (g \circ f)^{-1}(C) = f^{-1}(g^{-1}(C)) \end{align*}\]

It is a good exercise to try and prove a few of the above properties. We omit the proof. We can now define continuous functions between topological spaces.

Definition 62: Continuous function

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. Let \(f \colon X \to Y\) be a function.

1. Let \(x_0 \in X\). We say that \(f\) is continuous at \(x_0\) if it holds: \[ \forall \, V \in \mathcal{T}_Y \, \text{ s.t. } \, f(x_0) \in V \,, \,\, \exists \, U \in \mathcal{T}_X \, \text{ s.t. } \, x_0 \in U \,, \,\, f(U) \subseteq V \,. \]

2. We say that \(f\) is continuous from \((X,\mathcal{T}_X)\) to \((Y,\mathcal{T}_Y)\) if \(f\) is continuous at each point \(x_0 \in X\).

The following proposition presents a useful characterization of continuous functions in terms of pre-images.

Proposition 63

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. Let \(f \colon X \to Y\) be a function. They are equivalent:

1. \(f\) is continuous from \((X,\mathcal{T}_X)\) to \((Y,\mathcal{T}_Y)\).
2. It holds: \(f^{-1}(V) \in \mathcal{T}_X\) for all \(V \in \mathcal{T}_Y\).

Important

In other words, a function \(f \colon X \to Y\) is continuous if and only if the pre-image of open sets in \(Y\) are open sets in \(X\).

The proof of Proposition 63 is simple, but very tedious. We choose to skip it.

Example 64

Question. Let \(X\) be a set and \(\mathcal{T}_1\), \(\mathcal{T}_2\) be topologies on \(X\). Define the identity map \[ {\mathop{\mathrm{Id}}}_{X} \colon (X,\mathcal{T}_1) \to (X,\mathcal{T}_2) \,, \quad {\mathop{\mathrm{Id}}}_{X} (x):= x \,. \] Prove that they are equivalent:

1. \({\mathop{\mathrm{Id}}}_{X}\) is continuous from \((X,\mathcal{T}_1)\) to \((X,\mathcal{T}_2)\).
2. \(\mathcal{T}_1\) is finer than \(\mathcal{T}_2\), that is, \(\mathcal{T}_2 \subseteq \mathcal{T}_1\).

Solution. \({\mathop{\mathrm{Id}}}_{X}\) is continuous if and only if \[ {\mathop{\mathrm{Id}}}_{X}^{-1} (V) \in \mathcal{T}_1 \,, \quad \forall \, V \in \mathcal{T}_2 \,. \] But \({\mathop{\mathrm{Id}}}_{X}^{-1} (V) = V\), so that the above reads \[ V \in \mathcal{T}_1 \,, \quad \forall \, V \in \mathcal{T}_2 \,, \] which is equivalent to \(\mathcal{T}_2 \subseteq \mathcal{T}_1\).

Let us compare our new definition of contiuity with the classical notion of continuity in \(\mathbb{R}^n\). Let us recall the definition of continuous function in \(\mathbb{R}^n\).

Definition 65: Continuity in the classical sense

Let \(f \colon \subseteq \mathbb{R}^n \to \mathbb{R}^m\). We say that \(f\) is continuous at \(\mathbf{x}_0\) if it holds: \[ \forall \, \varepsilon> 0 \,, \, \exists \, \delta >0 \, \text{ s.t. } \, \|f(\mathbf{x}) - f(\mathbf{x}_0)\|< \varepsilon\,\, \mbox{ if } \,\, \| \mathbf{x}- \mathbf{x}_0 \| < \delta \,. \]

Proposition 66

Let \(f \colon \mathbb{R}^n \to \mathbb{R}^m\) and suppose \(\mathbb{R}^n,\mathbb{R}^m\) are equipped with the Euclidean topology. Let \(\mathbf{x}_0 \in \mathbb{R}^n\). They are equivalent:

1. \(f\) is continuous at \(\mathbf{x}_0\) in the topological sense.
2. \(f\) is continuous at \(\mathbf{x}_0\) in the classical sense.

Proof

Part 1. Suppose that \(f\) is continuous at \(\mathbf{x}_0\) in the topological sense. Let \(\varepsilon>0\) and consider the set \[ V := B_{\varepsilon}(f(\mathbf{x}_0))\,. \] We have that \(V \subset \mathbb{R}^m\) is open and \(f(\mathbf{x}_0) \in V\). As \(f\) is continuous in the topological sense, there exists \(U \subset \mathbb{R}^n\) open with \(\mathbf{x}_0 \in U\) and such that \[ f(U) \subset V = B_{\varepsilon}(f(\mathbf{x}_0)) \,. \tag{3.9}\] Since \(U\) is open and \(\mathbf{x}_0 \in U\), there exists \(\delta>0\) such that \[ B_{\delta}(\mathbf{x}_0) \subset U \,. \] By the above inclusion and (3.9) we conclude that \[ f(B_{\delta}(\mathbf{x}_0)) \subset f(U) \subset V = B_{\varepsilon}(f(\mathbf{x}_0)) \,. \] This is equivalent to \[ \mathbf{x}\in B_{\delta}(\mathbf{x}_0) \quad \implies \quad f(\mathbf{x}) \in B_{\varepsilon}(f(\mathbf{x}_0))\,, \] which reads \[ \| \mathbf{x}- \mathbf{x}_0 \| < \delta \quad \implies \quad \| f(\mathbf{x}) - f(\mathbf{x}_0) \| < \varepsilon\,. \] Therefore \(f\) is continuous at \(\mathbf{x}_0\) in the classical sense.

Part 2. Suppose \(f\) is continuous at \(x_0\) in the classical sense. Let \(V \subset \mathbb{R}^m\) be open and such that \(f(\mathbf{x}_0) \in V\). Since \(V\) is open, there exists \(\varepsilon>0\) such that \[ B_{\varepsilon} (f(\mathbf{x}_0)) \subset V \,. \tag{3.10}\] Since \(f\) is continous in the classical sense, there exists \(\delta>0\) such that \[ \| \mathbf{x}- \mathbf{x}_0 \| < \delta \quad \implies \|f(\mathbf{x}) - f(\mathbf{x}_0)\|< \varepsilon\,. \] The above is equivalent to \[ \mathbf{x}\in B_{\delta}(\mathbf{x}_0) \quad \implies \quad f(\mathbf{x}) \in B_{\varepsilon} (f (\mathbf{x}_0)) \,. \tag{3.11}\] Set \[ U:= B_{\delta}(\mathbf{x}_0) \] and note that \(U\) is open in \(\mathbb{R}^n\) and \(\mathbf{x}_0 \in U\). By definition of image of a set, (3.11) reads \[ f(U) = f( B_{\delta}(\mathbf{x}_0)) \subseteq B_{\varepsilon} (f(\mathbf{x}_0)) \,. \] Recalling (3.10) we conclude that \[ f(U) \subset V \,. \] In summary, we have shown that given \(V \subset \mathbb{R}^m\) open and such that \(f(\mathbf{x}_0) \in V\), there exists \(U\) open in \(\mathbb{R}^n\) such that \(\mathbf{x}_0 \in U\) and \(f(U) \subset V\). Therefore \(f\) is continuous at \(\mathbf{x}_0\) in the topological sense.

A similar proof yields the characterization of continuity in metric spaces. The proof is left as an exercise.

Proposition 67

Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. Denote by \(\mathcal{T}_X\) and \(\mathcal{T}_Y\) the topologies induced by the metrics. Let \(f \colon X \to Y\) and \(x_0 \in X\). They are equivalent:

1. \(f\) is continuous at \(x_0\) in the topological sense.
2. It holds: \[ \begin{aligned} & \forall \, \varepsilon> 0 \,, \, \exists \, \delta >0 \, \text{ s.t. } \, \\ & d_Y(f(x),f(x_0))<\varepsilon\,\, \mbox{ if } \,\, d_X(x , x_0) < \delta \,. \end{aligned} \]

Let us examine continuity in the cases of the trivial and discrete topologies.

Example 68

Question. Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be a topological space. Suppose that \(\mathcal{T}_Y\) is the trivial topology, that is, \[ \mathcal{T}_Y = \{ \emptyset, Y \} \,. \] Prove that every function \(f \colon X \to Y\) is continuous.

