Quantum Homotopy Type Theory

Theorem 32.1 Every second countable regular space is normal. Proof Let \(X\) be a second countable regular space. Pick a countable basis \(\mathcal{B}\). Let \(A\) and \(B\) be disjoint closed sets in \(X\). Pick a countable covering \(\{U_1, U_2, \ldots\}\) of \(A\) by open sets whose closures are disjoint from \(B\). Let \(\{U_1, U_2, \ldots\} = \{U \in \mathcal{B} \mid \overline{U} \cap B = \emptyset\}\). Prove: this covers \(A\). Let \(x \in A\). Prove: there exists \(U \in \mathcal{B}\) such that \(x \in U\) and \(\overline{U} \cap B = \emptyset\). Pick disjoint open sets \(U\) and \(V\) such that \(x \in U\) and \(B \subseteq V\). By regularity. Pick \(U' \in \mathcal{B}\) such that \(x \in U' \subseteq U\). \(x \in U'\) and \(\overline{U'} \cap B \subseteq \overline{U} \cap B \subseteq (X - V) \cap B = \emptyset\) Pick a countable covering \(\{V_1, V_2, \ldots\}\) of \(B\) by open sets whose ...

Definition A topological space \(X\) is regular iff it is \(T_1\) and, for any point \(x \in X\) and closed set \(B\) such that \(x \notin B\), there exist disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(B \subseteq V\). A topological space \(X\) is normal iff it is \(T_1\) and, for any disjoint closed sets \(A\) and \(B\), there exist disjoint sets \(U\) and \(V\) with \(A \subseteq U\) and \(B \subseteq V\). Lemma 31.1 Let \(X\) be a \(T_1\) space. \(X\) is regular if and only if, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\). \(X\) is normal if and only if, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\). Proof If \(X\) is regular then, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such th...

Definition A topological space is second countable iff it has a countable basis. Example 1 \(\mathbb{R}\), \(\mathbb{R}^n\) for \(n \in \mathbb{Z}_+\) and \(\mathbb{R}^\omega\) are all second countable. Proposition In a second countable space, any discrete subspace is countable. Proof Let \(X\) be a second countable space. Pick a countable basis \(\mathcal{B}\). Let \(A \subseteq X\) be discrete. For all \(a \in A\), pick \(B_a \in \mathcal{B}\) such that \(B_a \cap A = \{a\}\). The function that maps \(a\) to \(B_a\) is an injection \(A \rightarrow \mathcal{B}\). \(A\) is countable. \(\Box\) Example 2 \(\mathbb{R}^\omega\) under the uniform topology is not second countable. The set of sequences consisting of all 0s and 1s is an uncountable discrete subspace. Theorem 30.2 A subspace of a first countable space is first countable. A countable product of first countable spaces is first countable. A subspace of a second countable space is sec...

Definition A poset \(J\) is directed iff, for all \(x,y \in J\), there exists \(z \in J\) such that \(x \leq z\) and \(y \leq z\). Exercise 1 Show that the following are directed posets: (a) Any linearly ordered set. Solution For any \(x\), \(y\), we either have \(x \leq y\) or \(y \leq x\). In the first case, take \(z = y\). In the second case, take \(z = x\). (b) \(\mathcal{P} S\) for any set \(S\). Solution For any \(x,y \in \mathcal{P}S\) we have \(x \subseteq x \cup y\) and \(y \subseteq x \cup y\). (c) A set \(\mathcal{A}\) of subsets of \(S\) that is closed under finite intersection, partially ordered by \(\supseteq\). Solution For any \(X,Y \in \mathcal{A}\) we have \(X \supseteq X \cap Y\) and \(Y \supseteq X \cap Y\). (d) The set of all closed subsets of a topological space under \(\subseteq\). Solution For any \(C\), \(D\) closed we have \(C \cup D\) is closed and \(C, D \subseteq C \cup D\). Exercise 2 A subset \(K\) of \(J\) is...

Definition A space \(X\) is said to be locally compact at \(x\) if there is some compact subspace \(C\) of \(X\) that contains a neighbourhood of \(x\). If \(X\) is locally compact at each of its points, \(X\) is said to be locally compact . Example 1 \(\mathbb{R}\) is locally compact. For any point \(x\), the compact subspace \([x-1,x+1]\) includes the neighbourhood \((x-1,x+1)\). \(\mathbb{Q}\) is not locally compact. Assume for a contradiction \(C\) is a compact subspace of \(\mathbb{Q}\) that includes \((-\epsilon,\epsilon) \cap \mathbb{Q}\) for some \(\epsilon > 0\). Pick an irrational \(\xi \in (-\epsilon, \epsilon)\). Pick a strictly increasing sequence of rationals \((q_n)\) in \((-\epsilon, \epsilon)\) that converges to \(\xi\). \(\{(-\infty, q_n) \cap \mathbb{Q} \mid n \in \mathbb{Z}_+\} \cup \{(\xi, +\infty) \cap \mathbb{Q}\}\) is a sequence of open sets in \(\mathbb{Q}\) that covers \(C\). No finite subset covers \(C\). This is a contradic...