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



Does every locally compact Hausdorff space admit a locally finite open covering by 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 $A\_n$.