Solution. \(f\) is continuous if \(f^{-1}(V) \in \mathcal{T}_X\) for all \(V \in \mathcal{T}_Y\). We have two cases:

• \(V=\emptyset\): Then \(f^{-1}(V) = f^{-1}(\emptyset) = \emptyset \in \mathcal{T}_X\).

• \(V=Y\): Then \(f^{-1}(V) = f^{-1}(Y) = X \in \mathcal{T}_X\).

Therefore \(f\) is continuous.

Example 69

Question. Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. Suppose that \(\mathcal{T}_Y\) is the discrete topology, that is, \[ \mathcal{T}_Y = \{ V \, \text{ s.t. } \, V \subseteq Y \} \,. \] Let \(f \colon X \to Y\). Prove that they are equivalent:

1. \(f\) is continuous from \(X\) to \(Y\).
2. \(f^{-1}(\{y\}) \in \mathcal{T}_X\) for all \(y \in Y\).

Solution. Suppose that \(f\) is continuous. Then \[ f^{-1}(V) \in \mathcal{T}_X \,, \quad \forall \,\, V \in \mathcal{T}_Y \,. \] As \(V=\{y\} \in \mathcal{T}_Y\), we conclude that \(f^{-1}(\{y\}) \in \mathcal{T}_X\).

Conversely, assume that \(f^{-1}(\{y\}) \in \mathcal{T}_X\) for all \(y \in Y\). Let \(V \in \mathcal{T}_Y\). Trivially, we have \(V = \cup_{y \in V} \, \{ y \}\). Therefore \[ f^{-1}(V) = f^{-1}\left( \bigcup_{y \in V}\, \{ y \} \right) = \bigcup_{y \in V} \, f^{-1}( \{y \} ) \,. \] As \(f^{-1}( \{y \}) \in \mathcal{T}_X\) for all \(y \in Y\), by property (A2) we conclude that \(f^{-1}(V) \in \mathcal{T}_X\). Therefore \(f\) is continuous.

It is useful to introduce the notion of sequential continuity.

Definition 70: Sequential continuity

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, and \(f \colon X \to Y\). We say that \(f\) is sequentially continuous if the following condition holds: \[ \{x_n\} \subset X \,, \,\, x_n \to x_0 \,\, \mbox{ in }\, X \,\, \implies \,\, f(x_n) \to f(x_0) \,\, \mbox{ in }\, Y\,. \]

In other words, \(f\) is sequentially continuous if it takes convergent sequences into convergent sequences. In any topological space, continuity implies sequential continuity, as proven in the next Proposition.

Proposition 71

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. Let \(f \colon X \to Y\) be continuous. Then \(f\) is sequentially continuous.

Proof

Let \(\{x_n\} \subset X\) and suppose that \(x_n \to x_0\) in the topology \(\mathcal{T}_X\). We need to prove that \[ f(x_n) \to f(x_0) \\, \text{ in } \, Y \,. \] To this end, let \(V \in \mathcal{T}_Y\) be such that \(f(x_0) \in V\). Since \(f\) is continuous, there exists \(U \in \mathcal{T}_X\) with \(x_0 \in U\) such that \[ f(U) \subset V \,. \] Since \(U \in \mathcal{T}_X\) and \(x_n \to x_0\) in \(X\), there exists \(N \in \mathbb{N}\) such that \[ x_n \in U \,, \quad \forall \, n \geq N \,. \] Therefore \[ f(x_n) \in f(U) \,, \quad \forall \, n \geq N \,. \] Seeing that \(f(U) \subset V\), we conclude \[ f(x_n) \in V \,, \quad \forall \, n \geq N \,, \] showing that \(f(x_n) \to f(x_0)\) in \(Y\).

Warning

1. The reverse implication of Proposition 71 is false: \[ \text{sequential continuity} \quad \centernot\implies \quad \text{continuity} \] A counterexample is given in Example 73 below.

2. Continuity is equivalent to sequential continuity if the topologies on \(X\) and \(Y\) are first countable. This is the case for metrizable topologies, see Proposition 72 below.

Proposition 72

Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. They are equivalent:

1. \(f\) is continuous.
2. \(f\) is sequentially continuous.

Proof

Part 1. We have already proven that continuity implies sequential continuity in any topological space.

Part 2. Assume \(f\) is sequentially continuous. Suppose by contradiction \(f\) is not continuous at some point \(x_0 \in X\). Then there exists \({\varepsilon}_0 >0\) such that, for all \(\delta>0\) it holds \[ d_Y(f(x),f(x_0)) > \varepsilon_0 \,, \quad d_X(x,x_0)<\delta \,. \] We can therefore choose \(\delta = 1/n\) and construct a sequence \(\{x_n\} \subseteq X\) such that \[ d_Y(f(x_n),f(x_0)) > \varepsilon_0 \,, \quad d_X(x_n,x_0)< \frac{1}{n} \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore \(x_n \to x_0\) in \(X\). Define the sequence \[ y_n := \begin{cases} x_n & \,\, \mbox{ if } \, n \, \mbox{ even} \\ x_0 & \,\, \mbox{ if } \, n \, \mbox{ odd} \end{cases} \] As \(x_n \to x_0\), we have \(y_n \to x_0\). However \(\{f(y_n)\}\) does not converge to any point in \(Y\): Indeed \(\{f(y_n)\}\) cannot converge to \(f(x_0)\), since for \(n\) even we have \[ d_Y(f(y_n),f(x_0)) = d_Y(f(x_n),f(x_0)) > \varepsilon_0 \,. \] Also \(\{f(y_n)\}\) cannot converge to a point \(y \neq f(x_0)\), since for \(n\) odd \[ d_Y (f(y_n), y) = d_Y (f(x_0), y) > 0 \,. \] Hence, we have produced a sequence \(\{y_n\}\) which is convergent, but such that \(\{f(y_n)\}\) does not converge. This contradicts our assumption. Hence \(f\) must be continuous.

Example 73: Sequential continuity does not imply continuity

Question. Consider the co-countable and discrete topologies on \(\mathbb{R}\) \[\begin{gather*} \mathcal{T}_{\textrm{cc}}= \{ A \subseteq \mathbb{R}\, \colon \,A^c = \mathbb{R}\, \mbox{ or } \, A^c \, \mbox{ countable} \} \\ \mathcal{T}_{\textrm{discrete}}= \{ A \subseteq \mathbb{R}\} \end{gather*}\] Consider the identity function \[ f \colon (\mathbb{R},\mathcal{T}_{\textrm{cc}}) \to (\mathbb{R},\mathcal{T}_{\textrm{discrete}}) \,, \quad f(x):=x \,. \] Prove that

1. \(f\) is not continuous.
2. \(f\) is sequentially continuous.

Hint: You can use the following fact: Sequences in \((\mathbb{R},\mathcal{T}_{\textrm{cc}})\) and \((\mathbb{R},{\mathcal{T}_{\textrm{discrete}}})\) converge if and only if they are eventually constant.

Solution.

1. We have \(\{x\} \in {\mathcal{T}}_{\textrm{discrete}}\). However, \[ f^{-1}(\{ x\}) = \{x\} \notin {\mathcal{T}}_{\textrm{cc}} \,, \] since \(\{x\}^c = \mathbb{R}\smallsetminus \{x\}\) is neither equal to \(\mathbb{R}\), nor countable. Therefore \(f\) is not continuous.

2. Assume that \(\{x_n\}\) is convergent in \(\mathcal{T}_{\textrm{cc}}\). By the Hint, we have that \(\{x_n\}\) is eventually constant. Again by the Hint, we infer that \(\{x_n\}\) is convergent in \(\mathcal{T}_{\textrm{discrete}}\). Since \(f(x_n) = x_n\), we conclude that \(f\) is sequentially continuous.

Let us make an observation on continuity of compositions.

Proposition 74: Continuity of compositions

Let \((X,\mathcal{T}_X), (Y,\mathcal{T}_Y), (Z,\mathcal{T}_Z)\) be topological spaces. Assume \(f \colon X \to Y\) and \(g \colon Y \to Z\) are continuous. Then \((g \circ f) \colon X \to Z\) is continuous.

Proof

Let \(C \in \mathcal{T}_Z\). As \(g\) is continuous, we have that \[ g^{-1}(C) \in \mathcal{T}_Y \,. \] Since \(f\) is continuous, we also have \[ f^{-1} ( g^{-1}(C) ) \in \mathcal{T}_X \,. \] Therefore \[ (g \circ f)^{-1} ( C ) = f^{-1} ( g^{-1}(C) ) \in \mathcal{T}_X \,, \] so that \(g \circ f\) is continuous.

