# Does dense subset in a polish space contain a countable subset which is dense in the space?

Let $X$ be a separable complete metric space. Let $E$ be a dense subset of $X$ and fix $p\in X$.

I want to 'choose' an element for each $\overline{B(p,1/n)}\cap E$ in ZF. ($n\in \mathbb{Z}^+$)

Is it possible?

As you can see in the link; Constructing a choice function in a complete & separable metric space

It's possible to choose representatives for $\overline{B(p,1/n)}$ for each $n\in \mathbb{Z}^+$.

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I assume that $E$ is a countable dense subset, in which case it is clear that you can do that.

Write $E=\{e_n\mid n\in\omega\}$ and let $e\in\overline{B(p,\frac1n)}\cap E$ be $e_n$ such that $n=\min\{k\mid e_k\in B(p,\frac1n)\}$. By the density of $E$ you know that $E\cap B(p,\frac1n)$ is non-empty.

If $E$ is not well-ordered then this need not be possible. Suppose that $D$ is a Dedekind-finite dense set of real numbers.

We want to find $p_n\in D$ such that $|p-p_n|\leq\frac1n$. Because $p\notin D$ for every $x\in D$ you have some $n$ such that $|p-x|>\frac1n$, so you cannot choose the same $p_n$ more than a finite number of times.

Therefore such sequence would give us a sequence of elements from $D$, which is a contradiction to Dedekind-finiteness.

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I was asking about an arbitrary dense subset! –  Katlus Nov 12 '12 at 11:45
@Katlus: There you go... –  Asaf Karagila Nov 12 '12 at 11:55
Best. Thank you –  Katlus Nov 12 '12 at 11:59