Can someone help me understand the proof that every cauchy sequence is bounded?

Question

This proof is written by a user Batman as an answer to someone's question(just to give credit). Every proof that I've seen is the same idea, and I'm having trouble understanding it intuitively. (I don't see why to take n=N and why the max is used.) If someone could explain it in words it would be really helpful.

Choose ϵ>0. Then, there exists N such that for m,n ≥ N, |am−an|<ϵ. By the triangle inequality, |am|−|an|≤|am−an|<ϵ. Take n=N and we see |am|−|aN|<ϵ for all m≥N.

Rearranging, we have |am|<ϵ+|aN| for all m≥N. Thus, |am|≤max{|a0|,|a1|,…,|aN1|,ϵ+|aN|} for all m. Thus, am is bounded (it is sandwiched in ±max{|a0|,|a1|,…,|aN1|,ϵ+|aN|}).

Answer 1

There is a tail of the sequence such that the terms in the tail are close enough to a certain term in the sequence. Now we just have to take care of the terms in the sequence that are not in that tail, and this is simple because there are only finitely many such terms.

Answer 2

Intuitively the terms in a Cauchy sequence get arbitrarily close to themselves. This means that if you go out far enough into the sequence, all of the terms should be all bunched up together.

In particular, consider $\epsilon=1>0$. After some point, everything in the sequence has to be no further that $1$ unit apart.

Let's say this happen at $a_N$. Then after $N$, everybody is no bigger than $a_N+1$ and no smaller than $a_N-1$.

This means that the tail end of the sequence all lies below $a_N+1$. So $a_N,a_{N+1},\dots$ are bounded by $a_N+1$.

This just leaves the finite beginning of the sequence to worry about.

But that's a finite problem. Among $a_1,\dots,a_{N-1}$ we could just take the maximum (we can take the maximum of a finite list).

So to bound everyone (before and after $a_N$) we can take the maximum of $a_1,\dots,a_{N-1},$ and $a_N+1$.

Answer 3

Just to make things simpler, let's take $\epsilon=1$. Since the sequence is Cauchy, there is an integer N such that $m,n\ge N\implies|a_m-a_n|<1$. In particular, taking $n=N$, we have that $|a_m-a_N|<1$ for $m\ge N$;

so $|a_m|=|(a_m-a_N)+a_N|\le |a_m-a_N|+|a_N|<1+|a_N|$ for $m\ge N$ using the Triangle inequality.

Now we have a bound for all but finitely many of the terms of the sequence, but we still have to account for the terms $a_1,a_2,\cdots,a_{N-1}$.

If we take $M=\max\{|a_1|,\cdots,|a_{N-1}|, 1+|a_N|\}$, then $|a_n|\le M$ for all n

since $|a_n|\le M$ if $m\ge N$ by the first part, and $|a_n|\le M$ for $1\le m\le N-1$ by the definition of M.