We conclude the section by introducing homeomorphisms.

Definition 75: Homeomorphism

Let \((X,\mathcal{T}_X)\), \((Y,\mathcal{T}_Y)\) be topological space. A function \(f \colon X \to Y\) is called an homeomorphism if they hold:

1. \(f\) is continuous.
2. \(f\) admits continuous inverse \(f^{-1} \colon Y \to X\).

Therefore \(f\) is a homeomorphism if it is continuous and admits a continuous inverse. Homeomorphisms are used to study similarities between topological spaces: When 2 topological spaces are homeomorphic, they can be essentially considered to be the same space.

3.9 Subspace topology

Any subset \(Y\) in a topological space \(X\) naturally inherits a topological structure. Such structure is called subspace topology.

Definition 76: Subspace topology

Let \((X,\mathcal{T})\) be a topological space and \(Y \subseteq X\) a subset. Define the family of sets \[\begin{align*} \mathcal{S}& := \{ A \subseteq Y \, \colon \,\exists \,\, U \in \mathcal{T}\, \text{ s.t. } \, A = U \cap Y \} \\ & = \{ U \cap Y , \,\, U \in \mathcal{T}\} \,. \end{align*}\] The family \(\mathcal{S}\) is called subspace topology on \(Y\) induced by the inclusion \(Y \subseteq X\).

Proof: Well-posedness of Definition 76

We have to show that \((Y,\mathcal{S})\) is a topological space:

• (A1) \(\emptyset \in \mathcal{S}\) since \(\emptyset = \emptyset \cap Y\) and \(\emptyset \in \mathcal{T}\). Similarly, we have \(Y \in \mathcal{S}\), since \(Y = X \cap Y\) and \(X \in \mathcal{T}\).

• (A2) Let \(A_i \in \mathcal{S}\) for \(i \in I\). By definition there exist \(U_i \in \mathcal{T}\) such that \(A_i = U_i \cap Y\) for all \(i \in I\). Therefore \[ \cup_{i \in I} \, A_i = \cup_{i \in I} (U_i \cap Y ) = \left(\cup_{i \in I} U_i \right) \cap Y \,. \] The above proves that \(\cup_{i \in I} \, A_i \in \mathcal{S}\), since \(\cup_{i \in I} \, U_i \in \mathcal{T}\).

• (A3) Let \(A_1, A_2 \in \mathcal{S}\). By definition there exist \(U_1,U_2 \in \mathcal{T}\) such that \(A_1 = U_1 \cap Y\) and \(A_2 = U_2 \cap Y\). Therefore \[ A_1 \cap A_2 = ( U_1 \cap Y ) \cap (U_2 \cap Y) = (U_1 \cap U_2) \cap Y \] The above proves that \(A_1 \cap A_2 \in \mathcal{S}\), since \(U_1 \cap U_2 \in \mathcal{T}\).

If the set \(Y\) is open, the subspace topology coincides with the original topology, as see in the next Proposition.

Proposition 77

Let \((X,\mathcal{T})\) be a topological space and \(Y \in \mathcal{T}\). Let
\(A \subseteq Y\). Then \[ A \in \mathcal{S}\quad \iff \quad A \in \mathcal{T}\,. \]

Proof

Suppose \(A \in \mathcal{S}\). Then there exists \(U \in \mathcal{T}\) such that \[ A = U \cap Y \,. \] Since \(U, Y \in \mathcal{T}\), by property (A3) of topologies it follows that \[ A = U \cap Y \in \mathcal{T}\,. \]

Conversely, assume that \(A \in \mathcal{T}\). Then \[ A = A \cap Y \,, \] showing that \(A \in \mathcal{S}\).

Warning

Let \((X,\mathcal{T})\) be a topological space, \(A \subseteq Y \subseteq X\). In general we could have \[ A \in \mathcal{S}\quad \mbox{and} \quad A \notin \mathcal{T}\,. \]

Example. Let \(X=\mathbb{R}\) with \(\mathcal{T}_{\textrm{euclid}}\). Consider the subset \(Y = [0,2)\), and equip \(Y\) with the subspace topology \(\mathcal{S}\). Let \(A = [0,1)\). Then \(A \notin \mathcal{T}_{\textrm{euclid}}\) but \(A \in \mathcal{S}\), since \[ A = (-1,1) \cap Y \,, \qquad (-1,1) \in \mathcal{T}_{\textrm{euclid}}\,. \]

Example 78

Question. Let \(X=\mathbb{R}\) be equipped with \(\mathcal{T}_{\textrm{euclid}}\). Let \(\mathcal{S}\) be the subspace topology on \(\mathbb{Z}\). Prove that \[ \mathcal{S}= \mathcal{T}_{\textrm{discrete}}\,. \]

Solution. To prove that \(\mathcal{S}= \mathcal{T}_{\textrm{discrete}}\), we need to show that all the subsets of \(\mathbb{Z}\) are open in \(\mathcal{S}\).

1. Let \(z \in \mathbb{Z}\) be arbitrary. Notice that \[ \{z\} = \left( z-1 , z + 1 \right) \cap \mathbb{Z}\, \] and \((z - 1, z + 1) \in \mathcal{T}_{\textrm{euclid}}\). Thus \(\{z\} \in \mathcal{S}\).

2. Let now \(A \subseteq \mathbb{Z}\) be an arbitrary subset. Trivially, \[ A = \cup_{z \in A} \, \{ z \} \,. \] As \(\{z\} \in \mathcal{S}\), we infer that \(A \in \mathcal{S}\) by (A2).

3.10 Topological basis

We have seen that in metric spaces every open set is union of open balls, see Propostion 28. We can then regard the open balls as the building blocks for the whole topology. In this context, we call the open balls a basis for the topology.

We can generalize the concept of basis to arbitrary topological spaces.

Definition 79: Topological basis

Let \((X,\mathcal{T})\) be a topological space and let \(\mathcal{B} \subseteq \mathcal{T}\). We say that \(\mathcal{B}\) is a topological basis for the topology \(\mathcal{T}\), if for all \(U \in \mathcal{T}\) there exist open sets \(\{B_i\}_{i \in I} \subseteq \mathcal{B}\), with \(I\) family of indices, such that \[ U = \bigcup_{i \in I} \, B_i \,. \tag{3.12}\]

Example 80

Question. Prove the following statements.

1. Let \((X,\mathcal{T})\) be a topological space. Then \(\mathcal{B}:=\mathcal{T}\) is a basis for \(\mathcal{T}\).

2. Let \((X,d)\) be a metric space with topology \(\mathcal{T}_d\) induced by the metric. Then \[ \mathcal{B} :=\{ B_r(x) \, \colon \,x \in X \,, \,\, r>0 \} \] is a basis for \(\mathcal{T}_d\).

3. Let \(X\) be equipped with \(\mathcal{T}_{\textrm{discrete}}\). Then \[ \mathcal{B}:=\{ \{x\} \, \colon \,x \in X \} \] is a basis for \(\mathcal{T}_{\textrm{discrete}}\).

Solution.

1. This is true because one can just take \(B=U\) in (3.12).

2. This is true by Propostion 28.

3. This is true because for any \(U \in \mathcal{T}\) we have \[ U = \bigcup_{ x \in U } \, \{x\} \,. \]

Example 81

Question. Consider \(\mathbb{R}\) equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\). Which of the following collection of sets are basis for \(\mathcal{T}_{\textrm{euclid}}\)? Motivate your answer.

1. \(\mathcal{B}_1 = \{ (a,b) \,\colon \, a,b \in \mathbb{R}\}\).
2. \(\mathcal{B}_2 = \{ (a,b) \,\colon \, a,b \in \mathbb{Q}\}\).
3. \(\mathcal{B}_3 = \{ (a,b) \,\colon \, a,b \in \mathbb{Z}\}\).

Solution.

1. \(\mathcal{B}_1\) is a basis for \(\mathcal{T}_{\textrm{euclid}}\), for the following reason. Let \(d\) be the Euclidean distance on \(\mathbb{R}\), and \(\mathcal{T}_d\) the topology induced by \(d\). We know that \(\mathcal{T}_d = \mathcal{T}_{\textrm{euclid}}\), therefore \[ \mathcal{B} :=\{ B_r(x) \, \colon \,x \in \mathbb{R}\,, \,\, r>0 \} \] is a basis for \(\mathcal{T}_{\textrm{euclid}}\) by Propostion 28. Note that balls in \(\mathbb{R}\) are just open intervals \[ B_r(x) = (x-r,x+r) \,. \] Hence \(\mathcal{B}_1 = \mathcal{B}\), so that \(\mathcal{B}_1\) is a basis for \(\mathcal{T}_{\textrm{euclid}}\).

