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?

  • $\begingroup$ Existence might require AC, but uniqueness certainly doesn't. $\endgroup$ Jul 14, 2015 at 4:01
  • $\begingroup$ What is $C$? Is it the complex numbers? $\endgroup$ Jul 14, 2015 at 4:03

3 Answers 3


As Asaf says, Choice is not needed. Another way to see it.

Note that if $\bigcap_n \Omega_n = \varnothing$, then $\{ \mathbb C \setminus \Omega_n : n \in \mathbb N \}$ is a family of open subsets of $\mathbb C$ which covers $\Omega_1$. However no finite subfamily of this covers $\Omega_1$: if $n_1 < \ldots < n_k$ then $$( \mathbb C \setminus \Omega_{n_1} ) \cup \cdots \cup ( \mathbb C \setminus \Omega_{n_k} ) = \mathbb C \setminus ( \Omega_{n_1} \cap \cdots \cap \Omega_{n_k} ) = \mathbb C \setminus \Omega_{n_k},$$ and $\Omega_{n_k}$ is a nonempty subset of $\Omega_1$, contradicting the compactness of $\Omega_1$. This is a purely topological argument.

That $\bigcap_n \Omega_n$ contains a single point can be seen by supposing that $z \neq z^\prime$ are both in $\bigcap_n \Omega_n$, and taking an $n$ such that $\operatorname{diam} \Omega_n < | z - z^\prime |$.

  • $\begingroup$ Very nice.${}{}$ $\endgroup$
    – Asaf Karagila
    Jul 14, 2015 at 4:27

Fortunately, the axiom of choice can be circumvented here.

$\Bbb C$ is a Polish space, namely it is separable and completely metrizable. Moreover we are talking about compact sets and not sequentially compact sets. The equivalence between the two notion does require the axiom of choice, but it's not used here.

Now. Compact sets are closed in $\Bbb C$. And as luck would have it, we can choose from compact sets uniformly in $\Bbb C$, without using the axiom of choice. To see why, fix a countable dense subset $\{d_n\mid n\in\Bbb N\}$, and for each non-empty compact $E$ set define a sequence by induction, $x_n=d_k$ if either $k$ is the least for which $d_k\in E$, or else if it is the least index such that $\frac12d(x_{n-1},E)>d(d_k,E)$ (for $n=0$ take $d_0$).

This sequence is Cauchy, therefore it converges, and it converges to a unique point whose distance from $E$ is $0$. By compactness it means that the limit is in $E$.

In this definition we didn't use the axiom of choice because $\{d_k\mid k\in\Bbb N\}$ is a countable set.

Therefore, choosing from each $\Omega_n$ is possible even without assuming the axiom of choice holds.

  • $\begingroup$ Note, by the way, that this argument is not hinged on the family of compact sets being countable; and in fact we can push it to arbitrary non-empty closed sets. $\endgroup$
    – Asaf Karagila
    Jul 14, 2015 at 4:16
  • $\begingroup$ Did you mean $ d(d_k,E)<\frac12d(x_{n-1},E)$? $\endgroup$ Jul 14, 2015 at 6:43
  • $\begingroup$ Yes, thank you! $\endgroup$
    – Asaf Karagila
    Jul 14, 2015 at 6:44
  • $\begingroup$ I'd be interested to hear about the downvote. $\endgroup$
    – Asaf Karagila
    Jul 14, 2015 at 13:59

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 $\alpha,$ $SEP(K_\alpha).$

  • $\begingroup$ Exactly. This is what makes the real numbers so tame. $\endgroup$
    – user21820
    Jul 14, 2015 at 5:48

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