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Let $X$ be a separable complete metric space.

I wonder if following properties hold in ZF.

  1. Limit Compact ⇒ Compact
  2. Does there exist a function$f$ such that $f(E)$ is closed and $f(E)\subset E$, for every infinite set $E$ in $X$.

If 2 doesn't hold, what if $E$ is Dedekind-Infinite?

It seems if 2 holds, 1 holds immediately. (See Constructing a choice function in a complete & separable metric space.)

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Katlus
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  • There are two non-equivalent definitions for Tarski-finite sets. Which one are you using? – Asaf Karagila Oct 10 '12 at 06:40
  • @Asaf I don't know about Tarski-finite sets, but i'm referring Tarski-infinite set to a set of which cardinal is not smaller than $\aleph_0$. – Katlus Oct 10 '12 at 06:47
  • When we have a definition for $X$-finite, then we say that $A$ is $X$-infinite if it is not $X$-finite. So it is equivalent to ask how do you define finiteness. Either way, you should have just said an infinite set, this is the common interpretation of the term; whereas Tarski-infinite (especially in the context of AC) could mean a set that has a chain of subsets which is unbounded. – Asaf Karagila Oct 10 '12 at 06:49
  • Also, note that in a metric space every singleton is closed, so (2) holds trivially since $E$ is non-empty. – Asaf Karagila Oct 10 '12 at 06:50
  • The [third] edit is still trivial; take a constant function. – Asaf Karagila Oct 10 '12 at 06:55
  • @Asaf Thank you for the advice. Please let me know if it's still trivial. – Katlus Oct 10 '12 at 06:56
  • Maybe you want "$E$ is infinite then it has an infinite closed subset"? – Asaf Karagila Oct 10 '12 at 06:59
  • @Asaf It doesn't have to. I want to know that if there exists a choice function in a polish space. Even if $f(E)$ is finite, it's fine. – Katlus Oct 10 '12 at 07:03
  • But it's unclear how you wish to quantify this. Do you want a single $f$ which gives that out; or for every $E$ you want to find some $f$ which does that? – Asaf Karagila Oct 10 '12 at 07:05
  • If one can formulate a closed subset for a given arbitrary $E$, doesn't this mean that "there exists a function $f$ such that $f(E)$ is a closed subset of $E$, for every infinite set $E$ in$X$"? – Katlus Oct 10 '12 at 07:11
  • But do you want $\forall E\exists f$? This is trivially true. You could want $\exists f\forall E$ which requires some choice – Asaf Karagila Oct 10 '12 at 07:12
  • @Asaf I got it. I hope my final edit is fine :) – Katlus Oct 10 '12 at 07:16

1 Answers1

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The second answer is "no".

Consider a model in which there is an infinite Dedekind-finite set of real numbers, now consider all its subsets which are infinite.

We observe that every closed subset of a D-finite set is finite. (Not relatively closed, but really closed.) We also observe that we can always choose from finite sets of real numbers, since those are well-ordered by the natural order of the reals (every linear order on a finite set is a well-order). Therefore the existence of such $f$ implies that we can choose a point from every subset of our D-finite set, which means it is well-orderable, which means it is finite. Contradiction.

Asaf Karagila
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  • I believe the requirement that $E$ is D-infinite is still not enough, but I will have to think about it some more. – Asaf Karagila Oct 10 '12 at 07:36
  • If it's not true for D-infinite, what's the reason you think that makes it not possible? Choosing Countable subsets from D-infinite sets? Or Choosing Well-orderings of Countable sets? I would like to know how you think about 'existence of such function whose domain is a collection of Countable sets' – Katlus Oct 10 '12 at 08:19
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    @Katlus: Choosing a well-ordering. For example it is consistent that there is no choice of well-ordering for every countable set of real numbers, so the set ${A\subseteq\mathbb R\mid A\text{ is countably infinite}}$ has a cardinality strictly larger than $\mathbb R$. – Asaf Karagila Oct 10 '12 at 08:21