Vtyjkyuk

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 $barU$ 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?

For your first question, no, the open sets in a locally compact space are not relatively compact. The entire topological space is a clopen set, and is relatively compact if and only if the entire space is compact.

For example, $mathbbR$ with the usual topology, is locally compact, but not compact. Therefore, the subset $mathbbR$ 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 $mathcalU$ 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 $mathcalV$ of $x$ contained in $mathcalU$. All we do is use the definition of local compactness to find an open neighbourhood $mathcalW$ of $x$ with compact closure. Then $mathcalU cap mathcalW$ is an open set containing $x$. It has compact closure, because
$$mathcalU cap mathcalW subseteq mathcalW implies overlinemathcalU cap mathcalW subseteq overlinemathcalW,$$
and closed subsets of compact sets are compact.

So, if the text uses $mathcalU$ in such a way that we may always substitute $mathcalU$ for a sub-neighbourhood $mathcalV$, then we can safely replace $mathcalU$ with a relatively compact neighbourhood $mathcalV$. In most usages of neighbourhoods in topology, restricting neighbourhoods like this is acceptable (for example, showing convergence of a sequence/net).