2  Topology

Definition 1: Topological space

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

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

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

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

Further, we say:

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

Definition 2: Trivial topology
Let \(X\) be a set. The trivial topology on \(X\) is the collection of sets \[ \mathcal{T}_{\textrm{trivial}}:= \{ \emptyset , X \} \,. \]

Definition 3: Discrete topology
Let \(X\) be a set. The discrete topology on \(X\) is the collection of all subsets of \(X\) \[ \mathcal{T}_{\textrm{discrete}}:= \{ A \, \colon \,A \subseteq X \} \,. \]

Definition 4: Open set of \(\mathbb{R}^n\)
Let \(A \subseteq \mathbb{R}^n\). We say that the set \(A\) is open if it holds: \[ \forall \, \mathbf{x}\in A \,, \,\, \exists \, r > 0 \, \text{ s.t. } \, B_r(\mathbf{x}) \subseteq A \,, \tag{2.1}\] 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 \[ \| \mathbf{x}\| := \sqrt{ \sum_{i=1}^n x_i^2 } \,. \]

Definition 5: Euclidean topology of \(\mathbb{R}^n\)
The Euclidean topology on \(\mathbb{R}^n\) is the collection of sets \[ \mathcal{T}_{\textrm{euclid}}:= \{ A \, \colon \,A \subseteq \mathbb{R}^n \,, \,\, A \, \mbox{ is open} \} \,. \]

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

  • (A1) We have \(\emptyset , \mathbb{R}^n \in \mathcal{T}_{\textrm{euclid}}\): Indeed \(\emptyset\) is open because there is no point \(\mathbf{x}\) for which (2.1) needs to be checked. Moreover, \(\mathbb{R}^n\) is open because (2.1) holds with any radius \(r>0\).

  • (A2) Let \(A_i \in \mathcal{T}_{\textrm{euclid}}\) for all \(i \in I\). Define the union \(A =\bigcup_i A_i\). We need to check that \(A\) is open. Let \(\mathbf{x}\in A\). By definition of union, there exists an index \(i_0 \in I\) such that \(\mathbf{x}\in A_{i_0}\). Since \(A_{i_0}\) is open, by (2.1) there exists \(r>0\) such that \(B_r(\mathbf{x}) \subseteq A_{i_0}\). As \(A_{i_0} \subseteq A\), we conclude that \(B_r(\mathbf{x}) \subseteq A\), so that \(A \in \mathcal{T}_{\textrm{euclid}}\).

  • (A3) Let \(A, B \in \mathcal{T}_{\textrm{euclid}}\). We need to check that \(A \cap B\) is open. Let \(\mathbf{x}\in A \cap B\). Therefore \(\mathbf{x}\in A\) and \(\mathbf{x}\in B\). Since \(A\) and \(B\) are open, by (2.1) there exist \(r_1,r_2>0\) such that \(B_{r_1}(\mathbf{x}) \subseteq A\) and \(B_{r_2}(\mathbf{x}) \subseteq B\). Set \(r := \min\{ r_1,r_2\}\). Then \[ B_r(\mathbf{x}) \subseteq B_{r_1}(\mathbf{x}) \subseteq A \,, \quad B_r(\mathbf{x}) \subseteq B_{r_2}(\mathbf{x}) \subseteq B \,, \] Hence \(B_r(\mathbf{x}) \subseteq A \cap B\), showing that \(A \cap B \in \mathcal{T}_{\textrm{euclid}}\).

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

Proposition 6: \(B_r(\mathbf{x})\) is an open set of \(\mathcal{T}_{\textrm{euclid}}\)
Let \(\mathbb{R}^n\) be equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\). Let \(r>0\) and \(\mathbf{x}\in \mathbb{R}^n\). Then \(B_r(\mathbf{x}) \in \mathcal{T}_{\textrm{euclid}}\).

Definition 7: 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 \(X\).

