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
    @Asaf I got it. I hope my final edit is fine :)2012-10-10

1 Answers 1

1

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.

  • 1
    @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 $\mathbb R$.2012-10-10