# Does every locally compact Hausdorff space admit a locally finite open covering by relatively compact sets?

Let $X$ be a locally compact Hausdorff space. Does there exist a locally finite open covering consisting of relatively compact sets?

Not necessarily. The ordinal space $\omega_1 = [ 0 , \omega_1 )$ provides a counterexample.

To see that there is no locally finite cover by relatively compact sets, note that every compact subset — and therefore every relatively compact subset — is bounded. So if $\mathcal{A}$ is a cover by relatively compact sets, we may inductively pick $\alpha_n \in \omega_1$, $A_n \in \mathcal{A}$ ($n \in \omega$) satisfying

• $\alpha_n \in A_n$;
• $\alpha_{n+1} > \sup ( A_n )$.

Then each neighbourhood of $\alpha = \sup_n \alpha_n < \omega_1$ meets infinitely many $A_n$.

The condition that you are referring to when coupled with local compactness is precisely paracompactness. Not every locally compact space is paracompact as the example of Arthur Fischer illustrates and which one can look up here.

$\mathbf{Theorem}$

Suppose that $X$ is a locally compact Hausdorff space. Then the following are equivalent.

1. $X$ is paracompact.

2. $X$ is the disjoint union of $\sigma$-compact open sets.

3. $X$ has a locally finite open covering consisting of relatively compact open sets.

$\mathbf{Proof}$

$1\rightarrow 3$. Suppose that $X$ is paracompact. Then let $\mathcal{U}$ be the collection of all relatively compact open subsets of $X$. Then $\mathcal{U}$ has a locally finite open refinement $\mathcal{V}$ and clearly each $V\in\mathcal{V}$ is paracompact.

$3\rightarrow 2$. Suppose that $x$ has a locally finite open covering $\mathcal{C}$ consisting of relatively compact open sets. Let $(\mathcal{C},E)$ be the graph where we let $U,V$ be connected by an edge if $\overline{U}\cap\overline{V}\neq\emptyset.$ If $x\in\overline{U}$, then there is an open set $O_{x}$ which intersects only finitely many sets $\overline{W}$ for $W\in\mathcal{C}$, namely $W_{1,x},...,W_{n_{x},x}$. Therefore, there are $x_{1},...,x_{n}$ in $\overline{U}$ such that $\overline{U}\subseteq O_{x_{1}}\cup...\cup O_{x_{n}}$. In particular, the only sets $W$ in $\mathcal{C}$ such that it is possible that $\overline{U}\cap\overline{W}$ are the sets $W_{1,x_{1}},...,W_{n_{x_{1}},x_{1}},...,W_{1,x_{n}},...,W_{n_{x_{n}},x_{n}}$. Therefore, the graph $(\mathcal{C},E)$ is locally finite. Therefore let $P$ be the partition of $\mathcal{C}$ into connected components. Then $\{\bigcup\mathcal{R}|\mathcal{R}\in P\}$ is a partition of $X$ into open sets where each $\bigcup\mathcal{R}$ is $\sigma$-compact.

$2\rightarrow 1$. This follows from the well-known fact that every open cover $\mathcal{U}$ of $X$ has an open refinement $\bigcup_{n}\mathcal{A}_{n}$ such that each $\mathcal{A}_{n}$ is locally finite. $\mathbf{QED}$

• I agree that 1 and 3 are equivalent, but condition 2 is out of place here as Mirko gave a counterexample. – user1688 Apr 26 '15 at 13:45
• Here is a proof that 3 implies 1: Let $(V_i)$ be any open covering and $(U_j)$ a locally finite one by relatively compacts. Then each $U_j$ can be covered by finetely many $V_i$. Replace each $U_j$ by the finietely many intersections with those $V_i$ and you found a locally finite refinement of $(V_i)$. – user1688 Apr 26 '15 at 13:50
• @Corbennick My counterexample no longer works as the answerer corrected condition 2, replacing countable union with disjoint union. – Mirko Apr 26 '15 at 20:39