Lee Theorem 1.15
We wish to prove the following:
Theorem. Every topological manifold is paracompact.
Recall that paracompactness is the condition that for each open cover of a given space, there exists a locally finite refinement. Our strategy for proving this result hinges on some important lemmas, namely the following:
Lemma. Every topological manifold is locally compact.
This follow from the fact that we have a pre-compact basis, so each \(x \in X\) is contained in neighbourhood \(B\) such that \(B \subset C\) for some compact set \(C\) (here, \(C = \overline{B}\), as \(B\) is precompact).
Lemma. Every second-countable locally compact Hausdorff space has an exhaustion by compact sets.
By exhaustion, we mean a sequence of nested compact sets \(K_i\), such that \(K_i \subset \text{Int}(K_{i + 1})\). To show this, the idea will be to make use of the countable basis. In particular, since the space is locally compact, each \(x \in X\) has a compact neighbourhood \(x \in U_x \subset C_x\). Choose some element \(B_x\) of the countable basis which contains \(x\) and is in \(U_x\). Thus, the \(B_x\) are all compact neighbourhoods and are a countable cover of \(X\). We label these sets \(\mathcal{B} = \{B_1, B_2, \dots\}\), and let the corresponding compact sets be \(C_1, C_2, \dots\).
From here, note that \(\overline{B_k} \subset \overline{C_k} = C_k\) is closed in a compact set, and is thus compact (we know that \(C_k\) is closed as it is compact is a Hausdorff space). We let \(U_1 = B_1\). We then form a finite cover of \(\overline{U_1}\) (as this set is compact) with elements of \(\mathcal{B}\), which we call \(V_2\), and let \(U_2 = V_2 \cup B_2\). \(V\) is a finite union of precompact sets, and is thus precompact. Thus, we can repeat this process with \(U_k = V_k \cup B_k\), with \(V_k\) being a finite cover of \(\overline{U_{k - 1}}\).
It is easy to check that the sequence of compact sets \(K_i = \overline{U_i}\) is the desired exhaustion of \(X\).
Now, let us move forth, and prove the theorem. Of course, we will make use of the exhaustion. Let \(\mathcal{U}\) be an open cover for the space \(X\). The main idea will be to use the compact exhaustion to express the space as a union of a bunch of "finite" shells.
Let's be more specific: let \(K_i\) denote our exhaustion. We define \(W_i = (X - K_{i - 1}) \cap \text{Int}(K_{i + 1})\). Clearly, this is an open set. From \(\mathcal{U}\), we define our new refined cover as follows: we take a finite subcover \(\mathcal{V}_1\) for \(K_1\) from \(\mathcal{U} \cap K_2\). We then proceed inductively and finitely cover \(K_i\) using \(\mathcal{V}_{i - 1}\) and \(\mathcal{U} \cap W_i\), forming \(\mathcal{V}_i\).
We take \(\mathcal{V} = \mathcal{V}_1 \cup \mathcal{V}_2 \cup \cdots\). Clearly, this is a cover and a refinement of \(\mathcal{U}\). Moreover, it is locally finite: given \(x \in X\), note that \(x \in \text{Int}(K_i)\) for some \(i\). For some \(V \in \mathcal{V}_k\) for \(k \geq i + 1\), note that \(V \subset X - K_{i}\), implying that \(V \cap \text{Int}(K_i) = \emptyset\). Thus, \(x \in \text{Int}(K_i)\), which can only intersect elements of \(\mathcal{V}_1 \cup \cdots \cup \mathcal{V}_i\), a finite set.