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Let $X$ be a complete & separable metric space. Let $\{E_i\}_{i\in I}$ be a collection of closed and nonempty sets in $X$.

If $X$ is just a complete metric space, it seems not possible to construct a choice function $f$ such that $f(E_i)\in E_i$, without AC.

However, since $X$ is separable, it seems possible to construct a choice function.

How do I construct a choice function $f$ such that $f(E_i)\in E_i$ without AC$_\omega$?

EDIT; Since Hagen's argument doesn't require the condition 'boundedness of $E_i$, I removed it. Also $\{E_i\}$ doesn't need to be countable, so i changed it to $\{E_i\}_{i\in I}$ where $I$ is an arbitrary set.

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To be clear: by "complete" you mean "every Cauchy sequence converges"? There are other possible definitions which are equivalent given sufficient choice, but in ZF may be different. – Nate Eldredge Oct 8 '12 at 18:32
@Nate Yes, i meant "every Cauchy sequence converges". I would like to know what equivalent statements are there (in ZFC). – Katlus Oct 8 '12 at 18:41
For example, "any decreasing sequence of closed sets with diameters decreasing to 0 has nonempty intersection", which Hagen's answer used. Actually, if your claim holds in ZF then this statement will be equivalent after all. – Nate Eldredge Oct 8 '12 at 18:59
up vote 8 down vote accepted

Let $(x_n)_{n\in \mathbb N}$ be a dense sequence.

Given a closed non-empty set $E$, define $f(E)$ as follows:

a) If $E\cap \{x_n\mid n\in \mathbb N\}$ is non-empty, let $f(E)=x_m$ where $m=\min\{n\in\mathbb N\mid x_n\in E\}$.

b) If on the other hand $E\cap \{x_n\mid n\in \mathbb N\}=\emptyset$, define $A_0=E$, $r_0=1$ and recursively $$\tag1m_n=\min\{k\in\mathbb N\mid d(x_k,A_n)<2^{-n}\},$$ $$\tag2r_{n+1} = d(x_{m_n}, A_n),$$ $$\tag3A_{n+1}=\{a\in A_n\mid d(x_{m_n},a)\le 2r_{n+1}\}.$$ Note that we will always have $2^{1-n}\ge r_n>0$ and $E=A_0\supseteq A_1\supseteq\ldots$ is a descending chain of closed sets. Let $A=\bigcap_{n\in \mathbb N} A_n$. By completeness, $A$ is not empty. And because the diameter of $A_n$ is at most $4r_{n}\to0$, there can be at most one element in $A$. Define $f(E)$ as the only element of $A$.

Remark: Note that it is possible that no point $a\in A$ exists with $d(x,A)=d(x,a)$, but at least we have that $d(x,A)=0$ implies $x\in A$ for closed sets $A$, an dthat special case is covered in part a) so that we indeed have $r_n>0$ in part b).

Edit (after commments by Nate et. al.): The above proof used as the following definition of completeness: Every decreasing sequence of nonempty closed sets withvanishing diameter has nonempty intersection. The same construction also works with the definition Every Cauchy sequence converges.

In fact, the sequence $(y_n)_{n\in\mathbb N}$ with $y_n:=x_{m_n}$ as constructed above is Cauchy: By $(1)$ there exists $a\in A_{n+1}$ such that $d(y_{n+1},a)<2^{-(n+1)}$. On the other hand, $(3)$ and $(2)$ imply $d(y_n,a)\le 2r_{n+1}=2d(y_n,A_n)<2\cdot2^{-n}$ for this $a$. Consequently, $d(y_n,y_{n+1})<\frac5{2^{n+1}}$, which leads to $d(y_n,y_m)< \frac5{2^n}$ for $m>n$. Hence $y=\lim_{n\to\infty}y_n$ exists and from $d(y_n,E)\le d(y_n,A_n)\to 0$ we conclude $d(y,E)=0$ and by closedness $y\in E$ as required.

Edit: After a night of good sleep, the construction can be simplified (no cases, work directly with Cauchy sequences): Let $(x_n)_{n\in \mathbb N}$ be a dense sequence. Given a closed non-empty set $E$, define $f(E)$ as follows: Let $m_0=\min\{n\in \mathbb N\mid d(x_n,E)<1\}$ and $y_0=x_{m_0}$. Extend this recursively to a sequence $(y_n)_{n\in\mathbb N}$ with the properties $$\tag4d(y_n,y_{n+1})<2^{1-n}$$ and $$\tag5d(y_n,E)<2^{-n}.$$ To achieve this note that $d(y_n,E)<2^{-n}$ implies that the set $C_n=\{a\in E\mid d(y_n,a)<2^{-n}\}$ is nonempty, hence the union of open balls $U_n=\bigcup_{a\in C_n} B(a,2^{-1-n}\}$ is a nonempty open set. Let $m_{n+1}=\min\{n\in\mathbb N\mid x_n \in U_n\}$ and set $y_{n+1}=x_{m_{n+1}}$. From $y_{n+1}\in U_n$ we conclude that there exists $a\in C_n$ with $d(y_{n+1},a)<2^{-(n+1)}$. But for any such $a$ we see from $a\in C_n\subseteq E$ that $d(y_{n+1},A)<2^{-(n+1)}$ and $d(y_n,y_{n+1})\le d(y_n,a)+d(y_{n+1},a)<2^{-n}+2^{-(n+1)}<2^{1-n}$. Hence we can construct our sequence $(y_n)$ recursively such that $(1)$ and $(2)$ hold. By $(4)$ and the triangle inequality, it is a Cauchy sequence, hence has a limit $y$. From $(5)$ we conclude that $d(y,E)=0$, hence $y\in E$ because $E is closed.

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Nice. The construction does not require choice, we rely in the countability of the dense set. – Asaf Karagila Oct 8 '12 at 18:09
In your construction, how do you know $A_{n+1}$ is nonempty? Even in a complete separable metric space, given a point $x$ and a closed set $A$, there need not be any $y \in A$ with $d(x,y) = d(x,A)$. For example, let $X$ be a separable Hilbert space with orthonormal basis $\{e_n\}$ and let $A = \{ (1 + \frac{1}{n}) e_n \}$. $A$ is closed, indeed discrete, and $d(0,A) = 1$, but $d(0,y) > 1$ for every $y \in A$. – Nate Eldredge Oct 8 '12 at 18:16
The unmotivated $2^{-n}$ apparently can be relaxed - but I thought I might need something with nice convergence while I wrote this. One could replace it by any zero sequence - or (I think) ignore it completely and just take the nearmost points for all $x_n$ in sequence: If $a\ne b$, then some member $x_n$ will be closer to $a$ then $\frac12d(a,b)$, hence $a$ and $b$ cannot be both in $A$. – Hagen von Eitzen Oct 8 '12 at 18:16
... I think you can fix this problem by taking $A_{n+1} = \{a \in A_n \mid d(x_{m_n}, a) \le r_{n+1} + 2^{-n}\}$ or something like that. But please check me here. – Nate Eldredge Oct 8 '12 at 18:21
@NateEldredge: Thanks for the hint. I hope my last edit fixed that. – Hagen von Eitzen Oct 8 '12 at 18:24

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