2. \(\mathcal{B}_2\) is a basis for \(\mathcal{T}_{\textrm{euclid}}\). This is because any open interval \((a,b)\) with \(a,b \in \mathbb{R}\) can be written as \[ \bigcup_{q,r \in \mathbb{Q}, \, a< q , \, s<b} (q,s) = (a,b) \,. \] Therefore, since \(\mathcal{B}_1\) is a basis for \(\mathcal{T}_{\textrm{euclid}}\), we conclude that also \(\mathcal{B}_2\) is a basis for \(\mathcal{T}_{\textrm{euclid}}\).

3. \(\mathcal{B}_3\) is not a basis for \(\mathcal{T}_{\textrm{euclid}}\). Indeed, consider \(U = (0,1/2)\), which is open in \(\mathcal{T}_{\textrm{euclid}}\). It is clear that \(U\) cannot be obtained as the union of intervals \((q,s)\) with \(q,s \in \mathbb{Z}\).

Proposition 82

Let \((X,\mathcal{T})\) be a topological space, and \(\mathcal{B}\) a basis for \(\mathcal{T}\). They hold:

• (B1) We have \[ \bigcup_{B \in \mathcal{B}} \, B = X \,. \]

• (B2) If \(U_1,U_2 \in \mathcal{B}\) then there exist \(\{B_i \} \subseteq \mathcal{B}\) such that \[ U_1 \cap U_2 = \bigcup_{i \in I} \, B_i \,. \]

Proof

• (B1) This holds because \(X \in \mathcal{T}\). Therefore by definition of basis there exist \(B_i \in \mathcal{B}\) such that \[ X = \bigcup_{i \in I} \, B_i \,. \] Therefore taking the union over all \(B \in \mathcal{B}\) yields \(X\), and (B1) follows.

• (B2) Let \(U_1 , U_2 \in \mathcal{B}\). Then \(U_1 , U_2 \in \mathcal{T}\), since \(\mathcal{B} \subseteq \mathcal{T}\). By property (A3) we get that \(U_1 \cap U_2 \in \mathcal{T}\). Since \(\mathcal{B}\) is a basis we conclude (B2).

Properties (B1) and (B2) from Proposition 82 are sufficient for generating a topology.

Proposition 83

Let \(X\) be a set, and \(\mathcal{B}\) a collection of subsets of \(X\) satisfying (B1)-(B2). Define \[ \mathcal{T}:= \left\{ U \subseteq X \, \colon \,U = \bigcup_{i \in I} B_i \,, \,\, B_i \in \mathcal{B} \right\} \,. \] Then \(\mathcal{T}\) is a topology on \(X\), with basis given by \(\mathcal{B}\).

Proof

1. We need to verify that \(\mathcal{T}\) is a topology:

• (A1) We have that \(X \in \mathcal{T}\) by (B1). Moreover \(\emptyset \in \mathcal{T}\), since \(\emptyset\) can be obtained as empty union. Therefore (A1) holds.

• (A2) Let \(U_i \in \mathcal{T}\) for all \(i \in I\). By definition of \(\mathcal{T}\) we have \[ U_i = \bigcup_{k \in K_i} \, B_k^i \] for some family of indices \(K_i\) and \(B_k^i \in \mathcal{B}\). Therefore \[ U := \bigcup_{i \in I} \, U_i = \bigcup_{i \in I, \, k \in K_i} \, B_k^i \,, \] showing that \(U \in \mathcal{T}\).

• (A3) Suppose that \(U_1, U_2 \in \mathcal{T}\). Then \[ U_1 = \bigcup_{i \in I_1} \, B_i^1 \,, \quad U_2 = \bigcup_{i \in I_2} \, B_i^2 \] for \(B_i^1,B_i^2 \in \mathcal{B}\). From the above we have \[ U_1 \cap U_2 = \bigcup_{ i \in I_1 ,\, k \in I_2 } \, B_i^1 \cap B_k^2 \,. \] From property (B2) we have that for each pair of indices \((i,k)\) the set \(B_i^1 \cap B_k^2\) is the union of sets in \(\mathcal{B}\). Therefore \(U_1 \cap U_2\) is union of sets in \(\mathcal{B}\), showing that \(U_1 \cap U_2 \in \mathcal{T}\).

2. This trivially follows from defintion of \(\mathcal{T}\) and definition of basis.

3.11 Product topology

Given two topological spaces \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) we would like to equip the cartesian product \[ X \times Y = \{ (x,y) \, \colon \,x \in X \,, \,\, y \in Y \} \] with a topology. We proceed as follows.

Proposition 84

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. Define the family \(\mathcal{B}\) of subsets of \(X \times Y\) as \[ \mathcal{B} := \{ U \times V \, \colon \,U \in \mathcal{T}_X \,, \,\, V \in \mathcal{T}_Y \} \subseteq \mathcal{T}_X \times \mathcal{T}_Y \,. \] Then \(\mathcal{B}\) satisfies properties (B1) and (B2) from Proposition 82. In particular, \[ \mathcal{T}_{X \times Y} := \left\{ Z \, \, \colon \,\, Z = \bigcup_{i \in I} U_i \times V_i , \,\, U_i \times V_i \in \mathcal{B} \right\} \tag{3.13}\] is a topology on \(X \times Y\).

Proof

The proof that \(\mathcal{B}\) satisfies (B1)-(B2) is an easy check, and is left as an exercise. As \(\mathcal{B}\) satisfies (B1)-(B2), by Proposition 83 we know that \(\mathcal{T}_{X \times Y}\) is a topology on \(X \times Y\).

Definition 85: Product topology

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. The product topology on \(X \times Y\) is the collection of sets \(\mathcal{T}_{X \times Y}\) at (3.13).

Example 86

Let \(\mathbb{R}\) be equipped with the (one dimensional) Euclidean topology. The product topology on \(\mathbb{R}\times \mathbb{R}\) coincides with the topology on \(\mathbb{R}^2\) equipped with the (two dimensional) Euclidean topology.

Definition 87: Projection maps

Given two sets \(X,Y\) we define the projection maps as \[\begin{align*} & \pi_X \colon X \times Y \to X \,, \quad \pi_X (x,y):=x \,,\\ & \pi_Y \colon X \times Y \to Y \,, \quad \pi_Y (x,y):=y \,. \end{align*}\]

Proposition 88: Projections \(\pi_X\), \(\pi_Y\) are continuous for \(\mathcal{T}_{X \times Y}\)

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces and equip \(X \times Y\) with the product topology \(\mathcal{T}_{X \times Y}\). Then \(\pi_X\) and \(\pi_Y\) are continuous.

Proof

Let \(U \in \mathcal{T}_X\). Then \[ {\pi}_{X}^{-1} (U) = U \times Y \,. \] We have that \(U \times Y \in \mathcal{T}_{X \times Y}\) since \(U \in \mathcal{T}_X\) and \(Y \in \mathcal{T}_Y\). Therefore \(\pi_X\) is continuous. The proof that \(\pi_Y\) is continuous is similar, and is left as an exercise.

The following proposition gives a useful criterion to check whether a map into \(X \times Y\) is continuous.

Proposition 89

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces and equip \(X \times Y\) with the product topology \(\mathcal{T}_{X \times Y}\). Let \((Z,\mathcal{T}_Z)\) be a topological space and \[ f \colon Z \to X \times Y \] a function. They are equivalent:

1. \(f\) is continuous.
2. The compositions \[ \pi_X \circ f \colon Z \to X \,, \quad \pi_Y \circ f \colon Z \to Y \] are continuous.

The proof is left as an exercise.

3.12 Connectedness

Suppose that \((X,\mathcal{T})\) is a topological space. By property (A1) we have that \[ \emptyset \,, \,\, X \in \mathcal{T} \] Therefore \[ \emptyset^c = X \,, \quad X^c = \emptyset \] are closed. It follows that \(\emptyset\) and \(X\) are both open and closed.

