TITLE: Does this proposition from complex analysis depend on AC?
QUESTION [12 upvotes]: I was reading III vol. of Princeton lectures on analysis. Proposition 1.4:
"If $\Omega_{1}\supset\Omega_{2}\supset\ldots\supset\Omega_{n}\supset\ldots $ is a sequence of non-empty
compact sets in $\Bbb C$ with the property that:
$$\operatorname{diam}(\Omega_{n})\to 0\text{ as } n\to\infty,$$
then there exists a unique point $w\in\Bbb C$ such that $w \in \Omega_{n}$ for all $n$."
And in the proof:"Choose point $z_{n}$ in each $\Omega_{n}$"
Apparently this proof relies on Axiom of Choice. But I'm interested if it can be proved without reference to AC?

REPLY [3 votes]: Let $K\subset\mathbb {C}$ be compact. Let $a=\inf \{x: (x,y)\in K\}.$ Then set $b = \inf \{y: (a,y) \in K\}.$ Exercise: Show $(a,b)\in K.$ We can think of $(a,b)$ as the "southern-most eastern-most point" of $K.$ In fact, let's denote this point by $SEP(K).$ That solves our choosing problem: Given any family $\{K_\alpha\}$ of nonempty compact subsets of $\mathbb C,$ we simply choose, for each $\alpha,$ $SEP(K_\alpha).$