I am self-studying Real Analysis with S. Abbott's Understanding Analysis. In that book, there is a question (exercise 2.6.7 b) for the curious) that I am stumped on.
Use the Cauchy Criterion to prove the Bolzano–Weierstrass Theorem, and
find the point in the argument where the Archimedean Property is implicitly
required.
I have a good intuition as to why the Cauchy Criterion implies this, but my "proof" is not nearly as rigorous as I would like it to be.
My "proof" is by contradiction.
Proof:
Given a bounded sequence $a_n$, we can prove that it must contain at least one Cauchy subsequence. Since Cauchy = Convergent, this proves the Bolzano-Weirestrass theorem.
Assume that there exists no subsequence $a_{n_k}$ that is Cauchy ($|x_n – x_m| < \epsilon, \epsilon > 0, m, n \ge N$). In terms of the sequence and not subsequences, that means there exists no infinite "chain" of terms in $a_n$ where each the terms in this chain get closer and closer together (this chain is the cauchy subsequence).
Since there exists no infinite chain, we can find some point N in the sequence after which all terms get farther and farther apart, infinitely (since if it was finite, we could just move N all the way to when the increasing of distance between terms stopped).
The interval between terms can be getting larger at a very small rate, but it must be increasing in size (for example, you could have the interval between terms not grow in size for say, a million terms, as long as eventually you increase the interval by some amount, say 0.0001, like $\{ 5, 5, 5, 5, ….. 5, 5, 5.0001, 5.0001…..\}$) Therefore, the size of the interval between two terms will eventually outgrow any bound around the original sequence itself. This contradicts the fact that all terms must be within the bounds.
That's my intuition, but I have no idea how to prove it. Does anyone have any hints as to how I can mathematically formulate this?
Thanks,
A
Best Answer
(More or less what you said.) Assuming no Cauchy subsequence of $a_n$, there exists $\epsilon>0$ such that $|a_i-a_j|>\epsilon$ for all $i\neq j$. Without loss of generality, we can assume that $a_n$ is increasing (there is either an infinite increasing subsequence or an infinite decreasing subsequence) so that $a_{n+1}-a_n>\epsilon$. The Archimedean property implies that $a_n$ is unbounded.