Definition 90: Connected space

Let \((X,\mathcal{T})\) be a topological space. We say that:

1. \(X\) is connected if the only subsets of \(X\) which are both open and closed are \(\emptyset\) and \(X\).
2. \(X\) is disconnected if it is not connected.

The following proposition gives two extremely useful equivalent definitions of connectedness. Before stating it, we define the concept of proper set.

Definition 91: Proper subset

Let \(X\) be a set. A subset \(A \subseteq X\) is proper if \(A \neq \emptyset\) and \(A \neq X\).

Proposition 92: Equivalent definition for connectedness

Let \((X,\mathcal{T})\) be a topological space. They are equivalent:

1. \(X\) is disconnected.
2. \(X\) is the disjoint union of two proper open subsets.
3. \(X\) is the disjoint union of two proper closed subsets.

Proof

Part 1. Point 1 implies Points 2 and 3.

Suppose \(X\) is disconnected. Then there exists \(U \subseteq X\) which is open, closed, and such that \[ U \neq \emptyset \,, \quad U \neq X \,. \tag{3.14}\] Define \(A:= U\), \(B:= U^c\). By definition of complement we have \[ X = A \cup B \,, \quad A \cap B = \emptyset \,. \] Moreover:

• \(A\) and \(B\) are both open and closed, since \(U\) is both open and closed.

• \(A\) and \(B\) are proper, since (3.14) holds.

Therefore we conclude Points 2, 3.

Part 2. Point 2 implies Point 1. Suppose \(A,B\) are open, proper, and such that \[ X = A \cup B \,, \quad A \cap B = \emptyset \,. \] This implies \[ A^c = X \smallsetminus A = B \,, \] showing that \(A^c\) is open, and hence \(A\) is closed. Therefore \(A\) is proper, open and closed, showing that \(X\) is disconnected.

Part 3. Point 3 implies Point 1. Suppose \(A,B\) are closed, proper, and such that \[ X = A \cup B \,, \quad A \cap B = \emptyset \,. \] This implies \[ A^c = X \smallsetminus A = B \,, \] showing that \(A^c\) is closed, and hence \(A\) is open. Therefore \(A\) is proper, open and closed, showing that \(X\) is disconnected.

In the following we will use Point 2 and Point 3 in Proposition 92 as equivalent definitions of disconnected topological space.

Example 93

Question. Consider the set \(X = \{0,1\}\) with the subspace topology induced by the inclusion \(X \subseteq \mathbb{R}\), where \(\mathbb{R}\) is equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\). Prove that \(X\) is disconnected.

Solution. Note that \[ X = \{ 0 \} \cup \{ 1 \} \,, \quad \{ 0 \} \cap \{ 1 \} = \emptyset \,. \] The set \(\{ 0 \}\) is open for the subspace topology, since \[ \{ 0 \} = X \cap (-1,1) \,, \quad (-1,1) \in \mathcal{T}_{\textrm{euclid}}\,. \] Similarly, also \(\{ 1 \}\) is open for the subspace topology, since \[ \{ 1 \} = X \cap (0,2) \,, \quad (0,2) \in \mathcal{T}_{\textrm{euclid}}\,. \] Since \(\{ 0 \}\) and \(\{ 1 \}\) are proper subsets of \(X\), we conclude that \(X\) is disconnected.

Example 94

Question. Let \(\mathbb{R}\) be equipped with \(\mathcal{T}_{\textrm{euclid}}\), and let \(p \in \mathbb{R}\). Prove that the set \(X = \mathbb{R}\smallsetminus \{p\}\) is disconnected.

Solution. Define the sets \[ A = (-\infty,p) \,, \quad B = (p , \infty) \,. \] \(A\) and \(B\) are proper subsets of \(X\). Moreover \[ X = A \cup B \,, \quad A \cap B = \emptyset \,. \] Finally, \(A,B\) are open for the subspace topology on \(X\), since they are open in \((\mathbb{R},\mathcal{T}_{\textrm{euclid}})\). Therefore \(X\) is disconnected.

As it is easy to imagine, the conclusion of 94 is false if the dimension is \(n \geq 2\). In order to prove the statement rigorously, we need a technical lemma.

Lemma 95

Let \((X,\mathcal{T})\) be a topological space. Let \(A, U \subseteq X\) with \(A\) connected and \(U\) open and closed. Suppose that \(A \cap U \neq \emptyset\). Then \(A \subseteq U\).

Proof

The following set identities hold for any pair of sets \(U\) and \(A\): \[\begin{align*} A & = ( A \cap U ) \cup (A \cap U^c) \\ \emptyset & = ( A \cap U ) \cap (A \cap U^c) \end{align*}\] Now, suppose by contradiction \(A \not\subseteq U\). This means \(A \cap U^c \neq \emptyset\). By assumption we also have \(A \cap U \neq \emptyset\). Moreover the sets \(A \cap U\) and \(A \cap U^c\) are open for the subspace topology on \(A\), since \(U\) and \(U^c\) are open in \(X\). Hence \(A\) is the disjoint union of non-empty open sets, showing that \(A\) is disconnected. Contradiction. Thus \(A \subseteq U\).

Example 96

Question. Let \(n \geq 2\), and \(A \subseteq \mathbb{R}^n\) be open and connected. Let \(\mathbf{p}\in A\). Prove that \(X = A \smallsetminus \{\mathbf{p}\}\) is connected.

Solution. Assume that \[ X = U \cup V\,, \] with \(U,V\) disjoint and open in \(X\). If we show that \(U,V\) are not proper, we conclude that \(X\) is connected. In order to prove it, start by noting that \(X = A \smallsetminus \{\mathbf{p}\}\) is open, since \(A\) is open. As the sets \(U,V\) are open for the subspace topology on \(X\), and \(X\) is open in \(\mathbb{R}^n\), we conclude that \(U,V\) are open in \(\mathbb{R}^n\). As \(U,V\) are also closed for the subspace topology, we conclude that they are closed in \(\mathbb{R}^n\). As \(A\) is open, and \(\mathbf{p}\in A\), there exists \(r>0\) such that \(B_r(\mathbf{p}) \subseteq A\). Since \(X = A \smallsetminus \{\mathbf{p}\}\), we have \[ B_r(\mathbf{p}) \smallsetminus \{\mathbf{p}\} \subseteq X \,. \] As \(X = U \cup V\), we have \[ \left( B_r(\mathbf{p}) \smallsetminus \{\mathbf{p}\} \right) \cap U \neq \emptyset \quad \text{or} \quad \left( B_r(\mathbf{p}) \smallsetminus \{\mathbf{p}\} \right) \cap V \neq \emptyset \,. \] Without loss of generality, assume that \(\left( B_r(\mathbf{p}) \smallsetminus \{\mathbf{p}\} \right) \cap U \neq \emptyset\) (the argument is similar in the other case). Since \(B_r(\mathbf{p}) \smallsetminus \{\mathbf{p}\}\) is open, and \(U\) is open and closed, by Lemma 95 we conclude that \[ B_r(\mathbf{p}) \smallsetminus \{\mathbf{p}\} \subseteq U \quad \implies \quad B_r(\mathbf{p}) \subseteq U' := U \cup \{\mathbf{p}\} \,. \] Let \(\mathbf{q} \in U'\). We have two cases:

• \(\mathbf{q} \neq \mathbf{p}\): Then \(\mathbf{q} \in U\). As \(U\) is open, there exists \(\varepsilon>0\) such that \[ B_{\varepsilon}(\mathbf{q}) \subseteq U \subseteq U' \,. \]

• \(\mathbf{q} = \mathbf{p}\): We have shown that \(B_r(\mathbf{p}) \subseteq U'\).

This shows \(U'\) is open in \(\mathbb{R}^n\). In conclusion, we have \[ X = U \cup V \quad \implies \quad A = U' \cup V \,, \] with \(U',V\) disjoint and open in \(A\) (since they are open in \(\mathbb{R}^n\)). As \(\mathbf{p}\in U'\), we conclude that \(U' \neq \emptyset\). By connectedness of \(A\), we must have \(V = \emptyset\). This implies \(U = X\). Therefore \(U\) and \(V\) are not proper, implying that \(X\) is connected.

The next theorem shows that connectedness is preserved by continuous maps.

Theorem 97

Let \((X,\mathcal{T}_X)\), \((Y,\mathcal{T}_Y)\) be topological spaces. Suppose that \(f \colon X \to Y\) is continuous and let \(f(X) \subseteq Y\) be equipped with the subspace topology. If \(X\) is connected, then \(f(X)\) is connected.

