Topologies, Separation Axioms and a Glance at Cp(X)

Cvičení

Exercise 007

Suppose that \(X\) is a topological space and, for each \(x \in X\), we have a fixed local base \(\mathcal{B}_x\) at \(x\). Show that the family \(\{\mathcal{B}_x : x \in X\}\) has the following properties: Now, suppose that \(X\) is an arbitrary set without topology and \(\mathcal{B}_x\) is a family of subsets of \(X\) for any \(x \in X\) such that the collection \(\{\mathcal{B}_x : x \in X\}\) has the properties (LB1)-(LB3). Show that there exists a unique topology \(\tau\) on \(X\) such that \(\mathcal{B}_x\) is a local base of \((X, \tau)\) at \(x\) for each \(x \in X\). We will call \(\tau\) the topology generated by the family \(\{\mathcal{B}_x : x \in X\}\) as local bases.
Solution. Let us remind the definition of a local base at a point \(x\in X\).
Definition. Given an \(x\in X,\) a non-empty family \(\mathcal L \subset \tau(X)\) is called a local base of \(X\) at \(x\) if \(x \in \cap\mathcal L\) and \(x \in U \in \tau(X)\) implies \(V \subseteq U\) for some \(V \in \mathcal L\).
(LB1) is immediate from the definition of a local base. Indeed, for each \(x\in X\), we have \(x \in \cap\mathcal{B}_x\) and \(\mathcal{B}_x \neq \emptyset\).
(LB2) Let \(x\in X\) and \(U, V \in \mathcal{B}_x\). Then \(x \in U \cap V \in \tau(X).\) By the definition of a local base, there is a \(W \in \mathcal{B}_x\) such that \(W \subseteq U \cap V\).
(LB3) Let \(x \in U \in \mathcal{B}_y\). Then \(U \in \tau(X)\). By the definition of a local base, there is a \(V \in \mathcal{B}_x\) such that \(V \subseteq U\). \(\blacksquare\)

Exercise 008

Prove that \(\cup \mathcal{S} = X\) for any subbase \(\mathcal{S}\) of a topological space \(X\). Now, let \(X\) be an arbitrary set without topology. Prove that, for any family \(\mathcal{S} \subset \mathrm{exp}(X)\) with \(\cup \mathcal{S} = X\), there exists a unique topology \(\tau\) on \(X\) such that \(\mathcal{S}\) is a subbase of \((X, \tau)\). We will call \(\tau\) the topology generated by \(\mathcal{S}\) as a subbase.
Solution.
Definition. A family \(\mathcal{S} \subset \tau(X)\) is called a subbase of \(X\) if all finite intersections of elements of \(\mathcal{S}\) form a base of \(X\).
From the definition of a subbase it follows that if \(x \in X\), then there are \(U_1, \ldots, U_n \in \mathcal{S}\) such that \(x \in U_1 \cap \ldots \cap U_n\). Thus \(x \in U_1 \cap \ldots \cap U_n \subset \cup \mathcal{S}.\) Since \(x\) was arbitrary, we have \(\cup \mathcal{S} = X\).
Now suppose that \(X\) is an arbitrary set without topology and let \(\mathcal{S} \subset \mathrm{exp}(X)\) be a family such that \(\cup \mathcal{S} = X\). We will define a topology \(\tau\) on \(X\) as follows. A subset \(U \subset X\) is open if for each \(x \in U\) there are \(U_1, \ldots, U_n \in \mathcal{S}\) such that \(x \in U_1 \cap \ldots \cap U_n \subset U\). Next, we will show that \(\tau\) is a topology on \(X\). Indeed, \(\emptyset \in \tau\) obviously. Now suppose that \(\mathcal{U} \subset \mathrm{exp}(X)\) is a family of open sets. We need to show that \(\cup \mathcal{U}\) is open. Let \(x \in \cup \mathcal{U}\). Then there is \(U \in \mathcal{U}\) such that \(x \in U\). Since \(U\) is open, there are \(U_1, \ldots, U_n \in \mathcal{S}\) such that \(x \in U_1 \cap \ldots \cap U_n \subset U \subset \cup \mathcal{U}\). Thus \(\cup \mathcal{U}\) is open. Finally, let \(U_1, \ldots, U_n\) are open sets. We need to show that \(U_1 \cap \ldots \cap U_n\) is an open set. Let \(x \in U_1 \cap \ldots \cap U_n\). Then for every \(i = 1, \ldots, n\) there are \(V_{i1}, \ldots, V_{ik_i} \in \mathcal{S}\) such that \(x \in V_{i1} \cap \ldots \cap V_{ik_i} \subset U_i\). Thus \(x \in (V_{11} \cap \ldots \cap V_{1k_1}) \cap \ldots \cap (V_{n1} \cap \ldots \cap V_{nk_n}) \subset U_1 \cap \ldots \cap U_n\). Therefore, \(U_1 \cap \ldots \cap U_n\) is open. Thus \(\tau\) is a topology on \(X\).
From the definition of openness, it follows that \(\mathcal{S}\) is a subbase of this topology.
It remains to show that this topology is unique. Let us denote by \(\tau\) the topology we just defined. Suppose that \(\tau'\) is another topology on \(X\) such that \(\mathcal{S}\) is a subbase of \((X, \tau')\). Now \(U \in \tau'\) if and only if for each \(x \in U\) there are \(U_1, \ldots, U_n \in \mathcal{S}\) such that \(x \in U_1 \cap \ldots \cap U_n \subset U\) \(\iff\) \(U \in \tau\). Thus \(\tau = \tau'\). \(\blacksquare\)

