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) We have \(\emptyset, X \in \mathcal{T}\),

• (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 topology \(\mathcal{T}\) defined by \[ \mathcal{T}:= \{ \emptyset , X \} \,. \]

Let us check that \(\mathcal{T}\) is indeed a topology. We need to verify the 3 properties of a topology:

• (A1) We clearly have \(\emptyset , X \in \mathcal{T}\).
• (A2) The only non-trivial union to check is the one between \(\emptyset\) and \(X\). We have \[ \emptyset \cup X = X \in \mathcal{T}\,. \]
• (A3) The only non-trivial intersection to check is the one between \(\emptyset\) and \(X\). We have \[ \emptyset \cap X = \emptyset \in \mathcal{T}\,. \]

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

Definition 4: Discrete topology

Let \(X\) be a set. The discrete topology on \(X\) is the topology \(\mathcal{T}\) defined by \[ \mathcal{T}:= \{ A \, \colon \,A \subseteq X \} \,, \] that is, every subset of \(X\) is open.

Let us check that \(\mathcal{T}\) is a topology:

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

• (A2) The arbitrary union of subsets of \(X\) is still a subset of \(X\). Therefore \[ \bigcup_{i \in I} A_i \in \mathcal{T}\,, \] whenever \(A_i \in \mathcal{T}\) for all \(i \in I\).

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

Therefore \(\mathcal{T}\) 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 topology \(\mathcal{T}\) defined by \[ \mathcal{T}:= \{ A \, \colon \,A \subseteq \mathbb{R}^n \,, \,\, A \, \mbox{ is open} \} \,. \]

We need to check that the above definition is well-posed, in the sense that we have to prove that \(\mathcal{T}\) is a topology on \(\mathbb{R}^n\).

Proof: Well-posedness of Definition 6

Let us check that \(\mathcal{T}\) is a topology on \(\mathbb{R}^n\):

• (A1) We have \(\emptyset , \mathbb{R}^n \in \mathcal{T}\): 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 \(A_i \in \mathcal{T}\) for all \(i \in I\) and define the union set \[ A:=\bigcup_{i \in I} A_i \,. \] 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 \(B_r(\mathbf{x}) \subseteq A\). Thus \(A\) is open and \(A \in \mathcal{T}\).

• (A3) Let \(A, B \in \mathcal{T}\). 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 \(A \cap B\) open, so that \(A \cap B \in \mathcal{T}\).

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

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

Proposition 7

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

Proof

We need to shown 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). This proves that \(B_r(\mathbf{x})\) satisfies (3.1), and is therefore open.

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.

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. Such properties are 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:

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

• Suppose \(C_i\) are closed for all \(i \in I\). Therefore \(C_i^c\) are 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.

• 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 set 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 \[ f(x_1,\ldots,x_n) = \lambda_1 x_1 + \ldots + \lambda_n x_n \,, \] where \(\lambda_1, \ldots, \lambda_n\) are given elements of \(\mathbb{K}\). For \(I \subset \mathbb{K}[x_1,\ldots,x_n]\) define \[ V(I):= \{ (a_1,\ldots,a_n) \in \mathbb{K}^n \, \colon \, f(a_1,\ldots,a_n) = 0 \,, \,\, \forall \, f \in I \} \,. \] Define \[ \mathcal{C} := \{ V(I) \, \colon \,I \subset \mathbb{K}[x_1,\ldots,x_n] \} \,. \] Then \(\mathcal{C}\) satisfies (C1), (C2) and (C3). This is an easy check, and is left as exercise. \(\mathcal{C}\) is called the Zariski Topology on the field \(\mathbb{K}^n\). This is used in algebraic geometry to study Affine Varieties, an algebraic version of surfaces, see Wikipedia page.

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: Finer and coarser topology

Let \(X\) be a set and let \(\mathcal{T}_1\), \(\mathcal{T}_2\) be topologies on \(X\). Suppose that \[ \mathcal{T}_2 \subseteq \mathcal{T}_1 \,. \] We say that:

• \(\mathcal{T}_1\) is finer than \(\mathcal{T}_2\).
• \(\mathcal{T}_2\) is coarser than \(\mathcal{T}_1\).

If it holds \[ \mathcal{T}_2 \subsetneq \mathcal{T}_1 \,, \] we say that:

• \(\mathcal{T}_1\) is strictly finer than \(\mathcal{T}_2\).
• \(\mathcal{T}_2\) is strictly coarser than \(\mathcal{T}_1\).

We say that \(\mathcal{T}_1\) and \(\mathcal{T}_2\) are the same topology if \[ \mathcal{T}_1 = \mathcal{T}_2 \,. \]

Example 12

Let \(X\) be a set and consider the trivial and discrete topologies \[ \mathcal{T}_{\textrm{trivial}} = \{ \emptyset, X\} \,, \quad \mathcal{T}_{\textrm{discrete}} = \{ A \, \colon \,A \subseteq X \} \,. \] Then \[ \mathcal{T}_{\textrm{trivial}} \subsetneq \mathcal{T}_{\textrm{discrete}} \,, \] so that \(\mathcal{T}_{\textrm{discrete}}\) is strictly finer than \(\mathcal{T}_{\textrm{trivial}}\).

