Enumerating open sets around elements of an uncountable set in topology - how do we justify it? I was trying to prove for myself a basic result in point-set toplogy, namely that "Any compact subset of a Hausdorff space is also closed".
To be precise we're taking compact set here to mean - any open cover has a finite subcover that covers the set, and closed set to mean - its complement is open (which we also know holds iff it contains all of its limit points).
Now, one thing I kept trying to avoid is an argument that involves building sets around each element of the compact subset in question, since theoretically this subset can not only be infinite, but uncountable (e.g. think about the real numbers).
When I finally looked at some online proofs, it looks like they do exactly that though - i.e an argument of the form "for each element x in A there exists an open set such that..."
Do we just accept it as part of the definitions of topology that we can prove things using such an argument even if the set in question is countably infinite or even uncountable? it feels like a certain flavor of an "axiom of choice" at play but i'm curious if there is any formalism around this or we just take it for granted its allowed.
Again, The difficulty i'm having conceptualizing this is specifically the idea of enumerating elements of an uncountable set (e.g. some subset of $\mathbb{R}$) in such a "discrete" way as to build sets around each one as part of a proof..
 A: It’s completely standard, and we take it for granted; the cardinality of the set in question is irrelevant. Yes, in general one may need the axiom of choice to choose an open nbhd of each point of some set, but we routinely assume the axiom of choice; in fact, one of the most important theorems in general topology, the Tikhonov product theorem (which says that the arbitrary Cartesian product of compact spaces is compact) is actually equivalent to the axiom of choice.
To use your example, if $X$ is a Hausdorff space, $K$ is a compact subset of $X$, and $p\in X\setminus K$, it is completely routine to say that for each $x\in K$ there are disjoint open sets $U_x$ and $V_x$ such that $x\in U_x$ and $p\in V_x$; the only justification required here is justification for the assertion that $U_x$ and $V_x$ can be chosen to be disjoint, and the hypothesis that $X$ is Hausdorff takes care of that.
A: In the axioms of Set Theory, the Axiom Schema of Comprehension declares the existence of the set $\{x\in S: P(x)\}$ whenever $S$ is a set and $P(x)$ is some property (which any $x$ might or might not have). It is called a Schema because it is a collection of axioms, one for each $P(x)$ you can state. This is NOT the Axiom of Choice.
For example once we have the set $S=\Bbb R\times \Bbb R,$ we have the set $\{x=(u,v)\in S: v=u^2\}$ which, less formally, we call the real function $f(u)=u^2.$ (In Set Theory a function $is$ its graph.)
A topology on a set $X$ is a set $S$ of some or all of the subsets of $X$ (i.e. the set of open sets) that meets certain conditions.
There is a saying: If you can't (or don't want to) use Choice then choose everything.
Suppose $Y\subset X$ and $p\in \overline Y$ \ $Y.$ We have the set $V=\{x\in S: p\not \in \bar x\}.$ Suppose  $S$ is a Hausdorff topology. It readily follows that each $q\in X$ \ $\{p\}$ belongs to at least one member of $V.$ So $V$ is an open cover of $X$ \ $\{p\}$ so (a fortiori) $V$ is an open cover of $Y.$
It now follows readily that if $F$ is a finite subset of $V$ then $F$ cannot cover $Y,$ otherwise $X \setminus \cup  \{\bar x: x\in F\}$ would be an open set containing $p$ and disjoint from $Y,$ contrary to $p\in \overline Y.$ Therefore $Y$ is not compact.
A: As Brian M. Scott pointed out, in general we're using choice here, but in some common cases this can be avoided: e.g. if $(X,d)$ is a metric space, $K\subseteq X$ is compact and $p \notin K$ we can define $r_x = \frac{d(x,p)}{2} >0$, for $x \in K$ and we have an open cover $\{B(x, r_x): x \in K\}$ to consider, completely defined without the Axiom of Choice, thanks to the metric.
