Topology
Definition 1. A topology on a set \(X\) is a collection of subsets \(\tau \subseteq \mathcal{P}(X)\) such that
\(\{\emptyset, X\} \subseteq \tau\)
For any collection of subsets \(\mathcal{A} \subseteq \tau\) one has \(\cup\mathcal{A}\in \tau\)
For any finite collection of subsets \(\mathcal{A} \subseteq \tau\) one has \(\cap\mathcal{A}\in \tau\)
If \(\tau\) is a topology on \(X\), we call \((X,\tau)\) a topological space. We may occasionaly say \(X\) is a topological space with topology \(\tau\).
Definition 2. Let \((X,\tau)\) be a topological space. A subset \(U\subseteq X\) is an open set if \(U\in\tau\). A subset \(F\subseteq X\) is a closed set if \(X\setminus F\in\tau\). If \(x\in X\) and \(U\subseteq X\) is an open set with \(x\in U\), we call \(U\) a neighborhood of \(x\).
Theorem 1. Let \((X,\tau)\) be a topological space and let \(\rho\) be the collection of closed subsets of \(X\). Then
\(\{\emptyset, X\} \subseteq \rho\)
For any collection of subsets \(\mathcal{A} \subseteq \rho\) one has \(\cap\mathcal{A}\in \rho\)
For any finite collection of subsets \(\mathcal{A} \subseteq \rho\) one has \(\cup\mathcal{A}\in \rho\)
Proof. Since \(\tau\) is a topology, \(\emptyset\in\tau\) and \(X\in\tau\). Then \(X\setminus\emptyset = X\in \tau\) and \(X\setminus X = \emptyset\in\tau\) so that \(\{\emptyset, X\}\subseteq \rho\).
Let \(\mathcal{A} \subseteq \rho\) be any collection of closed subsets. Then \(\mathcal{C} = \{X\setminus A : A\in\mathcal{A}\} \subseteq \tau\). Since \(\tau\) is a topology \(\cup\mathcal{C}\in\tau\). By the laws of DeMorgan \[\cap\mathcal{A} = \bigcap_{A\in\mathcal{A}} A = X\setminus \bigcup_{A\in\mathcal{A}} (X\setminus A) = X\setminus \cup\mathcal{C}.\] Since \(\cup\mathcal{C}\) is open, \(\cap\mathcal{A}\) is closed.
Let \(\mathcal{A} \subseteq \rho\) be any finite collection of closed subsets. Then \(\mathcal{C} = \{X\setminus A : A\in\mathcal{A}\} \subseteq \tau\) is a finite collection of open sets. Since \(\tau\) is a topology \(\cap\mathcal{C}\in\tau\). By the laws of DeMorgan \[\cup\mathcal{A} = \bigcup_{A\in\mathcal{A}} A = X\setminus \bigcap_{A\in\mathcal{A}} (X\setminus A) = X\setminus \cap\mathcal{C}.\] Since \(\cap\mathcal{C}\) is open, \(\cup\mathcal{A}\) is closed. ◻
Definition 3. A topological space \((X,\tau)\) is Hausdorff if for any distinct points \(x,y\in X\) there are \(U,V\in\tau\) with \(x\in U\), \(y\in V\), and \(U\cap V = \emptyset\). That is, \((X,\tau)\) is Hausdorff if for any pair of distinct points in \(X\), one can find disjoint neighborhoods containing them.
Example 1. Let \(X\) be a set and let \(\tau\) be the collection of subsets \(A\subseteq X\) such that \(A=\emptyset\) or \(X\setminus A\) is finite. Then \(\tau\) is a topology on \(X\) and \((X,\tau)\) is Hausdorff if and only if \(X\) is finite. We call \(\tau\) the finite complement topology on \(X\).
Proof. We have \(\emptyset\in\tau\) and since \(X\setminus X = \emptyset\) is finite, \(X\in\tau\).
Let \(\mathcal{A} \subseteq \tau\). Then \[X\setminus \cup\mathcal{A} = \bigcap_{A\in\mathcal{A}} (X\setminus A)\] is contained in \(X\setminus A\) for each \(A\in\mathcal{A}\). Since each is finite, so is \(X\setminus\cup\mathcal{A}\). Then \(\cup\mathcal{A}\in\tau\).
