The proof I know for the Bolzano-Weierstrass Theorem uses Nested Intervals Theorem, for which we can find a infinitely small interval where infinitely many terms can be in that interval.

Now, the book asks me to prove that every sequence $x_n$ of real numbers must have a monotone subsequence. The book says after I prove this, I can discover an easy alternative proof for the Bolzano-Weierstrass Theorem but I can't understand how this is so.

Is it because for bounded sequence, we can find a monotone subsequence that converges to the least upper bound or greatest lower bound? And this is a convergent subsequence?

  • $\begingroup$ Then you have to use "A bounded monotone sequence is convergent" to prove Bolzano-Weierstrass. $\endgroup$ – Sungjin Kim Jun 3 '17 at 2:12

You cannot always find a sequence that converges to the least upper bound or greatest lower bound. For example, consider the sequence $a_n = (-1)^n n^{-1}$. Then $\sup \{a_n\} = 1/2$ and $\inf\{a_n\} = -1$ but $a_n \to 0$ so there are no subsequences that converge to anything but to $0$.

On the other hand, if you know that every bounded sequence has a monotone subsequence then the Monotone Convergence Theorem (every bounded monotone sequence converges) gives you what you want.

In fact, you can see that if $b_n$ is monotone increasing then $b_n \to \sup\{b_n\}$ (which is finite iff $\{b_n\}$ is bounded) and if $b_n$ is monotone decreasing then $b_n \to \inf\{b_n\}$.

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  • $\begingroup$ The sequence you just mentioned is bounded, so we can find monotone subsequence that is convergent. Actually, we can find two monotone subsequences where one is increasing and other is decreasing. Namely, if we take the smallests of the $T_n$ where $T_n$ is a $n$th tail of the sequence for all $n\in\mathbb{N}$. $\endgroup$ – user3000482 Jun 3 '17 at 2:17
  • $\begingroup$ Is this true? I am thinking that every bounded sequence must have two monotone subsequences. $\endgroup$ – user3000482 Jun 3 '17 at 2:18
  • $\begingroup$ @user3000482 Yes that's correct, the even and odd terms are respectively decreasing and increasing. They both converge to $0$ (their respective infimum and supremum). My point was is that the infimum and supremum of the original sequence may not be reachable by a subsequence. This is what I had assumed you meant by the least upper bound or greatest lower bound. If that's not what you meant than your reasoning is 100% correct. $\endgroup$ – Trevor Gunn Jun 3 '17 at 2:21
  • $\begingroup$ @user3000482 But it is not true that there are two monotone subsequences. For instance, a monotone increasing sequence only has monotone increasing subsequences, for obvious reasons. $\endgroup$ – Trevor Gunn Jun 3 '17 at 2:22

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