Connected, locally compact, paracompact Hausdorff space is exhaustible by compacts

Question

I'm trying to understand the proof of the following:

A connected, locally compact, paracompact Hausdorff space $X$ has an exhaustion by compact sets,

That is, there exists a sequence $(K_n)_n$ of compact subsets of $X$ whose union is $X$ itself, such that $K_n$ is included in the interior of $K_{n+1}$, for all $n$.

The proof goes as follows. Choose a locally finite open cover $(U_i)_{i∈I}$ of $X$, such that the closure of $U_i$ is compact, for all $i$. Then every compact subset of $X$ intersects only finitely many of the $U_i$ (why?).

Then $K_1$ is chosen as the closure of any nonempty $U_i$. $K_2$ is chosen as the union of the closures of the $U_i$ that intersect $K_1$ and so on. The rest is easy.

Answer 1

One starts by noting that all the set of all open $O$ with $\overline{O}$ compact is an open cover of $X$ by locally compactness and Hausdorffness.

The paracompactness of $X$ then gives us a locally finite refinement $(U_i)_{i \in I}$ of that cover. It's not the $U_i$ that need to be compact in this proof, but their closures and this follows as each $U_i$ is a subset of some $O$ with compact closure so the same holds for the $U_i$. The new thing is the local finiteness, which is used for the compact set fact:

Now if $K$ is compact, each $x \in K$ has a neighbourhood $W_x$ such that $\{i \in I: U_i \cap W_x\neq \emptyset \}$ is finite, as we have a locally finite refinement. Then $K$ being compact is covered by finitely many of these $W_x$, say $W_x, x \in F$ for some finite $F \subseteq K$.

But then $$\{i \in I: K \cap U_i \neq \emptyset \} \subseteq \{i \in I: ( \bigcup_{x \in F} W_x ) \cap U_i \neq \emptyset\} = \bigcup_{x \in F} \{i \in I: W_x \cap U_i\neq \emptyset\}$$ where the latter is a finite union of finite sets so finite.

So the $\{U_i: i \in I\}$ are as needed.

Answer 2

Suppose $K\subseteq X$ is compact. For each $x\in K$, choose an open set $V_x$ around $x$ that intersects only finitely many $U_i$ (by local finiteness). Since $K$ is compact, it is covered by only finitely many of these $V_x$. Since each $V_x$ intersects only finitely many $U_i$, so does $K$.

Answer 3

Say $X$ is locally Lindelöf if every $x\in X$ has Lindelöf neighbourhood, and that $X$ is para-Lindelöf when every open cover of $X$ has a locally countable open refinement. Say $X$ is locally compact if every $x\in X$ has a compact neighbourhood. Say $X$ is paracompact if every open cover of $X$ has a locally finite subcover.

Theorem. If $X$ is locally Lindelöf and para-Lindelöf, then $X$ is a union of pairwise disjoint clopen Lindelöf subsets.

Proof: From assumptions we can find a locally countable open cover $\mathcal{U} = \{U_i : i\in I\}$ of $X$ and a family of Lindelöf sets $\mathcal{K} = \{K_i : i\in I\}$ such that $U_i\subseteq K_i$. Since $\overline{U_i}\cap K_i$ is a closed subspace of Lindelöf space, it's Lindelöf so we can assume $U_i\subseteq K_i\subseteq \overline{U_i}$.

Suppose $L\subseteq X$ is Lindelöf. For each $y\in L$ there is open $V_y\ni y$ which intersects only countably many $U_i$. Since $L$ is Lindelöf there is countably many $V_y$ which cover $L$, and so $L$ intersects only countably many $U_i$.

Let $L_1 = \{x\}$ and by induction define $L_{n+1} = \bigcup\{K_i : U_i\cap L_n\neq\emptyset\}$. Then $L_n \subseteq \bigcup\{U_i : U_i\cap L_n\neq\emptyset\}\subseteq \text{int}(L_{n+1})$ since $\mathcal{U}$ is a cover of $X$, and each $L_n$ is Lindelöf as a countable union of Lindelöf sets. Then $\bigcup_n L_n = \bigcup_n \text{int}(L_n)$ is open. If $y\in U_i$ and $U_i\cap \bigcup_n L_n\neq \emptyset$ then $U_i\cap L_n\neq\emptyset$ for some $n$, and so $y\in U_i\subseteq L_{n+1}$. This shows that if $y\notin\bigcup_{n=1}^\infty L_n$ then $U_i\cap \bigcup_{n=1}^\infty L_n = \emptyset$. It follows that $\bigcup_n L_n$ is a clopen Lindelöf subset of $X$ containing $x$.

Lastly lets show that thus obtained clopen Lindelöf subsets are pairwise disjoint. Define $u\sim v$ if there are $U_{i_1}, ..., U_{i_n}$ such that $u\in U_{i_1}$, $U_{i_k}\cap U_{i_{k+1}}\neq \emptyset$ for $k = 1, ..., n-1$ and $v\in U_{i_n}$. If $x\sim y$ then one easily shows by induction that $U_{i_k}\subseteq L_{k+1}$ and so $y\in L_{n+1}$. Conversely if $y\in L_{n+1}$ for some $n$ and $y\in U_i$ then $U_i\cap K_j\neq\emptyset$ for some $j$ with $U_j\cap L_n\neq\emptyset$ and so $U_i\cap \overline{U_j}\neq\emptyset$, thus $U_i\cap U_j\neq\emptyset$. We can then find some $z\in U_j\cap L_n$ and it will follow that $y\sim z$. By induction on $n$ we have $y\sim x$. So $\bigcup_n L_n$ constructed above is the equivalence class of the equivalence relation $\sim$ containing $x$. $\square$

Corollary 1. A connected locally Lindelöf para-Lindelöf space is Lindelöf

The following proposition can be found here:

Proposition. Locally compact Lindelöf space is exhaustible by compact sets.

Since locally compact implies locally Lindelöf, and paracompact implies para-Lindelöf, in particular we see that

Corollary. A connected locally compact paracompact space is exhaustible by compact sets.