Definition 8: Comparing topologies

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

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

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

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

Example 9: Comparing \(\mathcal{T}_{\textrm{trivial}}\) and \(\mathcal{T}_{\textrm{discrete}}\)
Let \(X\) be a set. Then \(\mathcal{T}_{\textrm{trivial}} \subseteq \mathcal{T}_{\textrm{discrete}}\).

Example 10: Cofinite topology on \(\mathbb{R}\)
Question. The cofinite topology on \(\mathbb{R}\) is the collection of sets \[ \mathcal{T}_{\textrm{cofinite}}:= \{ U \subseteq \mathbb{R}\, \colon \,U^c \, \mbox{ is finite, } \mbox{or } U^c = \mathbb{R}\}\,. \]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2.1 Sequences

Definition 11: Convergent sequence
Let \((X,\mathcal{T})\) be a topological space. Consider a sequence \(\{x_n\} \subseteq X\) and a point \(x \in X\). We say that \(x_n\) converges to \(x_0\) in the topology \(\mathcal{T}\), if the following property holds: \[ \begin{aligned} & \forall \, U \in \mathcal{T}\, \text{ s.t. } \, x_0 \in U\,, \,\, \exists \, N = N(U) \in \mathbb{N}\, \text{ s.t. } \, \\ & x_n \in U \,, \, \forall \, n \geq N \,. \end{aligned} \tag{2.2}\]

The convergence of \(x_n\) to \(x_0\) is denoted by \(x_n \to x_0\).

Proposition 12: Convergent sequences in \(\mathcal{T}_{\textrm{trivial}}\)
Let \(X\) be equipped with \(\mathcal{T}_{\textrm{trivial}}\). Let \(\{x_n\} \subseteq X\), \(x_0 \in X\). Then \(x_n \to x_0\).

Proof
To show that \(x_n \to x_0\) we need to check that (2.2) holds. Let \(U \in \mathcal{T}_{\textrm{trivial}}\) with \(x_0 \in U\). We have two cases:

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

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

Thus (2.2) holds for all the sets \(U \in \mathcal{T}_{\textrm{trivial}}\), showing that \(x_n \to x_0\).

Warning
Proposition 12 shows the topological limit may not be unique!

Proposition 13: Convergent sequences in \(\mathcal{T}_{\textrm{discrete}}\)

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

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

Proof
Part 1. Assume that \(x_n \to x_0\). Let \(U = \{x_0\}\). Then \(U \in \mathcal{T}_{\textrm{discrete}}\). Since \(x_n \to x_0\), by (2.2) there exists \(N \in \mathbb{N}\) such that \[ x_n \in U \,, \quad \forall \, n \geq N \,. \] As \(U = \{x_0\}\), we infer \(x_n = x_0\) for all \(n \geq N\). Hence \(x_n\) is eventually constant.

Part 2. Assume that \(x_n\) is eventually equal to \(x_0\), that is, there exists \(N \in \mathbb{N}\) such that \[ x_n = x_0 \,, \quad \forall \, n \geq N \,. \tag{2.3}\] Let \(U \in \mathcal{T}\) be an open set such that \(x_0 \in U\). By (2.3) we have that \[ x_n \in U \,, \quad \forall \, n \geq N \,. \] Since \(U\) was arbitrary, we conclude that \(x_n \to x_0\).

Definition 14: Classical convergence in \(\mathbb{R}^n\)
Let \(\{\mathbf{x}_n\} \subseteq \mathbb{R}^n\) and \(\mathbf{x}_0 \in \mathbb{R}^n\). We say that \(\mathbf{x}_n\) converges \(\mathbf{x}_0\) in the classical sense if \(\left\| \mathbf{x}_n - \mathbf{x}_0 \right\| \to 0\), that is, \[ \forall \, \varepsilon>0 , \, \exists \, N \in \mathbb{N}, \, \, \text{ s.t. } \, \left\| \mathbf{x}_n - \mathbf{x}_0 \right\| < \varepsilon\,, \, \forall \, n \geq N \,. \]

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

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

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

2.2 Metric spaces

Definition 16: Distance and Metric space
Let \(X\) be a set. A distance on \(X\) is a function \(d \colon X \times X \to \mathbb{R}\) such that, for all \(x,y,z \in X\) they hold:

