The proofs I found of the Bolzano Weierstrass Theorem consist of 2 Lemmas:
Every sequence has a monotonic subsequence
Every bounded monotonic sequence converges.
I am confused about (2). Assume the sequence ($a_n$) is monotonic increasing, for the decreasing case it is similar. Since the sequence is bounded, there is a least upper bound, $l = \sup(a_n)$. They say this $l$ is the value the sequence converges to, but why…?
For example, if the sequence looks like a exponential function, it is monotonic increasing. Just cut off this function somewhere, and it is a bounded monotonic sequence, right? So then the least upper bound $l$, in this case the maximum value, should be the limit of the sequence?
Since the main sequence in the Bolzano-Weierstrass is bounded, it has a monotonic subsequence according to (1), and is bounded as well? Then this subsequence converges.
Finally, I read another proof where the "Nested Interval Theorem" is used. It was about dividing the interval until there is only one point (?), so this point would be the limit of the sequence?
Best Answer
It looks like you're quite on top of the first lemma you describe. Please clarify if you need any help with that.
For the second one, I think your confusion is caused by you misunderstanding "bounded", or at least getting the wrong mental picture when you see it. A bounded sequence still contains infinitely many points -- that it is bounded means that all of the points lie within the same interval on the $y$-axis.
One example is the sequence $a_n = \sin(n)$ for $0\le n\lt \infty$. Here $n$ can be as large as you please, but every $a_n$ is between $-1$ and $1$, which makes the sequence bounded.
Therefore, Bolzano-Weierstrass says that $(\sin(n))_n$ contains a convergent subsequence. It is not easy to write down one explicitly, though. (Since we're mathematicians here, $\sin(n)$ treats $n$ as an angle in radians, so there's no value in the sequence that repeats exactly).