Proof

Suppose that \(A,B\) are open in \(f(X)\) and such that \[ f(X) = A \cup B \,, \quad A \cap B = \emptyset \,. \] If we show that \[ A = \emptyset \,\, \mbox{ or } \,\, B = \emptyset \tag{3.15}\] the proof is concluded. Since \(A,B\) are open for the subspace topology, there exist \(\widetilde{A}, \widetilde{B} \in \mathcal{T}_Y\) such that \[ A = \widetilde{A} \cap f(X) \,, \quad B = \widetilde{B} \cap f(X) \,. \tag{3.16}\] Since \(f(X) = A \cup B\) we have \[\begin{align*} X & = {f}^{-1} (A \cup B) \\ & = {f}^{-1} (A ) \cup {f}^{-1} (B) \\ & = {f}^{-1} (\widetilde{A} ) \cup {f}^{-1} (\widetilde{B}) \end{align*}\] where in the last equality we used (3.16). Since \(A \cap B = \emptyset\), we also have that \[\begin{align*} {f}^{-1} (\widetilde{A} ) \cap {f}^{-1} (\widetilde{B}) & = {f}^{-1} (A ) \cap {f}^{-1} (B) \\ & = {f}^{-1} (A \cap B) \\ & = {f}^{-1} (\emptyset) \\ & = \emptyset \end{align*}\] where in the first equality we used (3.16). By continuity of \(f\) we have that \[ f^{-1}(\widetilde{A}) \,, \,\, f^{-1}(\widetilde{B}) \in \mathcal{T}_X \,. \] Therefore, using that \(X\) is connected, we deduce that \[ f^{-1}(\widetilde{A}) = \emptyset \,\, \mbox{ or } \,\, f^{-1}(\widetilde{B}) = \emptyset \,. \] The above implies \[ \widetilde{A} \cap f(X) = \emptyset \,\, \mbox{ or } \,\, \widetilde{B} \cap f(X) = \emptyset \,. \] Recalling (3.16), we obtain (3.15), ending the proof.

An immediate corollary of Theorem 97 is that connectedness is a topological invariant, e.g., connectedness is preserved by homeomorphisms.

Theorem 98: Connectedness is topological invariant

Let \((X,\mathcal{T}_X)\), \((Y,\mathcal{T}_Y)\) be homeomorhic topological spaces. Then \[ X \, \mbox{ is connected } \,\, \iff \,\, Y \, \mbox{ is connected } \]

The proof follows immediately by Theorem 97, and is left to the reader as an exercise.

Example 99

Question. Let \(n \geq 2\). Prove that \(\mathbb{R}^n\) is not homeomorphic to \(\mathbb{R}\).

Solution. Suppose by contradiction that there exists a homeomorphism \(f \colon \mathbb{R}^n \to \mathbb{R}\). Define \(p = f({\pmb{0}})\) and the restriction \[ g \colon \left(\mathbb{R}^n \smallsetminus \{{\pmb{0}}\} \right) \to \left(\mathbb{R}\smallsetminus \{p\} \right) \,, \quad g(x) = f(x) \,. \] Note that \(g\) is a homeomorphism, being restriction of a homeomorphism. By Example 96, we have that \(\mathbb{R}^n \smallsetminus \{{\pmb{0}}\}\) is connected. Hence, by Theorem 98, we infer that \(\mathbb{R}\smallsetminus \{p\}\) is connected. This is a contradiction, since \(\mathbb{R}\smallsetminus \{p\}\) is disconnected, as shown in Example 94.

A stronger version of the statement in Example 99 holds.

Theorem 100: Topological invariance of dimension

Let \(n \neq m\). Then \(\mathbb{R}^n\) is not homeomorphic to \(\mathbb{R}^m\).

Unfortunately, the argument in Example 99 cannot be extended to prove Theorem 90. The bottom line is that connectedness does not give enough information to tell apart \(\mathbb{R}^n\) from \(\mathbb{R}^m\). The right topological tool to prove Theorem 90 is called homology, which requires a serious effort to construct/define.

Let us give another example of spaces which are not homeomorphic.

Example 101

Question. Define the one dimensional unit circle \[ \mathbb{S}^1 := \{(x,y) \in \mathbb{R}^2 \, \colon \, x^2 + y^2 = 1 \} \,. \] Prove that \(\mathbb{S}^1\) and \([0,1]\) are not homeomorphic.

Solution. Suppose by contradiction that there exists a homeomorphism \[ f \colon [0,1] \to \mathbb{S}^1 \,. \] The restriction of \(f\) to \([0,1] \smallsetminus \{\frac12\}\) defines a homeomorphism \[ g \ \colon \left( [0,1] \smallsetminus \left\{\frac12\right\} \right) \to \left( \mathbb{S}^1 \smallsetminus \{\mathbf{p}\} \right) \,, \quad \mathbf{p} := f\left(\frac12 \right) \,. \] The set \([0,1] \smallsetminus \left\{ \frac12 \right\}\) is disconnected, since \[ [0,1] \smallsetminus \{1/2\} = [0,1/2) \, \cup \, (1/2,1] \] with \([0,1/2)\) and \((1/2,1]\) open for the subset topology, non-empty and disjoint. Therefore, using that \(g\) is a homeomorphism, we conclude that also \(\mathbb{S}^1 \smallsetminus \{\mathbf{p}\}\) is disconnected. Let \(\theta_0 \in [0,2\pi)\) be the unique angle such that \[ \mathbf{p} = (\cos (\theta_0),\sin(\theta_0)) \,. \] Thus \(\mathbb{S}^1 \smallsetminus \{\mathbf{p}\}\) is parametrized by \[ {\pmb{\gamma}}(t):=(\cos(t),\sin(t)) \,, \quad t \in (\theta_0,\theta_0 + 2\pi) \,. \] Since \({\pmb{\gamma}}\) is continuous and \((\theta_0,\theta_0 + 2\pi)\) is connected, by Theorem 97, we conclude that \(\mathbb{S}^1 \smallsetminus \{\mathbf{p}\}\) is connected. Contradiction.

3.13 Intermediate Value Theorem

Another consequence of Theorem 97 is a generalization of the Intermediate Value Theorem to arbitrary topological spaces. Before providing statement and proof of such Theorem, we need to characterize the connected subsets of \(\mathbb{R}\).

Definition 102: Interval

A subset \(I \subset \mathbb{R}\) is an interval if it holds: \[ \forall \, a,b \in I \,, \, x \in \mathbb{R}\, \, \text{ s.t. } \, a<x<b \quad \implies \quad x \in I \,. \]

Theorem 103: Intervals are connected

Let \(\mathbb{R}\) be equipped with the Euclidean topology and let \(I \subseteq \mathbb{R}\). They are equivalent:

1. \(I\) is connected.
2. \(I\) is an interval.

Proof

Part 1. Suppose \(I\) is connected. If \(I=\{p\}\) for some \(p \in \mathbb{R}\) then \(I\) is an interval and the thesis is achieved. Otherwise there exist \(a,b \in I\) with \(a<b\). Assume that \(x \in \mathbb{R}\) is such that \[ a < x < b \,. \] We need to show that \(x \in I\). Suppose by contradiction that \(x \notin I\) and define the open sets \[ A = (-\infty,x) \,, \quad B = (x,\infty) \,. \] Then \[ \widetilde{A} = (-\infty,x) \cap I \,, \quad \widetilde{B} = (x,\infty) \cap I \] are open in \(I\) for the subspace topology. Clearly \[ \widetilde{A} \cap \widetilde{B} = \emptyset \,. \] Moreover \[ I = \widetilde{A} \cup \widetilde{B} \] since \(x \notin I\). We have:

• Since \(a<x\) and \(a \in I\), we have that \(a \in \widetilde{A}\). Therefore \(\widetilde{A} \neq \emptyset\).
• Similarly, \(b>x\) and \(b \in I\), therefore \(b \in \widetilde{B}\). Hence \(\widetilde{B} \neq \emptyset\).

Therefore \(I\) is disconnected, which is a contradiction.