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 \(\mathbb{R}\)

Consider the following family \(\mathcal{T}_{\textrm{cofinite}}\) of subsets of \(\mathbb{R}\) \[ \mathcal{T}_{\textrm{cofinite}} := \{ U \subseteq \mathbb{R}\, \colon \,U^c \, \mbox{ is finite, } \mbox{or } U^c = \mathbb{R}\}\,. \] Then \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is a topological space, and \(\mathcal{T}_{\textrm{cofinite}}\) is called the cofinite topology. We have that \[ \mathcal{T}_{\textrm{cofinite}} \subsetneq \mathcal{T}_{\textrm{euclidean}} \,. \]

Exercise: Show that \(\mathcal{T}_{\textrm{cofinite}}\) is a topology on \(\mathbb{R}\) and that \(\mathcal{T}_{\textrm{cofinite}} \subsetneq \mathcal{T}_{\textrm{euclidean}}\).

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. Consider a sequence \(\{x_n\}_{n \in \mathbb{N}} \subseteq X\) and a point \(x \in X\). We say that \(x_n\) converges to \(x_0\) if the following property holds: \[ \forall \, U \in \mathcal{T}\, \text{ s.t. } \, x_0 \in U\,, \,\, \exists \, N = N(U) \in \mathbb{N}\, \text{ s.t. } \, x_n \in U \,, \, \forall \, n \geq N \,. \tag{3.4}\]

Notation

The convergence of \(x_n\) to \(x_0\) is denoted by \[ x_n \to x_0 \quad \mbox{ or } \quad \lim_{n \to \infty} x_n = x_0 \,. \]

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: Convergence for trivial topology

Let \((X,\mathcal{T})\) be topological space, with \(\mathcal{T}\) the trivial topology, that is, \[ \mathcal{T}= \{ \emptyset, X \} \,. \] Let \(\{x_n\} \subseteq X\) be a sequence and \(x_0 \in X\) a point. Then \[ x_n \to x_0 \,. \]

Proof

To show that \(x_n \to x_0\) we need to check that (3.4) holds. Therefore, let \(U \in \mathcal{T}\) with \(x_0 \in U\). We have two cases:

• \(U = \emptyset\): This case is not possible, since \(x_0\) cannot be in \(U\).

• \(U = X\): Take \(N=1\). Since \(U\) is the whole space, then \(x_n \in U\) for all \(n \geq 1\).

As these are all the open sets, we conclude that \(x_n \to x_0\).

Warning

This example is saying that in general the topological limit of a sequence is not unique!

Proposition 16: Convergence for discrete topology

Let \((X,\mathcal{T})\) be topological space, with \(\mathcal{T}\) the discrete topology, that is, \[ \mathcal{T}= \{ A \, \colon \,A \subseteq X \} \,. \] Let \(\{x_n\} \subseteq X\) be a sequence and \(x_0 \in X\) a point. They are equivalent:

1. \(x_n \to x_0\).
2. \(\{x_n\}\) is eventually constant, that is, there exists \(N \in \mathbb{N}\) such that \[ x_n = x_0 \,, \quad \forall \, n \geq N \,. \]

Proof

Part 1. Assume that \(x_n \to x_0\).

We have to prove that \(\{x_n\}\) is eventually constant. To this end, let \[ U = \{x_0\} \,. \] Then \(U \in \mathcal{T}\). 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 \(U = \{x_0\}\), the above is saying that \(x_n = x_0\) for all \(n \geq N\). Hence \(x_n\) is eventually constant.

Part 2. Assume that \(x_n\) is eventually equal to \(x_0\).

By assumption 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 \[ \lim_{n \to \infty} \left\| \mathbf{x}_n - \mathbf{x}_0 \right\| = 0 \,. \] The above is equivalent to: For all \(\varepsilon>0\) there exists \(N \in \mathbb{N}\) such that \[ \left\| \mathbf{x}_n - \mathbf{x}_0 \right\| < \varepsilon\,, \quad \forall \, n \geq N \,. \]

Proposition 18: Convergence for Euclidean topology

Let \(\mathbb{R}^n\) be equipped with \(\mathcal{T}\) the Euclidean topology. Let \(\{\mathbf{x}_n\} \subseteq \mathbb{R}^n\) be a sequence and \(\mathbf{x}_0 \in \mathbb{R}^n\) a point. They are equivalent:

1. \(\mathbf{x}_n \to \mathbf{x}_0\) with respect to \(\mathcal{T}\).
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 \(\mathcal{T}\).

Fix \(\varepsilon>0\) and consider the set \[ U := B_{\varepsilon}(\mathbf{x}_0) \,. \] By Proposition 7 we know that \(U \in \mathcal{T}\). Moreover \(\mathbf{x}_0 \in U\). By the convergence \(\mathbf{x}_n \to \mathbf{x}_0\) with respect to \(\mathcal{T}\), 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 \(U \in \mathcal{T}\) be 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 \[ \left\| \mathbf{x}_n - \mathbf{x}_0 \right\| < r \,, \quad \forall \, n \geq N \,. \] The above is equivalent to \[ \mathbf{x}_n \in B_{r}(\mathbf{x}_0) \,, \quad \forall \, n \geq N \,. \] Since \(B_r(\mathbf{x}_0) \subseteq U\), we have proven that \[ \mathbf{x}_n \in U \,, \quad \forall \, n \geq N \,. \] Since \(U\) is arbitrary, we conclude that \(\mathbf{x}_n \to \mathbf{x}_0\) with respect to \(\mathcal{T}\).

