Hausdorff condition: existence or choice?

Question

I was looking at the proof that a compact set $C$ in a Hausdorff space $X$ must be closed, and something confused me.

In essence, the argument assigns, to each $c,d$ with $c\in C$ and $d\in X-C$, disjoint open sets $U,V$ with $c\in U$ and $d\in V$. It is claimed that this step follows immediately from the definition of a Hausdorff space. If that is accepted, then of course pick $c_1,...,c_n \in C$ such that their corresponding open sets $U$ make a cover for $C$, then take the intersection of the corresponding $V$.

It seems, however, that this line of reasoning assumes that some sort of choice function between pairs of points and pairs of open sets can be found. It was my understanding that the Hausdorff condition only says that some pairs of disjoint open sets can be found for each pair of points.

Or have I misread the definition?

Answer 1

This does indeed use the axiom of choice.

What it means for $X$ to be Hausdorff is that for all $x,y \in X$ with $x \ne y$, there exist disjoint open $U,V \subseteq X$ with $x \in U$ and $y \in V$. In order to select such a pair $U,V$ for each $x \in C$ and $y \in D \setminus C$, you need the axiom of choice.

To avoid choice, you'd need there to be a function $(X \times X) \setminus \Delta_X \to \mathcal{O}(X) \times \mathcal{O}(X)$, where $\Delta_X = \{ (x,x) \mid x \in X \}$, sending a pair $(x,y)$ to a pair of disjoint open sets $(U,V)$. However, the definition of Hausdorff space doesn't give you such a function in absence of choice.

There are individual cases where choice may not be necessary. As a trivial example, suppose $X$ is discrete; then the function $(x,y) \mapsto (\{x\},\{y\})$ works. For a less trivial example, suppose $X=\mathbb{R}^n$; then the function $(x,y) \mapsto (B_{|x-y|/2}(x),B_{|x-y|/2}(y))$ works, where $B_{\varepsilon}(x)$ is an open $\varepsilon$-ball about $x$.

Answer 2

This is one of the many cases where choice is used, but only to make the argument intuitive and making sense.

Use the common non-AC motto: If you can't choose, just take all of them!

So instead of choosing an open set for every pair, take all the open sets which witness a separation of a point in $C$ from a fixed point outside (i.e. they are disjoint from some nonempty open set which is a neighborhood of your fixed point), again this constitutes a cover, so there is a finite subcover which gives you a way to separate all points in $C$ from your fixed point. Note that the separation implies that the fixed point is outside the closure of the open sets, so it is not in the closed set which is the union of the closure of the open sets.

It follows that every point outdise $C$ has an open neighborhood disjoint from $C$, so it is a closed set.