Part 2. Suppose \(I\) is an interval. Suppose by contradiction that \(I\) is disconnected. Then there exist \(A,B\) proper and closed, such that \[ I = A \cup B \,, \quad A \cap B = \emptyset \,. \] Since \(A\) and \(B\) are proper, there exist points \(a \in A\), \(b \in B\). WLOG we can assume \(a<b\). Define \[ \alpha = \sup \ S \,, \quad S:= \{ x \in \mathbb{R}\colon [a,x) \cap I \subseteq A \} \,. \]

Note that \(\alpha\) exists finite since \(b\) is an upper bound for the set \(S\).

Suppose by contradiction \(b\) is not an upper bound for \(S\). Hence there exists \(x \in \mathbb{R}\) such that \([a,x) \cap I \subseteq A\) and that \(x>b\). As \(b>a\), we conclude that \(b \in [a,x) \cap I \subseteq A\). Thus \(b \in A\), which is a contradiction, since \(b \in B\) and \(A \cap B = \emptyset\).

Moreover we have that \(\alpha \in A\).

This is because the supremum \(\alpha\) is the limit of a sequence in \(S\), and hence of a sequence in \(A\). Therefore \(\alpha\) belongs to \(\overline{A}\). Since \(A\) is closed, we infer \(\alpha \in A\).

Note that \(A^c = B\), which is closed. Therefore \(A^c\) is closed, showing that \(A\) is open. As \(\alpha \in A\) and \(A\) is open in \(I\), there exists \(\varepsilon>0\) such that \[ (\alpha - \varepsilon, \alpha + \varepsilon) \cap I \subseteq A \,. \] In particular \[ [a , \alpha + \varepsilon) \cap I \subseteq A \,, \] showing that \(\alpha + \varepsilon\in S\). This is a contradiction, since \(\alpha\) is the supremum of \(S\).

We are finally ready to prove the Intermediate Value Theorem.

Theorem 104: Intermediate Value Theorem

Let \((X,\mathcal{T})\) be a connected topological space. Suppose that \(f \colon X \to \mathbb{R}\) is continuous. Suppose that \(a,b \in X\) are such that \(f(a)<f(b)\). It holds: \[ \forall \, c \in \mathbb{R}\, \text{ s.t. } \, f(a)< c < f(b) \,, \,\, \exists \, \xi \in X \, \text{ s.t. } \, f(\xi) = c \,. \]

Proof

As \(f\) is continuous and \(X\) is connected, by Theorem 97 we know that \(f(X)\) is connected in \(\mathbb{R}\). By Theorem 103 we have that \(f(X)\) is an interval. Since \(a,b \in X\) it follows \(f(a), f(b) \in f(X)\). Therefore, if \(c \in \mathbb{R}\) is such that \[ f(a) \leq c \leq f(b) \] we conclude that \(c \in f(X)\), since \(f(X)\) is an interval. Hence there exists \(\xi \in X\) such that \(f(\xi) = c\).

There is an alternative proof to the fact that intervals are connected. It makes use of the classical Intermediate Value Theorem in \(\mathbb{R}\). It is an interesting exercise.

Example 105: Intervals are connected - Alternative proof

Question. Prove the following statements.

1. Let \((X,\mathcal{T})\) be a disconnected topological space. Prove that there exists a function \(f \colon X \to \{0,1\}\) which is continuous and surjective.

2. Consider \(\mathbb{R}\) equipped with the Euclidean topology. Let \(I \subseteq \mathbb{R}\) be an interval. Use point (1), and the Intermediate Value Theorem in \(\mathbb{R}\) (see statement below), to show that \(I\) is connected.

Intermediate Value Theorem in \(\mathbb{R}\): Suppose that \(f \colon [a,b] \to \mathbb{R}\) is continuous, and \(f(a) < f(b)\). Let \(c \in \mathbb{R}\) be such that \(f(a) \leq c \leq f(b)\). Then, there exists \(\xi \in [a,b]\) such that \(f(\xi) = c\).

Solution. Part 1. Since \(X\) is disconnected, there exist \(A,B \in \mathcal{T}\) proper and such that \[ X = A \cup B \,, \quad A \cap B = \emptyset \,. \] Define \(f \colon X \to \{0,1\}\) by \[ f(x) = \begin{cases} 0 & \,\, \mbox{ if } \, x \in A \\ 1 & \,\, \mbox{ if } \, x \in B \\ \end{cases} \] Since \(A\) and \(B\) are non-empty, it follows that \(f\) is surjective. Moreover \(f\) is continuous: Indeed suppose \(U \subseteq \mathbb{R}\) is open. We have 4 cases:

• \(0,1 \notin U\). Then \(f^{-1}(U) = \emptyset \in \mathcal{T}\).
• \(0 \in U, \, 1 \notin U\). Then \(f^{-1}(U) = A \in \mathcal{T}\).
• \(0 \notin U, \, 1 \in U\). Then \(f^{-1}(U) = B \in \mathcal{T}\).
• \(0,1 \in U\). Then \(f^{-1}(U) = X \in \mathcal{T}\).

Then \(f^{-1}(U) \in \mathcal{T}\) for all \(U \subseteq \mathbb{R}\) open, showing that \(f\) is continuous.

Part 2. Let \(I \subseteq \mathbb{R}\) be an interval. Suppose by contradiction \(I\) is disconnected. By Point (1), there exists a map \(f \colon I \to \{0,1\}\) which is continuous and surjective. As \(f\) is surjective, there exist \(a,b \in I\) such that \[ f(a) = 0 \,, \quad f(b) = 1 \,. \] Since \(f\) is continuous, and \(f(a) = 0 < 1 = f(b)\), by the Intermediate Value Theorem in \(\mathbb{R}\), there exists \(\xi \in [a,b]\) such that \(f(\xi)=1/2\). As \(I\) is an interval, \(a,b \in I\), and \(a\leq \xi \leq b\), it follows that \(\xi \in I\). This is a contradiction, since \(f\) maps \(I\) into \(\{0,1\}\), and \(f(\xi) = 1/2 \notin \{0,1\}\). Therefore \(I\) is connected.

3.14 Path-connectedness

Definition 106: Path-connectedness

Let \((X,\mathcal{T})\) be a topological space. We say that \(X\) is path-connected if for every \(x,y \in X\) there exist \(a,b \in \mathbb{R}\) with \(a<b\), and a continuous function \[ \alpha \colon [a,b] \to X \,\, \, \text{ s.t. } \, \,\, \alpha (a) = x \,, \quad \alpha(b) = y \,. \]

It turns out that path-connectedness implies connectedness.

Theorem 107: Path-connectedness implies connectedness

Let \((X,\mathcal{T})\) be a path-connected topological space. Then \(X\) is connected.

Proof

Suppose that \(X = A \cup B\) with \(A, B \in \mathcal{T}\) and non-empty. In order to conclude that \(X\) is connected, we need to show that \[ A \cap B \neq \emptyset \,. \] Since \(A\) and \(B\) are non-empty, we can find two points \(x \in A\) and \(b \in B\). As \(X\) is path-connected, there exists \(\alpha \colon [0,1] \to X\) continuous such that \(\alpha(0) = x\) and \(\alpha(1) = y\). In particular, \[ \alpha^{-1} (A) \neq \emptyset \, , \quad \alpha^{-1} (B) \neq \emptyset \,. \] Moreover \[\begin{align*} [0,1] & = \alpha^{-1}(X) \\ & = \alpha^{-1}(A \cup B) \\ & = \alpha^{-1}(A) \cup \alpha^{-1}(B) \,. \end{align*}\] As \(\alpha\) is continuous, \(\alpha^{-1}(A)\) and \(\alpha^{-1}(B)\) are open in \([0,1]\). Suppose by contradiction that \(A \cap B = \emptyset\). Then \[ \alpha^{-1}(A) \cap \alpha^{-1}(B) = \alpha^{-1}(A \cap B) = \alpha^{-1}(\emptyset) = \emptyset \,. \] Hence \([0,1]\) is disconnected, which is a contradiction. Therefore \(A \cap B \neq \emptyset\) and \(X\) is connected.

Example 108

Question. Let \(A \subseteq \mathbb{R}^n\) be convex. Show that \(A\) is path-connected, and hence connected.

Solution. A is convex if for all \(x,y \in A\) the segment connecting \(x\) to \(y\) is contained in \(A\), namely, \[ [x,y] := \{ (1-t)x + t y \, \colon \,t \in [0,1] \} \subseteq A \,. \] Therefore we can define \[ \alpha \colon [0,1] \to A \,, \quad \alpha(t):=(1-t)x + t y \,. \] Clearly \(\alpha\) is continuous, and \(\alpha(0)=x, \alpha(1)=y\).

