3
$\begingroup$

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.)

  • 0
    There are two non-equivalent definitions for Tarski-finite sets. Which one are you using?2012-10-10
  • 0
    @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 $\aleph_0$.2012-10-10
  • 0
    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.2012-10-10
  • 0
    Also, note that in a metric space every singleton is closed, so (2) holds trivially since $E$ is non-empty.2012-10-10
  • 0
    The [third] edit is still trivial; take a constant function.2012-10-10
  • 0
    @Asaf Thank you for the advice. Please let me know if it's still trivial.2012-10-10
  • 0
    Maybe you want "$E$ is infinite then it has an infinite closed subset"?2012-10-10
  • 0
    @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.2012-10-10
  • 0
    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?2012-10-10
  • 0
    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$X$"?2012-10-10
  • 0
    But do you want $\forall E\exists f$? This is trivially true. You could want $\exists f\forall E$ which requires some choice2012-10-10
  • 0
    @Asaf I got it. I hope my final edit is fine :)2012-10-10

1 Answers 1