I just came across the Bolzano-Weierstrass Theorem (every bounded sequence has a converging subsequence), it was stated somewhere without proof. So curious as I am, I looked for a proof in the Internet.

The proofs I found consist of 2 Lemmas: (1) Every sequence has a monotonic subsequence, and (2) Every bounded monotonic sequence converges.

I think I understand (1): a sequence either has a finite or infinite amount of peaks (all values after a peak are less than the peak). If there is an infinite amount of peaks, then the peaks form a monotonic decreasing sequence. If there is only a finite amount of peaks (so also 0 peaks would be valid), then a value $n_1$ after the last peak is less than the last peak (by definition -- and in case of 0 peaks, $n_1$ is the first value). So there must be a $n_2$ somewhere after $n_1$ which is greater, otherwise $n_1$ would be a peak. Then there should be a $n_3$ greater than $n_2$, otherwise $n_2$ would be a peak. Repeat, and find a monotonic increasing sequence.

Now 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?

Exponential function, bounded

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?

By the way, my background is Mechanical Engineering, so I'm not that used to reading and understanding proofs.

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    $\begingroup$ Welcome to math.stackexchange! Could you just clarify - are you looking for rigorous proofs, or more a sort of intutition for these facts? Your post suggests you have already seen numerous proofs, so I think it is the latter but please make this more clear. $\endgroup$ – Ragib Zaman Oct 22 '11 at 12:25
  • $\begingroup$ Thanks! An intuitive proof would be very nice, but otherwise a rigorous proof is fine too, as long as I will be able to understand this Theorem. By the way, in some proofs they talk about infinite sequences, is the BZ-Theorem only valid for infinite sequences? $\endgroup$ – Ailurus Oct 22 '11 at 12:31
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    $\begingroup$ @Ailurus, limits and convergence are always about infinite sequences. (If you really want to, you can concoct a generalization to finite sequences, but they become completely uninteresting concepts in that case -- every finite sequence converges, and its limit is exactly its last element. The ones in front of it never makes any difference). $\endgroup$ – hmakholm left over Monica Oct 22 '11 at 13:01

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).

  • $\begingroup$ Ah the "same interval on the $y$-axis" makes it clear, thanks! As for the sequence, there are no peaks (or if you define it a little different, every $n$ for which $sin(n)=1$ could be a peak). In this case, the subsequence consists of only ones, which is monotonic, and it converges to 1. If there are no peaks, I'm not sure where this monotonic subsequence would be. $\endgroup$ – Ailurus Oct 22 '11 at 13:36
  • $\begingroup$ Wait, I think I see my mistake. I interpreted the sequence $sin(n)$ as a continuous function, but of course only discrete values (infinitely many of them) of this function are part of the sequence, so $f:\mathbb{N}\rightarrow\mathbb{R}$. Since no value repeats, the only peak is at $n=0$ where $sin(n)=1$, right. $\endgroup$ – Ailurus Oct 22 '11 at 13:44
  • $\begingroup$ Right -- almost, since $\sin(0)=0$ rather than $1$, so there are no peaks at all. Your construction will then produce an increasing sequence that converges towards $1$. $\endgroup$ – hmakholm left over Monica Oct 22 '11 at 13:52
  • $\begingroup$ Oops, that was foolish of me. Ok, so to summarize: when there are infinitely many peaks, the monotonic sequence is a decreasing one. Its limit is unknown (could be the lower bound of the main sequence, doesn't has to be). If there are finitely many peaks, then the monotonic function is an increasing one. Its limit is the value of the last peak. Special case, 0 peaks. You can take a random value in the sequence, and find a higher value, continue this way. The limit of this monotonic increasing sequence is the upper bound of the main sequence. $\endgroup$ – Ailurus Oct 22 '11 at 14:11
  • $\begingroup$ For finitely many peaks, the limit can be smaller than the last peak value -- such as for example $a_0=2$ and $a_n=\sin(n)$ for $n\ge 1$. In the no-peak case, the limit is the least upper bound of the sequence (every larger value also qualifies formally as an upper bound). Except for that, I think you got it. $\endgroup$ – hmakholm left over Monica Oct 22 '11 at 15:24

If $l$ is the supremum of your sequence $(a_n)$ then, by the very definition of the supremum we have the following:

For every $\varepsilon >0$ there is some $n$ such that $a_n\geq l-\varepsilon$.

But since your sequence is monotonically increasing, you have $a_m \geq l-\varepsilon$ for every $m\geq n$. This is exactly what it means for the sequence $(a_n)$ to converge to $l$.

  • $\begingroup$ Ok, that's clear. But what about my graph, which stops after n = 15 terms, $a_{15} \approx 4$. Do you then say that this monotonic bounded sequence converges to $4$? $\endgroup$ – Ailurus Oct 22 '11 at 12:53
  • $\begingroup$ @Ailurus: please see Henning Makholm's answer. $\endgroup$ – Rasmus Oct 22 '11 at 13:21

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