# Every $\sigma$-compact LCH space is para-compact

Let $(X,\mathcal{T})$ be a $\sigma$-compact locally compact $T_2$ space, and $\mathcal{C} \subseteq \mathcal{T}$ an open cover. $\newcommand{\seq}[1]{\left( #1 \right)}$

1. There exists a locally finite open cover $\seq{V_j}_{j \in M}$ refining $\mathcal{C}$ such that $V_j^-$ is compact.

2. There exists a locally finite open cover $\seq{W_j}_{j \in M}$ refining $\mathcal{C}$ such that $W_j^- \subseteq V_j$.

I have no trouble proving (1). Proof: By the hypothesis on $(X,\mathcal{T})$, there exists an increasing $\seq{U_k}_{k=0}^\infty$ sequence of pre-compact open sets with $U_k^- \subseteq U_{k+1}$ and $X = \bigcup_{k=0}^\infty U_k$. Define $U_{-1} = \emptyset$. Observe that $\{S \cap (U_{k+2} \setminus U_{k-1}^-)\mid S \in \mathcal{C}\}$ is an open cover for $U_{k+1}^- \setminus U_k$, which as a closed subset of a compact set $U_{k+1}^-$ is compact.. Hence there exists a finite subcover $\seq{S_r \cap (U_{k+2} \setminus U_{k-1}^-)}_{r=0}^{p_k-1}$. Then the subcover $\seq{S_r \cap (U_{k+2} \setminus U_{k-1}^-)}_{r=0,k=0}^{p_k-1,\infty}$ is a locally finite refinement of $\mathcal{C}$. Let this subcover be $\seq{V_{k,r}}_{r=0,k=0}^{p_k-1,\infty}$. Evidently $V_{k,r}^-$ is compact.

How do I show (2)? I considered closed sets of the form $$K_{k_0,r_0} = \bigcap_{\substack{k \in \mathbb{W} \\ r \in [0\,..\,p_k-1] \\ k \neq k_0 \lor r \neq r_0}} V_{r,k}^c$$ where $k_0 \in \mathbb{W}, r_0 \in [0\,..\,p_k-1]$. The above is a intersection of complements of all $V_{r,k}$ except for one. It is necessarily contained in $V_{k_0,r_0}$ and compact. Hence there exists a pre-comact open set $W_{k_0,r_0}$ with $K_{k_0,r_0} \subseteq W_{k_0,r_0} \subseteq W_{k_0,r_0}^- \subseteq V_{k_0,r_0}$. I don't know how to show the collection of $W_{k,r}$ cover $X$, and is the argument above correct?

I think showing (2) must involve the use of $\mathcal{C}$ since otherwise the above argument can be successively repeated to obtain "smaller" locally finite refinements of $\mathcal{C}$.

HINT: $X$ is $T_3$ and Lindelöf, so it’s normal. Thus, (2) is a consequence of the following theorem, which is Theorem $\mathbf{15.10}$ in Willard’s General Topology.

Theorem. A space $X$ is normal if and only if every point-finite open cover is shrinkable.

To prove the direction that we need here, suppose that $X$ is normal, and let $\mathscr{U}$ be a point-finite open cover of $X$. If $\kappa=|\mathscr{U}|$, we can index $\mathscr{U}$ as $\mathscr{U}=\{U_\xi:\xi<\kappa\}$. Let $F_0=X\setminus\bigcup_{\xi>0}U_\xi$; $F_0$ is a closed subset of $U_0$, so there is an open $V_0$ such that $F_0\subseteq V_0\subseteq\operatorname{cl}V_0\subseteq U_0$. Now suppose that $\eta<\kappa$, and $V_\xi$ has been defined for each $\xi<\eta$. Let

$$F_\eta=X\setminus\left(\bigcup_{\xi<\eta}V_\xi\cup\bigcup_{\xi>\eta}U_\xi\right)\;;$$

$F_\eta$ is a closed subset of $U_\eta$, so there is an open $V_\eta$ such that $F_\eta\subseteq V_\eta\subseteq\operatorname{cl}V_\eta\subseteq U_\eta$. To complete the argument we need only show that $\mathscr{V}=\{V_\xi:\xi<\kappa\}$ is a cover of $X$. I’ll leave that to you for now, but feel free to ask questions if you get stuck.

The other direction is the easy one. If $H$ and $K$ are disjoint closed subsets of $X$, then $\{X\setminus H,X\setminus K\}$ is a point-finite open cover of $X$, so it has a shrinking $\{U,V\}$: $U\cup V=X$, $\operatorname{cl}U\subseteq X\setminus H$, and $\operatorname{cl}V\subseteq X\setminus K$. But then $X\setminus\operatorname{cl}U$ and $X\setminus\operatorname{cl}V$ are disjoint open nbhds of $H$ and $K$, so $X$ is normal.

• By $\kappa = |\mathscr{U}|$ do you mean the cardinality of $\mathscr{U}$? Commented Nov 7, 2016 at 2:33
• @Henry: Yes, that's right. The indexing is really just a convenient way to get a well-ordering of the cover. Commented Nov 7, 2016 at 2:51
• but such an indexing requires AoC, which should not be needed for this theorem. Commented Nov 7, 2016 at 2:52
• @Henry: Why do you think that it shouldn't be necessary? (And why would you care? $\mathsf{AC}$ is a normal part of everyday mathematics -- especially if one is doing topology, since it's equivalent to one of the most important results, the Tikhonov product theorem.) Commented Nov 7, 2016 at 3:01
• It came from this post. Does this argument still hold if we remove the axiom that every infinite set has a total order? Commented Nov 7, 2016 at 3:05