Notation

Since classical convergence in \(\mathbb{R}^n\) agrees with topological convergence with respect to \(\mathcal{T}\), 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 \[ x_n \in U \,, \quad \forall \, n \geq N \,. \] Since \(U \in \mathcal{T}_2\), the above proves \(x_n \to x_0\) in \(\mathcal{T}_2\).

3.4 Metric spaces

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

Definition 20: Distance

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: The distance is non-negative \[ d(x,y) \geq 0 \,. \] Moreover \[ d(x,y) = 0 \quad \iff x=y \,. \]

• (M2) Symmetry: The distance is symmetric \[ d(x,y) = d(y,x) \,. \]

• (M3) Triangle Inequality: It holds \[ d(x,z) \leq d(x,y) + d(y,z) \,. \]

Definition 21: Metric space

Let \(X\) be a set and \(d \colon X \times X \to \mathbb{R}\) be a distance on \(X\). We say that the pair \((X,d)\) is a metric space.

Example 22: \(\mathbb{R}^n\) as metric space

The Euclidean norm naturally induces a distance over \(\mathbb{R}^n\) by setting \[ d(\mathbf{x},\mathbf{y}) := \left\| \mathbf{x}- \mathbf{y} \right\| \,. \] Then \((\mathbb{R}^n,d)\) is a metric space.

It is trivial to check that the Euclidean distance satisfies (M1) and (M2). To show (M3), recalling the triangle inequality in \(\mathbb{R}^n\): \[ \| \mathbf{x}+ \mathbf{y}\| \leq \| \mathbf{x}\| + \| \mathbf{y}\| \,, \] for all \(\mathbf{x}, \mathbf{y}\in \mathbb{R}^n\). Using the above we obtain \[\begin{align*} d(\mathbf{x},\mathbf{y}) & = \| \mathbf{x}- \mathbf{y}\| \\ & = \| (\mathbf{x}- \mathbf{z}) + (\mathbf{z}- \mathbf{y}) \| \\ & \leq \| \mathbf{x}- \mathbf{z}\| + \| \mathbf{z}- \mathbf{y}\| \\ & = d (\mathbf{x},\mathbf{z}) + d (\mathbf{z},\mathbf{y}) \,, \end{align*}\] proving that \(d\) satisfies (M3). This prove that \((\mathbb{R}^n,d)\) is a metric space.

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

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

Indeed 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 \(p \geq 1\). 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 \[ \| \mathbf{x}+ \mathbf{y}\|_p \leq \| \mathbf{x}\|_p + \| \mathbf{y}\|_p \,, \] for all \(\mathbf{x},\mathbf{y}\in \mathbb{R}^n\). 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 24: Topology induced by the metric

Let \((X,d)\) be a metric space. We define the topology \(\mathcal{T}_d\) induced by the metric \(d\) as the collection of sets \(U \subseteq X\) that satisfy the following property: \[ \forall \, x \in U \,, \, \exists \, r \in \mathbb{R}, \, r > 0 \, \, \text{ s.t. } \, \, B_r(x) \subseteq U \,, \] where \(B_r(x)\) is the ball centered at \(x\) of radius \(r\). This is defined by \[ B_r(x) := \{ y \in X \, \colon \, d(x,y)<r \} \,. \]

We need to check that the above definition is well-posed, that is, we need to show that \(\mathcal{T}_d\) is actually a topology on \(X\). The proof follows, line by line, the proof that the Euclidean topology is indeed a topology, see proof immediately below Definition 6. This is left as an exercise.

Example 25: Topology induced by Euclidean distance

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

Exercise: Prove the above statement. It is an immediate consequence of definitions.

Example 26: Discrete distance

Let \(X\) be a set. Define the function \(d \colon X \times X \to \mathbb{R}\) by \[ d(x,y) := \begin{cases} 0 & \mbox{ if } \, x = y \\ 1 & \mbox{ if } \, x \neq y \end{cases} \] Then \((X,d)\) is a metric space, and \(d\) is called the discrete distance. Moreover \[ \mathcal{T}_d = \mathcal{T}_{\textrm{discrete}} \] where \(\mathcal{T}_{\textrm{discrete}}\) is the discrete topology on \(X\).

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

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 27

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

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

We now define the concept of equivalent metrics.

Definition 28: 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 29

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 29 is trivial, and is left as an exercise.

Example 30

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

This follows from Proposition 29 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 3.3.

