# Does the proof of Bolzano-Weierstrass theorem require axiom of choice?

When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me.
[edited]
I'm sorry for the lack of explanation. I want to prove this statement:
Let $a_1, a_2, \ldots \in \mathbf{R}$, and $(a_n)_{n\in\mathbf{N}}$ is bounded, then $(a_n)$ has some convergent subsequence.

The proof is as follows. Since $(a_n)$ is bounded, for all $n\in\mathbf{N}$, $a_n \in I = [b, c]$.
Now, let $I_0 = I$ and if $I_n = [b_n, c_n]$, we define $d_n = (b_n+c_n)/2$ and if infinite terms of $(a_n)$ is included in $[b_n, d_n]$(resp. $[d_n, c_n]$), we will define $I_{n+1} = [b_n, d_n]$(resp. $[d_n, c_n]$).If both intervals contain infinite terms, let $I_{n+1}$ be $[d_n, c_n]$.
For all $n\in \mathbf{N}$, infinite numbers of $m \in \mathbf{N}$ exist such that $a_m \in I_n$ suffices. We take the sequence of natural numbers $(n(k))_{k\in\mathbf{N}}$which suffices $n(0) < n(1) < \cdots < n(k) < \cdots$ following this procedure:
Now we have already selected $a_{n(1)}, \ldots, a_{n(k)}$, there are infinite numbers of $m\in \mathbf{N}$ which suffices $n(k)<m, a_m \in I_{k+1}$, so let's take the minimum m out of it. Applying this process recursively, we obtain a infinite convergent subsequence(?).
I think intuitively, by only repeating this process we can't obtain countable infinite terms of subsequence because we have to repeat infinite times.

• I am not the downvoter (yet). But please state your problem more carefully by giving an outline of the proof of B-W you are using. (At your mathematical level, most users on this site will expect you to be much more precise and detailed.) – Simon S May 11 '15 at 12:49

• That's why in my proof the choice of $I_k$ is indicated precisely so that the way of selecting $a_{n(k)}$ is specified uniquely. Thank you for your kind answer considering my level of understanding the foundation of mathematics. I have a question. Why we have to use the special weapon, axiom of choice, when the way of choice is not unique? I can't comprehend the importance. – dazaga May 11 '15 at 16:10
• I have another question. In the example of Bertrand Russel, the rule of choice("left from each pair") makes it possible to select infinite left shoes 'at once'. In contrast, my proof requires to decide $a_{n(k)}$ sequentially(for k=1, 2, \ldots) I think this might be a big difference, but I'm not sure. – dazaga May 11 '15 at 16:28