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If a metric space has the limit point property, is it separable? (ZF + AC$_\omega$)

I'm struggling with this problem for a week.

I'm talking about this in Metric space.

Here's the part of argument in Rudin PMA p.45; Let $X$ be limit point compact. Fix $\epsilon>0$ and $x_0 \in X$. Having chosen $x_0,\ldots,x_j \in X$, choose $x_{j+1}$, if possible, so that $d(x_i,x_{j+1})≧\epsilon$ for $i=0,\ldots,j$. Then this process must stop after a finite number of steps.

Here, Dependent Choice is used.

I know that ‘Limit point property $\Rightarrow$ separable’ is unprovable in ZF. (You can see this: If every infinite subset has a limit point in a metric space $X$, then $X$ is separable (in ZF))

I want to prove this in ZF+AC$_\omega$. Help.

So far, I proved that [Limit point compact $\Rightarrow$ separable] $\Rightarrow$ [Limit point compact $\Rightarrow$ Compact].

I need this to complete my proof.

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@t.b. Dependent Choice. I'll edit my post –  Katlus Aug 10 '12 at 13:40
    
When you post a question that's closely related to earlier questions, please link to the earlier questions (as you did in your previous question) to avoid unnecessary duplication of efforts. –  joriki Aug 10 '12 at 13:40
    
@joriki i'll edit. Sorry –  Katlus Aug 10 '12 at 13:45
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3 Answers

up vote 5 down vote accepted

Countable choice is sufficient. Separability is an easy consequence of total boundedness. Suppose that $X$ is not totally bounded. Then there is an $r>0$ such that for each $n\in\Bbb Z^+$,

$$X_n=\left\{\langle x_1,\dots,x_n\rangle\in X^n:d(x_i,x_j)\ge r\text{ for }1\le i<j\le n\right\}\ne\varnothing\;.$$

By countable choice there is some $\langle a_n:n\in\Bbb Z^+\rangle\in\prod_nX_n$. For $n\in\Bbb Z^+$ let $a_n=\langle x^n_i,\dots,x^n_n\rangle$. Now concatenate the $a_n$ to form an infinite sequence

$$\sigma=\langle x^1_1,x^2_1,x^2_2,x^3_1,x^3_2,x^3_4,\dots\rangle$$

in $X$. Note that if $1\le k\le m<n$, there is at most one $s\in\{1,\dots,n\}$ such that $d(x^m_k,x^n_s)<\frac{r}2$: this is an easy consequence of the triangle inequality and the fact that the points $x^n_1,\dots,x^n_n$ are mutually at least $r$ apart.

For $n\in\Bbb Z^+$ let $B_n=\{x^n_1,\dots,x^n_n\}$. Let $y_1=x^1_1$. Suppose that $n\in\Bbb Z^+$ and that we’ve chosen $y_k\in B_k$ for $1\le k\le n$ so that the points $y_1,\dots,y_n$ are mutually at least $r/2$ apart. For $k=1,\dots,n$ let $A_k=\varnothing$ if $d(y_k,x)\ge r/2$ for each $x\in B_{n+1}$, and let $A_k=\{x\}$ if $x$ is the unique member of $B_k$ whose distance from $y_k$ is less than $r/2$. Let $A=\bigcup_{k=1}^nA_k$; clearly $|A|\le n$, so $B_{n+1}\setminus A\ne\varnothing$, and we may set $y_{n+1}=x^{n+1}_k$, where $k$ is minimal such that $x^{n+1}_k\notin A$. (Note that this choice of $y_{n+1}$ is well-defined and does not require any part of the axiom of choice.) Then $d(y_{n+1},y_k)\ge r/2$ for $k=1,\dots,n$, and the construction goes through to produce an infinite sequence $\langle y_n:n\in\Bbb Z^+\rangle$ whose terms are mutually at least $r/2$ apart.

But then the set $\{y_n:n\in\Bbb Z^+\}$ is an infinite subset of $X$ with no limit point.