Figure 3.3: 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 31: Convergence in metric space

Suppose \((X,d)\) is a metric space and denote by \(\mathcal{T}_d\) the topology induce 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 32: Interior of a set

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

Remark 33

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 34

Let \((X,\mathcal{T})\) be a topological space and \(A \subseteq X\) a set. 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

We have:

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 35: Closure of a set

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

Remark 36

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 37

Let \((X,\mathcal{T})\) be a topological space and \(A \subseteq X\) a set. 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

We have:

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 38

Let \((X,\mathcal{T})\) be a topological space and \(A \subseteq X\) a set. 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. We have \[ A \cap U = A \cap (\overline{A})^c = \emptyset \,, \] since \(A \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\).

• 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 39: Boundary of a set

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

Proposition 40

Let \((X,\mathcal{T})\) be a topological space and \(A \subseteq X\) a set. 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 41

Let \((X,\mathcal{T})\) be a topological space and \(A \subseteq X\). 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 42

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

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

The converse of Proposition 42 is false in general, that is, \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \not\subset L(A) \,. \] We show a counterexample of the above in Example 43. The above relation holds in the so-called first countable topological spaces, such as metric spaces, see Proposition 44 below.

Example 43: Co-countable topology

Let \(X=\mathbb{R}\) with the co-countable topology \[ \mathcal{T}:= \{ A \subseteq \mathbb{R}\, \colon \,A^c = \mathbb{R}\,\, \mbox{ or } \,\, A^c \, \mbox{ countable }\} \,. \] The set \[ A = (-\infty,0] \] is not closed and \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = \mathbb{R}\). Moreover, convergent sequences in \((X,\mathcal{T})\) are eventually constant. Therefore \(L(A) = A\), showing that \({}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \not\subset L(A)\).

Exercise: Prove all the above statements.

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

Proposition 44

Let \((X,d)\) be a metric space. Denote by \(\mathcal{T}_d\) the topology induced by \(d\). Let \(A \subseteq X\). We have \[ \mathop{\mathrm{Int}}{A} = \{ x \in A \, \colon \,\exists \,\, r>0 \, \text{ s.t. } \, B_r(x) \subseteq A \} \,. \tag{3.6}\] and \[ {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = L(A):= \{ x \in X \, \text{ s.t. } \, \exists \,\, \{ x_n \} \subseteq A \, \text{ s.t. } \, x_n \to x \} \,. \tag{3.7}\]

Proof

The proof of (3.6) is left as an exercise. Let us prove (3.7). The inclusion \(L(A) \subseteq {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {}\) holds by Proposition 42. 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 38 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 \[ d(x_n,x_0) < \frac{1}{n} \to 0 \] as \(n \to \infty\). Thus \(x_n \to x_0\), showning that \(x_0 \in L(A)\).

Example 45

Consider \(\mathbb{R}\) with the Euclidean topology and \(A :=[0,1)\). We have that \[ \mathop{\mathrm{Int}}{A} = (0,1) \,, \quad {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} = [0,1] \,, \quad \partial A = \{0,1\} \,. \] In particular \[ \mathop{\mathrm{Int}}{A} \neq A \,, \quad {}\mkern 2mu \overline{\mkern-2mu A \mkern-2mu}\mkern 2mu {} \neq A \,, \] showing that \(A\) is neither open, nor closed.

The proof of the above statements is left as an exercise.

3.6 Density

Definition 46: Density

Let \((X,\mathcal{T})\) be a topological space and \(A \subseteq X\) a set. We say that \(A\) 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 47

Let \((X,\mathcal{T})\) be a topological space and \(A \subseteq X\) a set. 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 48

Consider \(\mathbb{R}\) with the Euclidean topology.

1. We have that the set of integers \(\mathbb{Z}\) is closed in \(\mathbb{R}\). Indeed, \[ \mathbb{Z}^c = \bigcup_{ z \in \mathbb{Z}} \, (z , z + 1 ) \,. \] 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 \[ {}\mkern 2mu \overline{\mkern-2mu \mathbb{Z} \mkern-2mu}\mkern 2mu {} = \mathbb{Z}\,, \] showing that \(\mathbb{Z}\) is not dense in \(\mathbb{R}\).

2. The rational numbers \(\mathbb{Q}\) are instead dense in \(\mathbb{R}\), as proven in the Analysis module. Therefore \[ {}\mkern 2mu \overline{\mkern-2mu \mathbb{Q} \mkern-2mu}\mkern 2mu {} = \mathbb{R}\,. \] It is also easy to check that \[ \mathop{\mathrm{Int}}{\mathbb{Q}} = \emptyset \,. \] Therefore \[ \mathop{\mathrm{Int}}{\mathbb{Q}} \neq \mathbb{Q}\,, \quad {}\mkern 2mu \overline{\mkern-2mu \mathbb{Q} \mkern-2mu}\mkern 2mu {} \neq \mathbb{Q}\,, \] showing that \(\mathbb{Q}\) is neither open, nor closed.

Example 49

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

Proof. Suppose \(C\) is a closed set such that \(\mathbb{Z}\subseteq C\). By definition of \(\mathcal{T}_{\textrm{cofinite}}\) we have \(C = \mathbb{R}\) or \(C\) finite. Since \(\mathbb{Z}\subseteq C\) and \(\mathbb{Z}\) is not finite, we conclude \(C = \mathbb{R}\). This proves that \(\mathbb{R}\) is the only closed set containing \(\mathbb{Z}\), and so \({}\mkern 2mu \overline{\mkern-2mu \mathbb{Z} \mkern-2mu}\mkern 2mu {} = \mathbb{R}\).

