Question on proof of the Bolzano-Weierstrass theorem I was looking at the proof of the Bolzano-Weierstrass theorem and got a little lost (it's just one detail). The proof uses the nested intervals property and it talks about chopping the interval into halves over and over. My question is: Why does one half of the intervals contain infinitely many terms of the sequence and the other half contain finitely many? Is it from the monotonic subsequence theorem that this follows?
 A: To expand on what @DonAntonio said:
Assume there are only a finite number of sequence elements in one half of the interval. Then there must be an infinite number in the other half.
So in the context of the proof, you pick one half (doesn't matter which one). If it already has an infinite number of the sequence elements, move on to the next step. Otherwise, the other half has an infinite number of the sequence elements, so take that one instead. All we wanted to do is prove that at least one of the halves contains an infinite sub-sequence ... done.
Doing this step repeatedly lets you get infinite sub-sequences within smaller and smaller intervals. If, for example, we then choose an element from each of the intervals we obtain in this manner, we can derive an infinite sub-sequence that converges.
The statement you were taking in your question is actually not true (that one of them has to contain a finite number of sequence elements). Consider the bounded sequence: $$a_n=(-1)^n$$
When we perform our halving process the first time (starting with $[-1,1]$), both of the resulting intervals have an infinite number of sequence elements.
A: As Tim says your statement is not true.  
Here is an alternative proof. 
Let call $n\in \mathbb{N}$ "nice" if $a_n >a_m$ for all $m> n$. 
(1) Suppose that $(a_n)$ contain infinitely many "nice" points. In this case if $n_1<n_2<\ldots<n_i< \ldots$ are the "nice" points, we have $a_{n_1}>a_{n_2}> \ldots>a_{n_i}> \ldots$, then $(a_{n_i})$ is a decreasing subsequence.
(2) The sequence contains finitely many "nice" points. Let $n_1$ be greater that all the "nice" points. Since $n_1$ is not "nice" there is a  $n_2>n_1$ such that $a_{n_2}\ge a_{n_1}$. Since $n_2$ is also not "nice", is greater than $n_1$ and for instance greater than all the "nice" points, there is a point $n_3>n_2$ such that $a_{n_3}\ge a_{n_2}$. Continuous in this manner, we obtain a non-decreasing subsequence $(a_{n_i})$
Then if the sequence is bounded we obtain a monotonic subsequence of $a_n$ which is bounded and then by the monotonic convergence principle must converge.