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

  • (M2) Symmetry: \(d(x,y) = d(y,x)\)

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

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

Definition 17: Euclidean distance on \(\mathbb{R}^n\)
The Euclidean distance over \(\mathbb{R}^n\) is defined by \[ d(\mathbf{x},\mathbf{y}) := \left\| \mathbf{x}- \mathbf{y} \right\| = \left( \sum_{i=1}^n |x_i - y_i|^2 \right)^{1/2} \,, \quad \forall \mathbf{x},\mathbf{y}\in \mathbb{R}^n \,. \]

Proposition 18
Let \(d\) be the Euclidean distance on \(\mathbb{R}^n\). Then \((\mathbb{R}^n,d)\) is a metric space.

Definition 19: Topology induced by the metric
Let \((X,d)\) be a metric space. The set \(A \subseteq X\) is open if it holds \[ \forall \, x \in U \,, \, \exists \, r \in \mathbb{R}, \, r > 0 \, \, \text{ s.t. } \, \, B_r(x) \subseteq U \,, \] where \(B_r(x)\) is the ball centered at \(x\) of radius \(r\), defined by \[ B_r(x) = \{ y \in X \, \colon \, d(x,y)<r \} \,. \] The topology induced by the metric \(d\) is the collection of sets \[ \mathcal{T}_d = \{ U\, \colon \, U \subseteq X, \, U \text{ open} \}\,. \]

Remark 20: Topology induced by Euclidean distance
Consider the metric space \((\mathbb{R}^n,d)\) with \(d\) the Euclidean distance. Then \[ \mathcal{T}_d = \mathcal{T}_{\textrm{euclid}}\,, \] where \(\mathcal{T}_{\textrm{euclid}}\) is the Euclidean topology on \(\mathbb{R}^n\).

Example 21: Discrete distance
Question. Let \(X\) be a set. The discrete distance is the function \(d \colon X \times X \to \mathbb{R}\) defined by \[ d(x,y) := \begin{cases} 0 & \mbox{ if } \, x = y \\ 1 & \mbox{ if } \, x \neq y \end{cases} \]

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

Solution. See Question 3 in Homework 3.

Proposition 22: Convergence in metric space

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

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

2.3 Hausdorff spaces

Definition 23: Hausdorff space
We say that a topological space \((X,\mathcal{T})\) is Hausdorff if for every \(x,y \in X\) with \(x \neq y\), there exist \(U,V \in \mathcal{T}\) such that \[ x \in U \,, \quad y \in V \,, \quad U \cap V = \emptyset \,. \]

Proposition 24
Let \((X,d)\) be a metric space, \(\mathcal{T}_d\) the topology induced by \(d\). Then \((X, \mathcal{T}_d)\) is a Hausdorff space.

Proof
Let \(x,y \in X\) with \(x \neq y\). Define \[ U := B_{\varepsilon}(x)\,, \quad V := B_{\varepsilon}(y) \,, \quad \varepsilon:= \frac12 \, d(x,y) \,. \] By Proposition 24, 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.

Definition 25: 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 26
Let \((X,\mathcal{T})\) be a metrizable space. Then \(X\) is Hausforff.

Example 27: \((X,\mathcal{T}_{\textrm{trivial}})\) is not Hausdorff
Question. Let \(X\) be equipped with the trivial topology \(\mathcal{T}_{\textrm{trivial}}\). Then \(X\) is not Hausdorff.

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

Example 28: \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is not Hausdorff
Question. Consider the cofinite topology on \(\mathbb{R}\) \[ \mathcal{T}_{\textrm{cofinite}}= \{ U \subseteq \mathbb{R}\, \colon \,U^c \, \mbox{ is finite, } \mbox{or } U^c = \mathbb{R}\} \,. \] Prove that \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is not Hausdorff.