3.7 Hausdorff spaces

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

Definition 50

Let \((X,\mathcal{T})\) be a topological space. We say that \(X\) is a Hausdorff space if for every two points \(x,y \in X\) 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 51

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

Proof

Let \(x,y \in X\) with \(x \neq y\). Set \[ \varepsilon:= \frac12 \, d(x,y) \,, \] and define \[ U := B_{\varepsilon}(x)\,, \quad V := B_{\varepsilon}(y) \,. \] By Proposition 27 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 52: 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 53

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 51 we know that \((X,\mathcal{T}_d)\) is Hausdorff. Hence \((X,\mathcal{T})\) is Hausdorff.

As a conseuqence of Corollary 53 we have that spaces which are not metrizable are not Hausdorff. Let us make a few examples.

Example 54: Trivial topology is not Hausdorff

Let \((X,\mathcal{T})\) be a topological space with \(\mathcal{T}\) trivial topology. Assume that \(X\) has more than one element. Then \(X\) is not Hausdorff.

Indeed, let \(x,y \in X\) with \(x \neq y\). Suppose by contradiction that \(X\) is Hausdorff. Then there exist \(U,V \in \mathcal{T}\) such that \[ x \in U \,, \quad y \in V \,, \quad U \cap V = \emptyset \,. \] Recall that \[ \mathcal{T}= \{ \emptyset, X \} \,. \] Since \(x \in U\) and \(y \in V\), we deduce that \(U\) and \(V\) are non-empty. Since \(U\) and \(V\) are open, the only possibility is that \[ U = V = X \,. \] In this case we have \[ U \cap V = X \cap X = X \neq \emptyset \,, \] leading to a contradiciton. Hence \(X\) is not Hausdorff.

Example 55: Cofinite topology on \(\mathbb{R}\)

Consider the following family \(\mathcal{T}\) of subsets of \(\mathbb{R}\) \[ \mathcal{T}:= \{ U \subseteq \mathbb{R}\, \colon \,U^c \, \mbox{ is finite, } \mbox{or } U^c = \mathbb{R}\}\,. \] Then \((\mathbb{R},\mathcal{T})\) is a topological space which is not Hausdorff. The topology \(\mathcal{T}\) is called the cofinite topology.

Exercise: Show that \((\mathbb{R},\mathcal{T})\) is not Hausdorff.

Example 56

Consider the following family \(\mathcal{T}\) of subsets of \(\mathbb{R}\) \[ \mathcal{T}:= \{ U = (-\infty,a) \, \colon \,- \infty \leq a \leq \infty \}\,. \] Then \((\mathbb{R},\mathcal{T})\) is a topological space which is not Hausdorff.

We start by showing that \((\mathbb{R},\mathcal{T})\) is a topological space. We need to check the properties of topologies:

• (A1) We have that \[ (\infty,\infty) = \emptyset \in \mathcal{T}\,, \quad (-\infty,\infty) = \mathbb{R}\in \mathcal{T}\,. \]

• (A2) Suppose that \(A_i \in \mathcal{T}\) for all \(i \in I\). By definition \[ A_i = (-\infty, a_i) \,, \quad - \infty \leq a_i \leq \infty \,. \] Set \[ a := \sup_{i \in I} \ a_i \,, \quad A := (-\infty, a) \,. \] Note that \(a\) always exists, and possibly \(a=\infty\). Moreover \(A \in \mathcal{T}\). We claim \[ A = \bigcup_{i \in I} \, A_i \,. \tag{3.8}\] To prove (3.8) first suppose that \(x \in A\). Then \(x < a\). Set \(\varepsilon:= a - x\), so that \(\varepsilon>0\). By definition of supremum there exists \(i_{0} \in I\) such that \[ a - \varepsilon< a_{i_0} \,. \] From the above, and from the definition of \(\varepsilon\), we deduce \[ a_{i_0} > a - \varepsilon= a - a + x = x \,, \] showing that \(x \in (-\infty, a_{i_0}) = A_{i_0}\). Therefore \[ A \subseteq \bigcup_{i \in I} \, A_i \,. \] Conversely, assume that \(x \in \cup_{i \in I} \, A_i\). Therefore there exists \(i_0 \in I\) such that \(x \in A_{i_0} = (-\infty, a_{i_0})\). In particular \[ x < a_{i_0} \leq \sup_{i \in I} a_i = a \,, \] showing that \(x \in (-\infty,a) = A\). Therefore \[ \bigcup_{i \in I} \, A_i \subseteq A\,, \] and (3.8) is proven.

• (A3) Let \(A, B \in \mathcal{T}\). Therefore \[ A = (-\infty, a)\,, \quad B = (-\infty, b)\,, \] for some \(a,b \in [-\infty, \infty]\). Set \[ U := A \cap B \,, \quad z:= \min \{a,b\} \,. \] It is immediate to check that \[ U = (-\infty, z) \,, \] showing that \(U \in \mathcal{T}\).

