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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.

Katlus
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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 08 '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 08 '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 08 '12 at 18:59

1 Answers1

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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.

Hagen von Eitzen
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  • Nice. The construction does not require choice, we rely in the countability of the dense set. – Asaf Karagila Oct 08 '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 08 '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 08 '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 08 '12 at 18:21
  • @NateEldredge: Thanks for the hint. I hope my last edit fixed that. – Hagen von Eitzen Oct 08 '12 at 18:24
  • With @Nate's fix you get that $A_{n}$ is a descending sequence of non-empty sets and vanishing diameter as $n \to \infty$. But... How do you prove that the intersection is non-empty? The usual proof of the Cantor nested sets theorem proceeds by choosing elements in $A_n$ and showing they form a Cauchy sequence, hence converge to an element of the intersection, but you shouldn't choose... – commenter Oct 08 '12 at 18:27
  • @commenter: Good point. But it should be possible to arrange that the $x_{m_n}$ themselves are a Cauchy sequence. – Nate Eldredge Oct 08 '12 at 18:31
  • @Nate: right, this should be possible. Or you could choose the least index $k$ such that $x_k \in A_{n} \cap B_{2r_n}(x_{m_n})$, so $x_k \in A_{n+1}$. This gives a definite sequence that should turn out to be Cauchy by the usual proof of the nested sets theorem. – commenter Oct 08 '12 at 18:42
  • @Hagen Beautiful Constructuon. Thank you! – Katlus Oct 09 '12 at 00:24
  • @commenter: It seems that my $f(E)$ is constructed without choice, but the proof that it exists requires Choice. However, the subsequence $(x_{m_n})n$ itself is Cauchy (an of course its limit is in $E$): $d(x{m_n},x_{m_{n+1}})$ is bounded by $\frac5{2^{n+1}}$: There is $a\in A_{n+1}$ with $d(x_{m_{n+1}}, a)<2^{-n-1}$ and $d(x_{m_n},a)<2r_{n+1}<2^{1-n}$. – Hagen von Eitzen Oct 09 '12 at 16:16
  • Sorry, I didn't understand the first sentence of your comment. Where does the existence proof require Choice? (side note: There's a small omission in your first definition of completeness: you need the vanishing diameter condition.) – commenter Oct 09 '12 at 16:50
  • I thought Hagen intended ${x_{m_n}}$ to be a Cauchy sequence. – Katlus Oct 10 '12 at 06:00
  • @commenter: What my first sentence should express was: While - in the presence of AC - different people may "choose" different basis of $\mathbb R$ as a $\mathbb Q$-linear space, they would not obtain different choice functions with my (original) construction; one needs AC only to prove that $A$ is empty in the first place, not to take the only element of $A$ once you know its non-empty. I thought that was weird. $\quad$ For the side note: I edited accordingly, but do you have an example of a space where nonempty intersection follows if the diameters tend to 0, but not in general? – Hagen von Eitzen Oct 10 '12 at 06:19
  • Silly example: $A_n = [n,\infty)$ in $\mathbb{R}$. If you want the $A_n$ to have bounded diameter you need something that isn't locally compact, for example: Take the unit sphere $S = {x \in \ell^2 : \sum x_{n}^2 = 1}$ in the Hilbert space $l^2$ and consider the right shift $T(x_0,x_1,\dots) = (0,x_0,x_1,\dots)$. Then $A_n = T^n(S)$ is a nested sequence of closed sets of diameter $2$ and $\bigcap_{n=1}^\infty A_n = \emptyset$. Similar examples can be found in the algebraic closure $\mathbb{C}_p = \overline{\mathbb{Q}_p}$. – commenter Oct 10 '12 at 06:59