# Cluster point of a sequence $\{x_n\}$ is the limit of some subsequence - Axiom of Choice? [duplicate]

This question is an exact duplicate of:

In a metric space, a cluster point of a sequence $\{x_n\}$ is the limit of some subsequence. The only proof that I know works like this:

Construct a sequence $\delta _k \to 0$. For each $\delta _k$ find a point $x_{n_k}$ in the sequence which is within $\delta_k$ of the cluster point. Make sure that $n_{k} \lt n_{k+1}$, so that you really have a subsequence, and you can do this because there are only finitely many terms before index $n_{k}$.

Question: Is this using Countable Choice? If so is it because we're "picking" a sequence element for each $\delta_k$? If we are using Countable Choice, then is there a way to prove this without using CC?

I apologize if this is a duplicate question - I know there have been previous posts where it's explained why Sequential Continuity Implies $\delta - \epsilon$ Continuity requires Countable Choice. The argument used in that proof seems similar to this one.

Thank you very much.

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## marked as duplicate by Asaf Karagila, Julian Kuelshammer, azimut, Lord_Farin, O.L.Jul 5 '13 at 7:57

Suppose that $p$ is a cluster point of $\sigma=\langle x_n:n\in\Bbb N\rangle$. For each $k\in\Bbb N$ let
$$n_k=\min\{n\in\Bbb N:d(p,x_n)<2^{-n}\text{ and }\forall i<k(n_i<n)\}\;;\tag{1}$$
then $\langle x_{n_k}:k\in\Bbb N\rangle$ is a subsequence of $\sigma$ converging to $p$. The definition in $(1)$ does not require choice: it uniquely specifies $n_k$, and the hypothesis ensures that the set over which the minimum is taken is non-empty.
Yes. Also, note that rather than having to pick the $\delta_k\to0$, you can just work with the $1/2^n$. (So everything is rather explicit.) –  Andres Caicedo Jul 5 '13 at 5:16