Therefore \((\mathbb{R},\mathcal{T})\) is a topological space. We now show that \((\mathbb{R},\mathcal{T})\) is not Hausdorff. Suppose by contradiction that \((\mathbb{R},\mathcal{T})\) is Hausdorff. Let \(x,y \in \mathbb{R}\) with \(x \neq y\). By assumption there exist \(U,V \in \mathcal{T}\) such that \[ x \in U \,, \quad y \in V \,, \quad U \cap V = \emptyset \,. \] By definition of \(\mathcal{T}\) there exist \(a,b \in [-\infty,\infty]\) such that \[ U = (-\infty,a) \,, \quad V = (- \infty, b) \,. \] Since \(x \in U\) and \(y \in V\), in particular \(U\) and \(V\) are non-empty. Therefore \(a,b>-\infty\). Set \[ z := \min\{a , b \}\,, \quad Z:= U \cap V = (-\infty,z) \,. \] As \(a,b>-\infty\), we have \(z > - \infty\). Therefore \(Z \neq \emptyset\). This is a contradiction, since \(U \cap V = \emptyset\). Therefore \((\mathbb{R},\mathcal{T})\) is not Hausdorff.

In Hausdorff spaces the limit of a sequence is unique.

Proposition 57: 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 a convergent sequence. Suppose by contradiction that \[ x_n \to x_0 \,, \quad x_n \to y_0 \] in \(X\), for some \(x_0,y_0 \in X\) with \(x_0 \neq y_0\). 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 of \(x_n\) 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 58: Images and Pre-images

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

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

• 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 59

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 60: Continuous function

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

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

• 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 61

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: \[ f^{-1}(V) \in \mathcal{T}_X \,, \quad \forall \, V \in \mathcal{T}_Y \,. \]

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 61 is simple, but very tedious. We choose to skip it.

Example 62

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

Indeed, \({\mathop{\mathrm{Id}}}_{X}\) 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 63: 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 64

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 65

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: \[ \forall \, \varepsilon> 0 \,, \, \exists \, \delta >0 \, \text{ s.t. } \, \, d_Y(f(x),f(x_0))<\varepsilon\,\, \mbox{ if } \,\, d_X(x , x_0) < \delta \,. \]

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

Example 66

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, \[ \mathcal{T}_Y = \{ \emptyset, Y \} \,. \] Then every function \(f \colon X \to Y\) is continuous.

Indeed, we know that \(f\) is continuous if and only if it holds: \[ f^{-1}(V) \in \mathcal{T}_X \,, \quad \forall \,\, V \in \mathcal{T}_Y \,. \] 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 67

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 \(f \colon X \to Y\). They are equivalent:

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

Indeed, 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 \[ V = \bigcup_{y \in V} \, \{ y \} \,. \] 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.

In a topological space, continuity preserves limits of sequences.

Proposition 68

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. Let \(f \colon X \to Y\) be continuous. Let \(\{x_n\} \subset X\) and \(x_0 \in X\). We have \[ x_n \to x_0 \,\, \mbox{ in }\, X \quad \implies \quad f(x_n) \to f(x_0) \,\, \mbox{ in }\, Y\,. \]

Proof

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

The converse implication of Proposition 68 is false. That is, even if it holds \[ x_n \to x_0 \,\, \mbox{ in }\, X \quad \implies \quad f(x_n) \to f(x_0) \,\, \mbox{ in }\, Y\,. \] for all sequences \(\{x_n\} \subset X\), the function \(f\) might not be continuous. A counterexample is given in Example 70 below.

For the above to hold, it is necessary for the topologies on \(X\) and \(Y\) to be first countable, as for example is the case for metrizable topologies, see Proposition 69 below.

Proposition 69

Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. Let \(f \colon X \to Y\) and suppose that for all convergent sequences \(\{x_n\} \subseteq X\), the sequence \(\{f(x_n)\}\) is convergent in \(Y\). Then \(f\) is continuous.

Proof

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 70

Consider \(\mathbb{R}\) with the co-countable topology: \[ {\mathcal{T}}_{\textrm{cc}} := \{ A \subseteq \mathbb{R}\, \colon \,A^c = \mathbb{R}\, \mbox{ or } \, A^c \, \mbox{ countable} \} \,. \] Sequences in \((\mathbb{R},{\mathcal{T}}_{\textrm{cc}})\) converge if and only if they are eventually constant. Also consider the discrete topology on \(\mathbb{R}\), denoted by \({\mathcal{T}}_{\textrm{discrete}}\). We have seen that sequences in \((\mathbb{R},{\mathcal{T}}_{\textrm{discrete}})\) converge if and only if they are eventually constant. Consider the identity function \[ f \colon (\mathbb{R},{\mathcal{T}}_{\textrm{cc}}) \to (\mathbb{R},{\mathcal{T}}_{\textrm{discrete}}) \,, \quad f(x):=x \,. \] We have that:

• \(f\) is not continuous: Indeed \(\{x\} \in {\mathcal{T}}_{\textrm{discrete}}\) but \[ f^{-1}(\{ x\}) = \{x\} \notin {\mathcal{T}}_{\textrm{cc}} \,, \] since \(\{x\}^c\) is neither \(\mathbb{R}\), nor countable.

