# Are open sets in locally compact spaces relatively compact?

As the title suggests, I am investigating whether all open sets in locally compact metric spaces are relatively compact. To me, this seems intuitive but I appear to be unable to produce a formal proof, that is, for an open set $U$ I cannot show that $\bar{U}$ is compact. I would therefore appreciate some help.

EDIT: The exact phrasing of the book from which this claim comes is : "If (X,d) is locally compact, we can even assume that $U$ is relatively compact.". Does this make sense?

• No. Pick any topological space that is locally compact, but not compact (e.g. $\mathbb{R}$ under the usual topology). Then the space itself is open, but its closure is not compact. – Theo Bendit Sep 5 '18 at 8:28
• @TheoBendit Great example, thank you. Do you perhaps know of any conditions that will make this true? – JohnK Sep 5 '18 at 8:37
• Unfortunately, no I don't. Even in metric spaces, it's not that easy; it's possible for bounded sets in a locally compact space to not be relatively compact. – Theo Bendit Sep 5 '18 at 8:40
• @TheoBendit Please see the edit for the exact statement. – JohnK Sep 5 '18 at 9:07
• The phrase "we can even assume" makes it context-dependent. It depends on what you're using $U$ for. If you're using it in a situation where another smaller neighbourhood would suffice (i.e. one where we can always use some neighbourhood $V \subseteq U$ instead of $U$), then yes this make sense. If your neighbourhood $U$ is not relatively compact, then replace it with $U \cap V$, where $V$ is a relatively compact neighbourhood (which must exist by local compactness). – Theo Bendit Sep 5 '18 at 9:11

For example, $\mathbb{R}$ with the usual topology, is locally compact, but not compact. Therefore, the subset $\mathbb{R}$ is open, but not relatively compact.
Another example: any infinite set equipped with the discrete metric. The space is locally compact, since for any point $x$ in the space, the open ball $B(x; 1/2) = \lbrace x \rbrace$ is open and compact. But, the open ball $B(x; 2)$ is the entire space, which is non-compact (all the singletons form an infinite open cover with no proper subcover). This illustrates how, in a metric space (unlike in a normed linear space), it's possible to have certain balls be relatively compact, and some not. It also shows an example of a bounded set in a locally compact space that isn't relatively compact.
As for your second question, in the edit, it may make sense depending on context. While not every open neighbourhood $\mathcal{U}$ of a point $x$ in a locally compact space is relatively compact, it is true that we can always find a smaller relatively compact neighbourhood $\mathcal{V}$ of $x$ contained in $\mathcal{U}$. All we do is use the definition of local compactness to find an open neighbourhood $\mathcal{W}$ of $x$ with compact closure. Then $\mathcal{U} \cap \mathcal{W}$ is an open set containing $x$. It has compact closure, because $$\mathcal{U} \cap \mathcal{W} \subseteq \mathcal{W} \implies \overline{\mathcal{U} \cap \mathcal{W}} \subseteq \overline{\mathcal{W}},$$ and closed subsets of compact sets are compact.
So, if the text uses $\mathcal{U}$ in such a way that we may always substitute $\mathcal{U}$ for a sub-neighbourhood $\mathcal{V}$, then we can safely replace $\mathcal{U}$ with a relatively compact neighbourhood $\mathcal{V}$. In most usages of neighbourhoods in topology, restricting neighbourhoods like this is acceptable (for example, showing convergence of a sequence/net).