Let \(\mathcal{A} \subseteq\tau\) be a finite set. Then \[X\setminus \cap\mathcal{A} = \bigcup_{A\in\mathcal{A}} (X\setminus A)\] is a finite union of finite sets and hence finite. Then \(\cap\mathcal{A}\in\tau\). Hence \(\tau\) is a topology on \(X\).
Suppose \(X\) is finite. Let \(x,y\in X\) be distinct, i.e., \(x\neq y\). Since \(X\) is finite, \(X\setminus \{x\}\) and \(X\setminus \{y\}\) are finite sets. Then \(\{x\}\) and \(\{y\}\) are disjoint neighborhoods of \(x\) and \(y\) respectively. Hence \(X\) is Hausdorff.
Suppose \(X\) is infinite. Let \(x,y\in X\) be distinct. Let \(U\) be a neighborhood of \(x\) and \(V\) be a neighborhood of \(y\). Then \(X\setminus U\) and \(X\setminus V\) are finite sets. Then \[X\setminus (U\cap V) = (X\setminus U) \cup (X\setminus V)\] is a finite set. Since \(X\) is infinite, \(U\cap V\) must be infinite as well. In particular, \(U\cap V\neq\emptyset\) and \(\tau\) is not Hausdorff. ◻
Definition 4. A collection of subsets \(\mathcal{B}\) of \(X\) is a basis for a topology on \(X\) if
For each \(x\in X\), there is \(B\in\mathcal{B}\) with \(x\in B\)
For each \(x\in X\), if there is \(B_1,B_2\in\mathcal{B}\) with \(x\in B_1\cap B_2\), then there is \(B\in\mathcal{B}\) with \(x\in B\subseteq B_1\cap B_2\)
We define \(\tau_\mathcal{B}\) to be the collection of subsets \(U\) of \(X\) such that for each \(x\in U\) there is \(B\in\mathcal{B}\) with \(x\in B\subseteq U\).
Lemma 2. Let \(\mathcal{B}\) be a basis for a topology on \(X\). Let \(n\in\mathbb{N}\). If \(\mathcal{A}\subseteq\tau_\mathcal{B}\) is a set of cardinality \(n\), then \(\cap\mathcal{A}\in\tau_{\mathcal{B}}\).
Proof. We proceed with induction on \(n\). If \(|\mathcal{A}| = 0\), then \(\mathcal{A}=\emptyset\) and \(\cap\mathcal{A}=\emptyset\in \tau_\mathcal{B}\). If \(|\mathcal{A}|=1\), then there is \(A\in\tau_\mathcal{B}\) with \(\mathcal{A}=\{A\}\). Then \(\cap\mathcal{A} = A\in\tau_\mathcal{B}\). Suppose for \(k\in\mathbb{N}\) that any subset \(\mathcal{A}\subseteq\tau_\mathcal{B}\) with \(|\mathcal{A}|=k\) has \(\cap\mathcal{A}\in\tau_\mathcal{B}\). Let \(\mathcal{A}\subseteq\tau_\mathcal{B}\) be a subset with \(|\mathcal{A}|=k+1\). Let \(A\in\mathcal{A}\). Then \(\mathcal{A}\setminus \{A\}\) is a subset of \(\tau_\mathcal{B}\) of cardinality \(k\). Let \(x\in\cap\mathcal{A}\). Then \(x\in A\in \tau_\mathcal{B}\) and \(x\in \cap (\mathcal{A}\setminus\{A\}) \in \tau_\mathcal{B}\). Then there is \(B_1\in\mathcal{B}\) with \(x\in B_1\subseteq A\) and there is \(B_2\in\mathcal{B}\) with \(x\in B_2\subseteq \cap(\mathcal{A}\setminus\{A\})\). Then \(x\in B_1 \cap B_2\). Since \(\mathcal{B}\) is a basis, there is \(B\in\mathcal{B}\) with \(x\in B\subseteq B_1\cap B_2\subseteq \cap\mathcal{A}\). Then \(\cap\mathcal{A} \in \tau_\mathcal{B}\). ◻
Theorem 3. If \(\mathcal{B}\) is a basis for a topology on \(X\), then \(\tau_\mathcal{B}\) is a topology on \(X\).
Proof. Let \(\mathcal{B}\) be a basis for a topology on \(X\). Since \(\emptyset\) has no elements, it follows that \(\emptyset\in\tau_\mathcal{B}\). Since \(B\) is a basis, for each \(x\in X\) there is \(B\in\mathcal{B}\) with \(x\in B\subseteq X\). Then \(X\in\tau_\mathcal{B}\).