• If \(\{x_n\}\) is convergent in \({\mathcal{T}}_{\textrm{cc}}\), then it is eventually constant. Therefore \(\{f(x_n)\}\) is eventually constant, and so it is convergent in \({\mathcal{T}}_{\textrm{discrete}}\).

Let us make an observation on continuity of compositions.

Proposition 71

Let \((X,\mathcal{T}_X), (Y,\mathcal{T}_Y), (Z,\mathcal{T}_Z)\) be topological spaces. Let \[ f \colon X \to Y \,, \quad g \colon Y \to Z \,, \] be given functions. If \(f\) and \(g\) 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 72: Homeomoprhim

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. There exists \(g \colon Y \to X\) continuous such that \[ g \circ f = {\mathop{\mathrm{Id}}}_{X} \,, \quad f \circ g = {\mathop{\mathrm{Id}}}_{Y} \,. \]

The above is saying that \(f\) is a homeomorphism if it is continuous and has continuous inverse. Homeomorphisms are the way we say that two topological spaces look the same.

3.9 Subspace topology

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

Definition 73: Subspace topology

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

Proof: Well-posedness of Definition 73

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 \[ A_i = U_i \cap Y \,, \quad \forall \, i \in I \,. \] Therefore \[ \bigcup_{i \in I} \, A_i = \bigcup_{i \in I} (U_i \cap Y ) = \left(\bigcup_{i \in I} U_i \right) \cap Y \,. \] 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 \[ A_1 = U_1 \cap Y \,, \quad A_2 = U_2 \cap Y \] 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, then sets are open in the subspace topology if and only if they are open in \(X\).

Proposition 74

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

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

Example 75

Let \(X=\mathbb{R}\) be equipped with \(\mathcal{T}\) the euclidean topology. Let \(\mathcal{S}\) be the subspace topology on \(\mathbb{Z}\). Then \(\mathcal{S}\) coincides with the discrete topology.

Proof. The set \(\{z\}\) is open in \(\mathcal{S}\) for all \(z \in \mathbb{Z}\). Indeed, \[ \{z\} = \left( z-1 , z + 1 \right) \cap \mathbb{Z}\, \] and \((z - 1, z + 1) \in \mathcal{T}\). Thus \(\{z\} \in \mathcal{S}\). Let now \(A \subseteq \mathbb{Z}\). Then \[ A = \bigcup_{z \in A} \, \{ z \} \,, \] and therefore \(A \in \mathcal{S}\) by (A2). This proves that \[ \mathcal{S}= \{ A \, \text{ s.t. } \, A \subseteq \mathbb{Z}\} \,, \] that is, \(\mathcal{S}\) is the discrete topology on \(\mathbb{Z}\).

3.10 Topological basis

We have seen that in metric spaces every open set is union of open balls, see Propostion 27. We can then regard the open balls as 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 76: 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 \(\{B_i\} \subseteq \mathcal{B}\), with \(I\) family of indices, such that \[ U = \bigcup_{i \in I} \, B_i \,. \tag{3.12}\]

Example 77

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

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

2. \((X,d)\) 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 \(\mathcal{T}_d\).

This is true by Propostion 27.

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

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

Proposition 78

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 78 are sufficient for generating a topology.

Proposition 79

Let \(X\) be a set and \(\mathcal{B}\) a collection of subsets of \(X\) such that (B1)-(B2) hold. Define \[ \mathcal{T}:= \left\{ U \subseteq X \, \colon \,U = \bigcup_{i \in I} B_i \,, \,\, B_i \in \mathcal{B} \right\} \,. \] Then:

1. \(\mathcal{T}\) is a topology on \(X\).

2. \(\mathcal{B}\) is a basis for \(\mathcal{T}\).

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 80

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 \[ \mathcal{B} := \{ U \times V \, \colon \,U \in \mathcal{T}_X \,, \,\, V \in \mathcal{T}_Y \} \subset X \times Y \,. \] Then \(\mathcal{B}\) satisfies properties (B1) and (B2) from Proposition 78.

The proof is an easy check, and is left as an exercise. As \(\mathcal{B}\) satisfies (B1)-(B2), by Proposition 79 we know that \[ \mathcal{T}_{X \times Y} := \left\{ U \times V \, \colon \,U \times V = \bigcup_{i \in I} B_i \,, ,\,\, B_i \in \mathcal{B} \right\} \tag{3.13}\] is a topology on \(X \times Y\).

Definition 81: Product topology

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. We call \(\mathcal{T}_{X \times Y}\) at (3.13) the product topology on \(X \times Y\).

Example 82

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

Consider the projection maps \[ \pi_X \colon X \times Y \to X \,, \quad \pi_X (x,y):=x \] and \[ \pi_Y \colon X \times Y \to Y \,, \quad \pi_Y (x,y):=y \]

Proposition 83

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 84

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 85: Connected space

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

• \(X\) is connected if the only subsets of \(X\) which are both open and closed are \(\emptyset\) and \(X\).
• \(X\) 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 86: Proper subset

Let \(X\) be a set. A subset \(A \subseteq X\) is proper if \[ A \neq \emptyset \,, \quad A \neq X \,. \]

Proposition 87: 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 \[ A:= U \,, \quad B:= U^c \,. \] 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 87 as equivalent definitions of disconnected topological space.