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It's extremely hard for me to understand this since there are some notions i've never heard of. It will take some time to understand this and accept this.. Thank you in advance! –  Katlus Aug 10 '12 at 16:37
    
@Katlus: That’s okay. If you have any specific questions about any of it, feel free to ask, and I’ll try to expand the explanation. –  Brian M. Scott Aug 10 '12 at 16:42
    
how do i write down 'cocatenating' in mathematical language? –  Katlus Aug 10 '12 at 17:12
    
@Katlus: The superscript is just another index. I could just as well have written $a_n=\{x_{n,1},\dots,x_{n,n}\rangle$. You could define $\sigma$ more formally by saying that $\sigma=\langle z_n:n\in\Bbb Z^+\rangle$, where $$x^n_k=z_{\binom{n}2+k}\;.$$ The correspondence $\langle n,k\rangle\leftrightarrow\binom{n}2+k$ really is a bijection between $\{\langle n,k\rangle:1\le k\le n\in\Bbb Z^+\}$ and $\Bbb Z^+$. –  Brian M. Scott Aug 10 '12 at 17:19
    
Now i got it. Construction is wonderful! Thank you :) –  Katlus Aug 10 '12 at 18:56
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Countable choice implies that a metric space is limit point compact if and only if it is compact.

From there it is not hard to deduce separability, as in ZFC.


$\newcommand{AC}{\mathsf{AC}}$

Let me point out some key points in a way for a proof of the above.

First we observe that under $\AC_\omega$ every infinite set has a countable subset. Let us see that it also implies that "every infinite set has an accumulation point" implies that every infinite sequence has a convergent subsequence:

Suppose that $\{x_n\mid n\in\omega\}$ is an infinite sequence. It has an accumulation point by the assumption. Namely a point $x$ such that for every open neighborhood $U$, there is some $x_n\neq x$ such that $x_n\in U$. Since we are in a metric space, without loss of generality this means that for every $k$ there is some $x_n\in B(x,\frac1k)$ and $x\neq x_n$.

Assume that $x\neq x_n$ for all $n$, just for simplicity, now it is clear that in every $B(x,\frac1k)$ there are infinitely many points from the sequence, otherwise there were only $m$ many points in $B(x,\frac1k)$ for some $k$, and we could then take $k'>k$ to be large enough so that none of these points appear in $B(x,\frac1{k'})$.

Since we have infinitely many, let $x_{n_k}$ be such that $n_k = \min\{n\mid x_n\in B(x,\frac1k)\}$. Note that if $k>j$ then $n_k\geq n_j$ since $x_{n_k}$ is closer to $x$ than $x_{n_j}$. If there are repetitions we can remove them by induction. We don't use the axiom of choice at all here, since this is already a well-ordered set.

Now it is not hard to see that $x_{n_k}\to x$, since for every $\varepsilon>0$ there exists $K$ such that for $k>K$ we have $x_{n_k}\in B(x,\varepsilon)$.

Conclusion I: $X$ is sequentially compact. Now let us show that it is compact.

The implication now is as in Brian's answer.

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isn't there a direct proof? As i mentioned in my post, i assumed 'limit point compact ⇒ separable' to prove 'limit point compact ⇒ compact', i cannot use 'limit point compact ⇒ compact' to prove this –  Katlus Aug 10 '12 at 14:03
    
You can actually use that. You proved that $\text{Limit point compact}\implies\text{separable}$ implies that $\text{Limit point compact}\implies\text{compact}$. In ZF you know that $\text{compact}+AC_\omega\implies\text{separable}$. You also know that $ZF+AC_\omega$ proves that $\text{Limit point compact}\iff\text{compact}$. This means that if you assume $AC_\omega$ then the two notions of compactness are in fact equivalent. This has nothing to do with what you have proven already. –  Asaf Karagila Aug 10 '12 at 15:00
    
I proved the first implication in ZF+AC$_\omega$. I don't know how to prove 'Limit point compact ⇒ compact' in ZF+AC$_\omega$. This is what i want to prove consequently. –  Katlus Aug 10 '12 at 15:21
    
Katlus, I will add a proof later tonight, I wrote the original answer from my phone and now I have to go. Please wait a few hours. (In the meantime Brian has answered your question) –  Asaf Karagila Aug 10 '12 at 15:39
    
Would you please write a proof later? I think understanding two different proofs would be really helpful to improve myself. Thank you –  Katlus Aug 10 '12 at 15:59
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Suppose this process never stops and you generate a sequence $\{x_n\}$ so $d(x_k, x_l) \ge \epsilon$ for all $k$ and $l$. Then such a sequence cannot have a convergent subsequence; it is a discrete subset of $X$. This contradicts the limit point compactness of $X$.

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How do i write down 'this process' in mathematical language? Isn't asumming existence of 'this process' a consequence of DC? –  Katlus Aug 10 '12 at 13:57
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