TITLE: How can we identify a set is infinite when proving Bolzano–Weierstrass theorem?
QUESTION [7 upvotes]: I'm proving the Bolzano–Weierstrass theorem. It says
$$
\text{If } A \text{ : bounded and infinite set}, \text{ then } A \text{ has at least one limit point.}
$$
So the proof goes like the following.
Since $A \subset \mathbb{R}^N$ is bounded, it can be a subset of a box $B_1  = I_1 \times \cdots \times I_N$, where $I_i$ is an interval in $\mathbb{R}$. Let's divide the $B_1$ into $2^N$ sub-boxes. Then, at least one sub-box has the infinite elements of $A$. Let's say that sub-box as $B_2$. Let's say we can repeat this procedure for $B_3, B_4, \ldots$
Well, so proof ends as we prove some $x$ exists in $\cap_{n=1}^{\infty}B_n$ and it is a limit point of $A$. 
But what I'm wondering is, how can we identify the $B_2$? (and $B_3$, and so on)
Yes, it's clear to me that such set exists, but the existence itself does not guarantee us to identify the set (that is, to select and label it as $B_2$), since we don't have any tools to discern if $B_2$ is infinite or not.
It sounds a little bit philosophical, but is there anyone to help?

REPLY [7 votes]: It seems that you understand why at least one box contains infinitely many elements of $A$, and your difficulty is only in finding which box that is (or finding one such box if there are several of them). That is, it's a question of existence versus finding. Notice, though, that the theorem you're proving says only that there exists a limit point.  It doesn't claim anything about finding a limit point.
To prove the existence of a limit point, it's enough to have the existence of an appropriate box at every stage of the process.  
If you wanted to prove a stronger theorem saying that one can find a limit point, then you'd have to worry about finding an appropriate box at every stage. Any alleged theorem of that sort would have to be preceded by information about (1) exactly how the set $A$ is presented and (2) exactly what it means to "find" a point.  So such a theorem would not only be harder to prove but considerably more complicated to state than the Bolzano-Weierstrass theorem.