Example 88

Consider the set \(X = \{0,1\}\) with the subspace topology induced by the inclusion \(X \subset \mathbb{R}\), where \(\mathbb{R}\) is equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclidean}}\). Then \(X\) is disconnected.

Proof. Note that \[ X = \{ 0 \} \cup \{ 1 \} \,, \quad \{ 0 \} \cap \{ 1 \} = \emptyset \,. \] The set \(\{ 0 \}\) is open for the subspace topology, since \[ \{ 0 \} = X \cap (-1,1) \,, \quad (-1,1) \in \mathcal{T}_{\textrm{euclidean}} \,. \] Similarly, also \(\{ 1 \}\) is open for the subspace topology, since \[ \{ 1 \} = X \cap (0,2) \,, \quad (0,2) \in \mathcal{T}_{\textrm{euclidean}} \,. \] Clearly \[ \{ 0 \} \neq \emptyset \,, \quad \{ 1 \} \neq \emptyset \,, \] showing that \(X\) is disconnected.

Example 89

Let \(p \in \mathbb{R}\). The set \(X = \mathbb{R}\smallsetminus \{p\}\) is disconnected.

Proof. Define the sets \[ A = (-\infty,p) \,, \quad B = (p , \infty) \,. \] Then \(A,B\) are proper subsets of \(X\), since \(p \notin X\). Moreover \[ X = A \cup B \,, \quad A \cap B = \emptyset \,. \] Finally we have that \(A,B\) are open for the subspace topology, since they are open in \(\mathbb{R}\). Therefore \(X\) is disconnected.

Example 90

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

Exercise: Prove that \(X\) is connected.

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

Theorem 91

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 91 is that connectedness is a topological invariant, e.g., connectedness is preserved by homeomorphisms.

Corollary 92

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 91, and is left to the reader as an exercise.

Example 93

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

Proof. Suppose by contradiction that there exists an omeomorphism \[ f \colon \mathbb{R}^n \to \mathbb{R}\,. \] Define \(p = f(0)\) and the restriction \[ g \colon \mathbb{R}^n \smallsetminus \{0\} \to \mathbb{R}\smallsetminus \{p\} \,, \quad g(x) = f(x) \,. \] Since \(g\) is a restriction of an omeomorphism, then \(g\) is an omeomorphism. We have that \(\mathbb{R}^n \smallsetminus \{0\}\) is connected, as a consequence of
Example 90. Hence, by Corollary 92, we infer that \(\mathbb{R}\smallsetminus \{p\}\) is connected. This is a contradiction, since \(\mathbb{R}\smallsetminus \{p\}\) is disconnected, as shown in Example 89.

Example 94

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

Proof. Suppose by contradiction that there exists and omeomorphism \[ f \colon [0,1] \to \mathbb{S}^1 \,. \] The restriction of \(f\) to \([0,1] \smallsetminus \{\frac12\}\) defines an omeomorphism \[ g \ \colon \left( [0,1] \smallsetminus \left\{\frac12\right\} \right) \to \left( \mathbb{S}^1 \smallsetminus \{\mathbf{p}\} \right) \,, \quad \mathbf{p} := f\left(\frac12 \right) \,. \] 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 an omeomorphism, 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 91, we conclude that \(\mathbb{S}^1 \smallsetminus \{\mathbf{p}\}\) is connected. Contradiction.

3.13 Intermediate Value Theorem

Another consequence of Theorem 91 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 95: 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 96

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 97: 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 91 we know that \(f(X)\) is connected in \(\mathbb{R}\). By Theorem 96 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 \[ f(a) < c < f(b) \] we conclude that \(c \in f(X)\), since \(f(X)\) is an interval. Hence there exists \(\xi \in X\) such that \(f(\xi) = c\).

3.14 Path connectedness

Definition 98: 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 \[ \alpha \colon [a,b] \to X \] such that \[ \alpha (a) = x \,, \quad \alpha(b) = y \,. \]

Example 99

Let \(A \subset \mathbb{R}^n\) be convex. Then \(A\) is path connected.

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 \(\alpha(0)=x, \alpha(1)=y\).

It turns out that path-connectedness implies connectedness.

Theorem 100

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 \[ \alpha(0) = x \,, \quad \alpha(1) = y \,. \] 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.

The converse of the above theorem does not hold. A counterexample is given by the so-called topologist curve, which will be examined in Proposition 102. Prior to this, we need a basic Lemma.

Lemma 101

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. We conclude that \(A \subseteq U\).

Proposition 102: Topologist curve

Consider \(\mathbb{R}^2\) with the Euclidean topology and define the sets \[ X := A \cup B \] where \[\begin{align*} A & := \left\{ \left( t, \sin \left( \frac{1}{t} \right) \right) \, \colon \,t>0 \right\} \\ B & := \{(0,t) \, \colon \,t \in [-1,1] \} \end{align*}\] 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 91 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 Corollary 92 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 101 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 101 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 101 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.