Exercise 009

Suppose that \(X\) and \(Y\) are topological spaces and we have a map \(f: X \to Y\). Prove that the following conditions are equivalent:
Solution. Let us recal the definition of a continuous map.
Definition. A map \(f: X \to Y\) between topological spaces is called continuous if \(f^{-1}(U)\) is open in \(X\) for every open set \(U \subset Y\).
Definition. Given \(x_0 \in X\), we say that a map \(f: X \to Y\) is continuous at \(x_0\) if \(f(x_0) \in U \in \tau(Y)\) implies \(f(V) \subset U\) for some \(V \in \tau(X)\) such that \(x_0 \in V\).
\((i) \Rightarrow (ii)\) Let \(\mathcal{B}\) is any fixed base of topology \(\tau(Y)\). Then since \(\mathcal{B} \subset \tau(Y),\) we have \(f^{-1}(U) \in \tau(X)\) for any \(U\in \mathcal{B}.\)
\((ii) \Rightarrow (iii)\) Let \(\mathcal{B}\) is a base of topology \(\tau(Y)\) such that \(f^{-1}(U)\) is open in \(X\). Now, if we put \(\mathcal{S} = \mathcal{B}\), then \(f^{-1}(U)\) is open in \(X\) for every \(U \in \mathcal{S}\). Moreover, the family \(\mathcal{S}\) is a subbase of \(\tau(Y)\).
\((iii) \Rightarrow (iv)\) Let us suppose that \(\mathcal{S}\) is a subbase of \(\tau(Y)\) such that \(f^{-1}(U)\) is open in \(X\) for every \(U \in \mathcal{S}\). Take an arbitrary \(x \in X\) and let \(U \in \tau(Y)\) be an open set such that \(f(x) \in U\). Since \(\mathcal{S}\) is a subbase of \(\tau(Y)\), there are \(U_1, \ldots, U_n \in \mathcal{S}\) such that \(f(x) \in U_1 \cap \ldots \cap U_n \subset U\). Since \(f^{-1}(U_i)\) is open in \(X\) for each \(i = 1, \ldots, n\), then \(V = f^{-1}(U_1) \cap \ldots \cap f^{-1}(U_n)\) is open in \(X\) and \(x \in V.\) Furthermore, \[ f(V) = f(f^{-1}(U_1) \cap \ldots \cap f^{-1}(U_n)) = f(f^{-1}(U_1)) \cap \ldots \cap f(f^{-1}(U_n)) \subset U_1 \cap \ldots \cap U_n\subset U. \]
\((iv) \Rightarrow (v)\) Suppose that \(f\) is continuous at each point \(x \in X\). Let \(F \subset Y\) be a closed set. Note that \(f^{-1}(Y \setminus F) = X \setminus f^{-1}(F)\). Put \(U = X \setminus f^{-1}(F)\) and let us show that \(U\) is open in \(X\). Take an arbitrary \(x \in U\). Note that then \(f(x) \in Y \setminus F\). Since \(f\) is continuous at \(x\), there is an open set \(V \subset X\) such that \(x \in V\) and \(f(V) \subset Y \setminus F\). Then, \[ V \subset f^{-1}(f(V)) \subset f^{-1}(Y \setminus F) = U. \] Hence, for each \(x \in U\) there is an open set \(V \subset X\) such that \(x \in V\) and \(V \subset U\). Therefore, \(U\) is open in \(X\), i.e., \(f^{-1}(F)\) is closed in \(X\).
\((v) \Rightarrow (vi)\) Suppose that \(f^{-1}(F)\) is closed in \(X\) whenever \(F\) is closed in \(Y\). Let \(A \subset X\) be an arbitrary set. Now let us recall the definition of the set-operator \(\mathrm{cl}_X.\)
Definition. Given a set \(A \subset X\), the closure of \(A\) is defined as \[ \mathrm{cl}_X(A) = \cap \{F \subset X : F \text{ is closed and } A \subset F\}. \]
Assume that \(x \in \mathrm{cl}_X(A)\). Now we need to show that \(f(x) \in \mathrm{cl}_Y(f(A))\). But, suppose on the contrary that \(f(x) \notin \mathrm{cl}_Y(f(A))\). This implies that there is a closed set \(F \subset Y\) such that \(f(A) \subset F\) and \(f(x) \notin F\). Then \[ A \subset f^{-1}(f(A)) \subset f^{-1}(F). \] Since \(f^{-1}(F)\) is closed in \(X\), we have \(\mathcal{cl}_X(A) \subset f^{-1}(F)\). Since \(f(x) \notin F\), we have \(x \notin f^{-1}(F)\). But this contradicts the assumption that \(x \in \mathrm{cl}_X(A)\). Therefore, \(f(x) \in \mathrm{cl}_Y(f(A))\).
\((vi) \Rightarrow (vii)\) Suppose that the condition \((vi)\) holds. That is for any \(A \subset X\) we have \(f(\mathrm{cl}_X(A)) \subset \mathrm{cl}_Y(f(A))\). Now we have to show that \[ \mathrm{cl}_X(f^{-1}(B)) \subset f^{-1}(\mathrm{cl}_Y(B)) \] for any \(B \subset Y\).
Lemma. Let \(X\) be a topological space and \(A \subset X\). Then \(A\) is closed in \(X\) if and only if \(\mathrm{cl}_X(A) = A\).

