Bolzano–Weierstrass theorem conclusion Can I conclude from Bolzano–Weierstrass theorem that there is more than one convergent subsequence, or the theorem tells me that there's only one ?
To be more clear, given a bounded sequence $X_n$, not ecessarily converges, can I conclude there are two different subsequences $X_{n_k}$ that converges to $L_1$ and $X_{n_l}$ that converges to $L_2$?
 A: This does not quite follow from BW, but we have the following:

Proposition: A sequence in $\Bbb R$ will converge if and only if all subsequences converge to the same limit.

So: for any non-convergent bounded sequence, you will be able to find subsequences with distinct limits.  For any convergent sequence, every subsequence will have the same limit as the original sequence.
A: The theorem states that, given a bounded sequence, one (or more) convergent subsequence/s exist/s.
Given a sequence that converges to $L \in \mathbb{R}$, all its subsequences converge to $L$.
Example:
$(-1)^n$ is our bounded sequence.
We can observe two convergent subsequences: $1^n$ and $-(1^n)$.
The first one converges to $1$, whereas the second one converges to $-1$.
Indeed, the bounded sequence is irregular (it doesn't converge nor diverge).
A: The Bolzano-Weierstrass the theorem says that every infinite, bounded sequence has a convergent sub-sequence.  I think the question here is whether such a sequence can have more than one such convergent subsequence, converging to a different limit.  The answer to that is clearly "yes".  Look at the sequence formed by interweaving two sequences converging to two different limits.  For example, the sequence 1, 1/2, 1/3, ..., 1/n converges to 0 while the sequence 2, 3/2, 4/3, ..., (n+1)/n converges to 0.  The sequence 1, 2, 1/2, 3/2, 1/3, 3/4, ..., alternating terms from the two sequences is a bounded sequence that has two convergent subsequence, one converging to 0, the other converging to 1.
A: An equivalent formulation of the BW theorem is that any bounded sequence has at least one accumulation point.
As noted in the answer of Omnomnomnm we also know that for a convergent sequence, all subsequences converge to the same limit.
So, given two convergent sequences $\{a_n\} \to a$ and $\{b_n\} \to b$ with $a \ne b$ , the sequece $\{c_n\}$ with 
$c_{2k}=a_k$ and $c_{2k+1}=b_k$ is bounded ( because the two starting sequeces are bounded) and has two different accumulation points. 
Genarizing this we can construct a bounded sequence with many accumulation points.
A: If you sequence $X_n$ is not convergent then you will indeed find at least two limits. Bolzano-Weierstrass ensures one, say $x$. As the sequence does not itself converge to $x$, there is an $\epsilon$, so that infinitely many elemets of the sequence are outside of $U_\epsilon(x)$. These elements make up new subsequence of $X_n$ which itself is bounded and by Bolzano-Weierstrass has a (sub-)limit which, now, cannot be $x$.
A: Just to add a counter-example: 
We know that we can enumerate all rationals between 0 and 1 in a sequence $x_1, \ldots, x_n, \ldots$
So $(x_n)$ is real and bounded and therefore BW.
But for every real $a \in (0,1)$ (rational or not), we can find a subsequence of $(x_n)$ that converges to $a$ (by a simple density argument).
A: All subsequences of a convergent sequence are also convergent. So if you found one, you found infinitely many.