Solution. Assume by contradiction \((\mathbb{R},\mathcal{T}_{\textrm{cofinite}})\) is Hausdorff and let \(x, y \in \mathbb{R}\) with \(x \neq y\). Then, there exist \(U,V \in \mathcal{T}_{\textrm{cofinite}}\) such that \[ x \in U \,, \quad y \in V \,, \quad U \cap V = \emptyset \,. \] Taking the complement of \(U \cap V = \emptyset\), we infer \[ \mathbb{R}= (U \cap V)^c = U^c \cup V^c \,. \tag{2.4}\] There are two possibilities:

  1. \(U^c\) and \(V^c\) are finite. Then \(U^c \cup V^c\) is finite, so that (2.4) is a contradiction.

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

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

Example 29: Lower-limit topology on \(\mathbb{R}\) is not Hausdorff
Question. The lower-limit topology on \(\mathbb{R}\) is the collection of sets \[ \mathcal{T}_{\textrm{LL}}=\{\emptyset, \mathbb{R}\} \cup \{ (a,+\infty) \, \colon \, a \in \mathbb{R}\} \,. \]

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

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

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

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

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

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

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

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

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

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

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

Proposition 30: 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.

2.4 Continuity

Definition 31: Images and Pre-images

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

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

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

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

Definition 32: Continuous function

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

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

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

Proposition 33

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

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

Example 34
Question. Let \(X\) be a set and \(\mathcal{T}_1\), \(\mathcal{T}_2\) be topologies on \(X\). Define the identity map \[ {\mathop{\mathrm{Id}}}_{X} \colon (X,\mathcal{T}_1) \to (X,\mathcal{T}_2) \,, \quad {\mathop{\mathrm{Id}}}_{X} (x):= x \,. \] Prove that they are equivalent:

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

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

Definition 35: 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 36

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.
Proposition 37

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

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

Example 38
Question. Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be a topological space. Suppose that \(\mathcal{T}_Y\) is the trivial topology, that is, \[ \mathcal{T}_Y = \{ \emptyset, Y \} \,. \] Prove that every function \(f \colon X \to Y\) is continuous.

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

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

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

Therefore \(f\) is continuous.

Example 39
Question. Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. Suppose that \(\mathcal{T}_Y\) is the discrete topology, that is, \[ \mathcal{T}_Y = \{ V \, \text{ s.t. } \, V \subseteq Y \} \,. \] Let \(f \colon X \to Y\). Prove that they are equivalent:

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

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

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

Proposition 40: Continuity of compositions
Let \((X,\mathcal{T}_X), (Y,\mathcal{T}_Y), (Z,\mathcal{T}_Z)\) be topological spaces. Assume \(f \colon X \to Y\) and \(g \colon Y \to Z\) are continuous. Then \((g \circ f) \colon X \to Z\) is continuous.

Definition 41: Homeomorphism

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

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

2.5 Subspace topology

Definition 42: Subspace topology
Let \((X,\mathcal{T})\) be a topological space and \(Y \subseteq X\) a subset. Define the family of sets \[\begin{align*} \mathcal{S}& := \{ A \subseteq Y \, \colon \,\exists \,\, U \in \mathcal{T}\, \text{ s.t. } \, A = U \cap Y \} \\ & = \{ U \cap Y , \,\, U \in \mathcal{T}\} \,. \end{align*}\] The family \(\mathcal{S}\) is the subspace topology on \(Y\) induced by the inclusion \(Y \subseteq X\).

Proposition 43
Let \((X,\mathcal{T})\) be a topological space and \(Y \in \mathcal{T}\). Let
\(A \subseteq Y\). Then \[ A \in \mathcal{S}\quad \iff \quad A \in \mathcal{T}\,. \]

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

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

Example 44

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

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

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

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

2.6 Connectedness

Definition 45: Connected space

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

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

Definition 46: Proper subset
Let \(X\) be a set. A subset \(A \subseteq X\) is proper if \(A \neq \emptyset\) and \(A \neq X\).

Proposition 47: 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.

Example 48
Question. Consider the set \(X = \{0,1\}\) with the subspace topology induced by the inclusion \(X \subseteq \mathbb{R}\), where \(\mathbb{R}\) is equipped with the Euclidean topology \(\mathcal{T}_{\textrm{euclid}}\). Prove that \(X\) is disconnected.