Now note, that \(f^{-1}(B) \subset f^{-1}(\mathrm{cl}_Y(B))\). To prove (vii), it suffices to show that \(A = f^{-1}(\mathrm{cl}_Y(B))\) is closed in \(X\). Using the condition (vi), we have \begin{align*} f(\mathrm{cl}_X(A)) & \subset \mathrm{cl}_Y(f(A)) = \\ \mathcal{cl}_Y(f(f^{-1}(\mathrm{cl}_Y(B))) & \subset \mathrm{cl}_Y(B)\implies \\ \mathrm{cl}_X(A) & \subset f^{-1}(\mathrm{cl}_Y(B)). \end{align*} So, we have \[ \mathcal{cl}_X(f^{-1}(B)) \subset f^{-1}(\mathrm{cl}_Y(B)) \subset \mathcal{cl}_X(f^{-1}(B)). \] Therefore, \(\mathcal{cl}_X(f^{-1}(B)) = f^{-1}(\mathrm{cl}_Y(B))\). Hence \(f^{-1}(\mathrm{cl}_Y(B))\) is closed in \(X\). This proves that \[ \mathrm{cl}_X(f^{-1}(B)) \subset f^{-1}(\mathrm{cl}_Y(B)). \]
\((vii) \Rightarrow (viii)\) Suppose that the condition (vii) holds. That is \[ \mathrm{cl}_X(f^{-1}(B)) \subset f^{-1}(\mathrm{cl}_Y(B)) \] for any \(B \subset Y\). Now we have to show that for any \(B \subset Y\) \[ f^{-1}(\mathrm{int}_Y(B)) \subset \mathrm{int}_X(f^{-1}(B)). \]