Example 109: Spaces of matrices

Let \(\mathbb{R}^{2 \times 2}\) denote the space of real \(2 \times 2\) matrices. Assume \(\mathbb{R}^{2 \times 2}\) has the euclidean topology obtained by identifying it with \(\mathbb{R}^4\).

1. Consider the set of orthogonal matrices \[ \mathrm{O}(2) = \{ A \in \mathbb{R}^{2 \times 2} \, \colon \, A^TA = I \} \,. \] Prove that \(\mathrm{O}(2)\) is disconnected.

2. Consider the set of rotations \[ \mathrm{SO}(2) = \{ A \in \mathbb{R}^{2 \times 2} \, \colon \, A^TA = I ,\,\, \det(A) = 1 \} \,. \] Prove that \(\mathrm{SO}(2)\) is path-connected, and hence connected.

Solution. Let \(A \in \mathrm{O}(2)\), and denote its entries by \(a,b,c,d\). By direct calculation, the condition \(A^T A = I\) is equivalent to \[ a^2 + b^2 = 1 \,, \qquad b^2 + c^2 = 1 \,, \qquad ac + bd = 0 \,. \] From the first condition, we get that \(a = \cos(t)\) and \(b = \sin(t)\), for a suitable \(t \in [0,2\pi)\). From the second and third conditions, we get \(c = \pm \sin(t)\) and \(d = \mp \cos(t)\). We decompose \(\mathrm{O}(2)\) as \[\begin{align*} & \mathrm{O}(2) = A \cup B \,, \\ & A = \mathrm{SO}(2) = \left\{ \left( \begin{array}{cc} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{array} \right) , \, t \in [0,2\pi) \right \} \\ & B = \left\{ \left( \begin{array}{cc} \cos(t) & \sin(t) \\ \sin(t) & -\cos(t) \end{array} \right) , \, t \in [0,2\pi) \right \} \,. \end{align*}\]

1. The determinant function \(\det \colon \mathrm{O}(2) \to \mathbb{R}\) is continuous. If \(M \in A\), we have \(\det(M) = 1\). If instead \(M \in B\), we have \(\det(M) = -1\). Moreover, \[ {\det}^{-1}(\{1\}) = A \,, \qquad {\det}^{-1}(\{0\}) = B \,. \] As \(\det\) is continuous, and \(\{0\},\{1\}\) closed, we conclude that \(A\) and \(B\) are closed. Therefore, \(A\) and \(B\) are closed, proper and disjoint. Since \(\mathrm{O}(2) = A \cup B\), we conclude that \(\mathrm{O}(2)\) is disconnected.

2. Define the function \(\psi \colon [0,2\pi) \to \mathrm{SO}(2)\) by \[ \psi(t) = \left( \begin{array}{cc} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{array} \right) \,. \] Clearly, \(\psi\) is continuous. Let \(R,Q \in \mathrm{SO}(2)\). Then \(R\) is determined by an angle \(t_1\), while \(Q\) by an angle \(t_2\). Up to swapping \(R\) and \(Q\), we can assume \(t_1 < t_2\). Define the function \(f \colon [0,1] \to \mathrm{SO}(2)\) by \[ f(\lambda) = \psi( t_1 (1-\lambda) + t_2 \lambda ) \,. \] Then, \(f\) is continuous and \[ f(0) = \psi(t_1) = R, \quad f(1) = \psi(t_2) = Q \, . \] Thus \(\mathrm{SO}(2)\) is path-connected.

Warning

In general connectedness does not imply path-connectedness, as seen in Proposition 110.

Proposition 110: Topologist curve

Consider \(\mathbb{R}^2\) with the Euclidean topology, and define the sets \[\begin{align*} A & := \left\{ \left( t, \sin \left( \frac{1}{t} \right) \right) \, \colon \,t>0 \right\} \\ B & := \{(0,t) \, \colon \,t \in [-1,1] \} \,, \qquad X := A \cup B \,. \end{align*}\] Then \(X\) is connected, but not path-connected.

Proof

Step 1. \(X\) is not path-connected.

Let \(x \in A\) and \(y \in B\). There is no continuous function \(\alpha \colon [0,1] \to X\) such that \(\alpha(0)=x\) and \(\alpha(1)=y\). If such \(\alpha\) existed, then we would obtain a continuous extension for \(t=0\) of the function \[ f(t) = \sin \left( \frac{1}{t} \right) \,, \quad x>0 \] which is not possible. Hence \(X\) is not path-connected.

Step 2. Preliminary facts.

• \(A\) is connected: Define the curve \({\pmb{\gamma}}\colon (0,\infty) \to \mathbb{R}^2\) by \[ {\pmb{\gamma}}(t):= \left( t, \sin \left( \frac{1}{t} \right) \right) \,. \] Clearly \({\pmb{\gamma}}\) is continuous. Since \((0,\infty)\) is connected, by Theorem 97 we have that \({\pmb{\gamma}}((0,\infty)) = A\) is connected.

• \(B\) is connected: Indeed \(B\) is homeomorphic to the interval \([-1,1]\). Since \([-1,1]\) is connected, by Theorem 98 we conclude that \(B\) is connected.

• \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = X\): This is because each point \(y \in B\) is of the form \(y = (0,t_0)\) for some \(t_0 \in [-1,1]\). By continuity of \(\sin\) and the Intermediate Value Theorem there exists some \(z>0\) such that \[ \sin(z) = t_0 \,. \] Therefore \(z_n := z + 2n\pi\) satisfies \[ z_n \to \infty \,, \quad \sin(z_n) = t_0 \,, \quad \forall \, n \in \mathbb{N}\,. \] Define \(s_n:=1/z_n\). Trivially \[ s_n \to 0 \,, \quad \sin \left( \frac{1}{s_n} \right) = t_0 \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore we obtain \[ \left( s_n , \sin \left( \frac{1}{s_n} \right) \right) \to (0,t_0) \,. \] Hence the set \(B\) is contained in the set \(L(A)\) of limit points of \(A\). Since we are in \(\mathbb{R}^2\), we have that \(L(A)={}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\), proving that \(B \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\). Thus \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}= A \cup B = X\).

Step 3. \(X\) is connected.

Let \(U \subseteq X\) be non-empty, open and closed. If we prove that \(U = X\), we conclude that \(X\) is connected. Let us proceed.
Since \(U\) is non-empty, we can fix a point \(x \in U\). We have two possibilities:

• \(x \in A\): In this case \(A \cap U \neq \emptyset\). Since \(A\) is connected and \(U\) is open and closed, by Lemma 95 we conclude \(A \subseteq U\). As \(U\) is closed and contains \(A\), then \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \subseteq U\). But we have shown that
  \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = X \,, \] and therefore \(U = X\).

• \(x \in B\): Then \(U \cap B \neq \emptyset\). Since \(B\) is connected and \(U\) is open and closed, we can invoke Lemma 95 and conclude that \(B \subseteq U\). Since \((0,0) \in B\), it follows that \[ (0,0) \in U \,. \] As \(U\) is open in \(X\), and \(X\) has the subspace topology induced by the inclusion \(X \subseteq \mathbb{R}^2\), there exists an open set \(W\) of \(\mathbb{R}^2\) such that \[ U = X \cap W \,. \] Therefore \((0,0) \in W\). As \(W\) is open in \(\mathbb{R}^2\), there exists a radius \(\varepsilon>0\) such that \[ B_{\varepsilon} (0,0) \subseteq W \,. \] Hence \[ X \cap B_{\varepsilon} (0,0) \subseteq X \cap W = U \,. \] The ball \(B_{\varepsilon} (0,0)\) contains points of \(A\), and therefore \[ A \cap U \neq \emptyset \,. \] Since \(A\) is connected and \(U\) is open and closed, we can again use Lemma 95 and obtain that \(A \subseteq U\). Since we already had \(B \subseteq U\), and since \(U \subseteq X = A \cup B\), we conclude hence \(U = X\).

Therefore \(U = X\) in all possible cases, showing that \(X\) is connected.

We conclude with the observation that connectedness and path-connectedness are equivalent for open sets of \(\mathbb{R}^n\).

Theorem 111

Let \(A \subseteq \mathbb{R}^n\) be open for the Euclidean topology. Then \(A\) is connected if and only if it is path-connected.

The proof of this theorem is a bit delicate, and we decided to omit it. We conclude with an interesting example concerning spaces of matrices.