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

Example 49
Question. Let \(\mathbb{R}\) be equipped with \(\mathcal{T}_{\textrm{euclid}}\), and let \(p \in \mathbb{R}\). Prove that the set \(X = \mathbb{R}\smallsetminus \{p\}\) is disconnected.

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

Theorem 50
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.

Theorem 51: Connectedness is topological invariant
Let \((X,\mathcal{T}_X)\), \((Y,\mathcal{T}_Y)\) be homeomorhic topological spaces. Then \[ X \, \mbox{ is connected } \,\, \iff \,\, Y \, \mbox{ is connected } \]

Example 52
Question. Define the one dimensional unit circle \[ \mathbb{S}^1 := \{(x,y) \in \mathbb{R}^2 \, \colon \, x^2 + y^2 = 1 \} \,. \] Prove that \(\mathbb{S}^1\) and \([0,1]\) are not homeomorphic.

Solution. Suppose by contradiction that there exists a homeomorphism \[ f \colon [0,1] \to \mathbb{S}^1 \,. \] The restriction of \(f\) to \([0,1] \smallsetminus \{\frac12\}\) defines a homeomorphism \[ g \ \colon \left( [0,1] \smallsetminus \left\{\frac12\right\} \right) \to \left( \mathbb{S}^1 \smallsetminus \{\mathbf{p}\} \right) \,, \quad \mathbf{p} := f\left(\frac12 \right) \,. \] The set \([0,1] \smallsetminus \left\{ \frac12 \right\}\) is disconnected, since \[ [0,1] \smallsetminus \{1/2\} = [0,1/2) \, \cup \, (1/2,1] \] with \([0,1/2)\) and \((1/2,1]\) open for the subset topology, non-empty and disjoint. Therefore, using that \(g\) is a homeomorphism, we conclude that also \(\mathbb{S}^1 \smallsetminus \{\mathbf{p}\}\) is disconnected. Let \(\theta_0 \in [0,2\pi)\) be the unique angle such that \[ \mathbf{p} = (\cos (\theta_0),\sin(\theta_0)) \,. \] Thus \(\mathbb{S}^1 \smallsetminus \{\mathbf{p}\}\) is parametrized by \[ {\pmb{\gamma}}(t):=(\cos(t),\sin(t)) \,, \quad t \in (\theta_0,\theta_0 + 2\pi) \,. \] Since \({\pmb{\gamma}}\) is continuous and \((\theta_0,\theta_0 + 2\pi)\) is connected, by Theorem 50, we conclude that \(\mathbb{S}^1 \smallsetminus \{\mathbf{p}\}\) is connected. Contradiction.

Definition 53: 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 54: Intervals are connected

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

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

Theorem 55: 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: \[ \forall \, c \in \mathbb{R}\, \text{ s.t. } \, f(a)< c < f(b) \,, \,\, \exists \, \xi \in X \, \text{ s.t. } \, f(\xi) = c \,. \]

Example 56: Intervals are connected - Alternative proof
Question. Prove the following statements.

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

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

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

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

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

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

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

2.7 Path-connectedness

Definition 57: Path-connectedness
Let \((X,\mathcal{T})\) be a topological space. We say that \(X\) is path-connected if for every \(x,y \in X\) there exist \(a,b \in \mathbb{R}\) with \(a<b\), and a continuous function \[ \alpha \colon [a,b] \to X \,\, \, \text{ s.t. } \, \,\, \alpha (a) = x \,, \quad \alpha(b) = y \,. \]

Theorem 58: Path-connectedness implies connectedness
Let \((X,\mathcal{T})\) be a path-connected topological space. Then \(X\) is connected.

Example 59
Question. Let \(A \subseteq \mathbb{R}^n\) be convex. Show that \(A\) is path-connected, and hence connected.

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

Example 60: Spaces of matrices

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

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

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

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

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

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

Warning
In general connectedness does not imply path-connectedness, as seen in Proposition 64.