Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
There are two non-equivalent definitions for Tarski-finite sets. Which one are you using? –  Asaf Karagila Oct 10 '12 at 6:40
    
@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 6:47
    
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 6:49
    
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 6:50
    
The [third] edit is still trivial; take a constant function. –  Asaf Karagila Oct 10 '12 at 6:55

1 Answer 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.

share|improve this answer
    
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 7:36
    
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 8:19
    
@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 8:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.