Let \(\mathcal{A} \subseteq \tau_\mathcal{B}\). Let \(x\in \cup\mathcal{A}\). Then there is some \(A\in\mathcal{A}\) with \(x\in A\). Since \(A\in\tau_\mathcal{B}\), there is \(B\in\mathcal{B}\) with \(x\in B\subseteq A\subseteq\cup\mathcal{A}\). Then \(\cup\mathcal{A} \in \tau_\mathcal{B}\).
Let \(\mathcal{A} \subseteq \tau_\mathcal{B}\) be finite. Then there is \(n\in\mathbb{N}\) with \(|A|=n\). By the preceeding lemma, \(\cap\mathcal{A}\in\tau_\mathcal{B}\). Then \(\tau_\mathcal{B}\) is a topology on \(X\). ◻
Theorem 4. Let \(\tau\) be a topology on \(X\). Let \(\mathcal{B}\subseteq \tau\). If for each open set \(U\in\tau\) and each \(x\in U\) there is \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\), then \(\mathcal{B}\) is a basis for a topology on \(X\) and \(\tau_\mathcal{B} = \tau\).
Proof. Suppose for each open set \(U\in\tau\) and each \(x\in U\), there is \(B\in\mathcal{B}\) with \(x\in B\subseteq U\).
Since \(\tau\) is a topology on \(X\), we have \(X\in\tau\). Let \(x\in X\). Then there is \(B\in\mathcal{B}\) with \(x\in B\).
Let \(x\in X\) and suppose there is \(B_1,B_2\in\mathcal{B}\) with \(x\in B_1\cap B_2\). Since \(\mathcal{B}\subseteq\tau\) and \(\tau\) is a topology, \(B_1\cap B_2\in\tau\). Then there is \(B\in\mathcal{B}\) such that \(x\in B\subseteq B_1\cap B_2\). Hence \(\mathcal{B}\) is a basis for a topology on \(X\).
Let \(U\in\tau\). Let \(x\in U\). Then there is \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). Hence \(U\in\tau_\mathcal{B}\) and \(\tau\subseteq\tau_\mathcal{B}\).
Let \(V\in\tau_\mathcal{B}\). For each \(x\in V\) there is \(B_x\in\mathcal{B}\) with \(x\in B_x\subseteq V\). Since \(\mathcal{B}\subseteq \tau\), we have \(B_x\in \tau\) for each \(x\in V\). Then \[V = \bigcup_{x\in V} B_x \in \tau\] and \(\tau_\mathcal{B}\subseteq \tau\). Hence \(\tau_\mathcal{B} = \tau\). ◻
Theorem 5. Let \(\mathcal{A}\) and \(\mathcal{B}\) be bases of topologies on \(X\). Then \(\tau_\mathcal{A} \subseteq \tau_\mathcal{B}\) if and only if for each \(x\in X\) and each \(A\in\mathcal{A}\) with \(x\in A\) there is \(B\in\mathcal{B}\) with \(x\in B\subseteq A\).
Proof. Suppose \(\tau_\mathcal{A} \subseteq \tau_\mathcal{B}\). Let \(x\in X\) and let \(A\in\mathcal{A}\) with \(x\in A\). Let \(a\in A\). Then \(a\in A\subseteq A\) with \(A\in\mathcal{A}\). Hence \(A\in\tau_\mathcal{A}\subseteq\tau_\mathcal{B}\). Then there is \(B\in\mathcal{B}\) such that \(x\in B\subseteq A\).
Suppose for each \(x\in X\) and each \(A\in\mathcal{A}\), there is \(B\in\mathcal{B}\) with \(x\in B\subseteq A\). Let \(U\in\tau_\mathcal{A}\). If \(U\) is empty, then \(U=\emptyset\in\tau_\mathcal{B}\) since \(\tau_\mathcal{B}\) is a topology. Suppose \(U\) is nonempty and let \(x\in U\). Then there is \(A\in\mathcal{A}\) such that \(x\in A\subseteq U\). Then there is \(B\in\mathcal{B}\) with \(x\in B\subseteq A\subseteq U\). Then \(U\in\tau_\mathcal{B}\). It follows that \(\tau_\mathcal{A}\subseteq \tau_\mathcal{B}\). ◻
Source File: jakemath.com/latex/topology/topological_spaces.tex