# Every limit point is the limit of a sequence

Assume we have a metric space $X$, a subset $E\subseteq X$, and a limit point $p$ of $E$.

Proofwiki and Rudin both "construct" a sequence that converges to $p$ using the fact that every neighborhood of radius $\frac1n$ ($n = 1, 2, 3, ...$) is nonempty because $p$ is a limit point, and picking an arbitrary point in $N_{\frac1n}(p)$ as $s_n$.

What I'm not sure about is that since there's infinitely many points in each neighborhood, is whether they've really proven a sequence exists that converges to $p$ exists. I understand you can keep getting points closer to $p$, we have pleanty of points to make a sequence out of.

Is that enough, or am I right in feeling like the Axiom of (countable, right?) Choice has been quietly used to actually say any such sequence exists?

You can use a similar condition that you let D$_1$ = 1, and then you know there is an element of E, x$_1$, distinct from p so d(x$_1$,p) < 1. Now you can choose D$_2$ = Min(d(x$_1$,p),1/2). Again, you know that since every nbd of p contains a point in E distinct from p, there is a point x$_2$ (distinct from x$_1$) in E, such that d(x$_2$,p)< D$_2$.
If you keep doing this, you will get a sequence of elements from E, (x$_n$) that converge to p since for every $\epsilon$ > 0, choose N to be a natural number such that N > 1/$\epsilon$ which guarantees that for all n>N, you will have d(x$_n$,p) < D$_n$ and we know by construction D$_n$ $\leq$ 1/n < 1/N < $\epsilon$.
• The axiom of choice is really not a necessity here, as the defn of limit point guarantees that for every nbd of p, a limit point of E, there exists a point x in E, distinct from p, in the nbd. You are using more the properties of open sets than AC. To answer your other question, yes you do get a sequence of points that converges to p. The definition of "limit(x$_n$) = p" says that for every $\epsilon$ >0, there exists a natural number N such that for all n>N, d(x$_n$,p) < $\epsilon$. If you go back and look at our construction you will see that the sequence (x$_n$) satisfies this condition. – Matt Brenneman Oct 14 '13 at 2:49