So I was brushing up on my Cauchy sequences the other day. Man what a fucking blast!

I was having a tough time trying to figure out how Cauchy sequences relate to convergent sequences. My conclusion is that they actually describe two different properties about sequences. The Cauchy criterion establishes a condition on how close terms must get to each other. Convergence establishes how close terms get to a destination.

My intuition tells me that in general, they are very much related. For example, convergent sequences are all Cauchy sequences. But sometimes you can’t go the other way around (check out *completeness*).

We'll use that we already have two Cauchy sequences of real numbers, \(\{a_n\}_{n \geq 1}\) and \(\{b_n\}_{n \geq 1}\).

## Problem 1

Show that each of \(\{a_n\}\) and \(\{b_n\}\) are bounded above and below

## Solution Intuition

As always, take some time to piece together WHY this is true. The way I like to do this is to start by trying to “break” it. In this case, I’ll look for a Cauchy sequence that isn’t bounded above.

My first try was the sequence \(0.9,1,1.99,2,2.999,3,...\)

I thought it was pretty clever because the terms, *consecutively*, get infinitely close together, and there is no upper bound to the sequence.

## Learning Opportunity:

Can you tell why my example isn’t actually a counter example?

Once you realise WHY my example doesn’t work, the solution to this problem becomes quite clear.

Terms is a Cauchy sequence are eventually ALL arbitrarily close. It’s not enough that some weird consecutive pairing happens to have terms which become arbitrarily near.

## Formalish Solution

Let \(\epsilon>0\) be randomly chosen by fairies, and let \(N \in \mathbb{N}\) be chosen such that \( |a_n-a_m|<\epsilon\) whenever \(n,m>N\).

All the \(a_n\) terms for \(n>N\) fall within an \(\epsilon\) stripe so they are obviously bounded. The terms before and including the \(N\)th term form a finite set, which admits both a maximum and a minimum.

Idea: set the upper bound to \(max\{a_1,a_2,...,a_{N+1}\}+\epsilon\) and the lower bound to \(min\{a_1,a_2,...,a_{N+1}\}-\epsilon\).

## Learning Opportunity

Why must we go to \(N+1\), and not just \(N\)? Provide an example of a sequence whose upper bound is NOT \(max\{a_1,a_2,...,a_{N}\}+\epsilon\)

## Learning Opportunity Answer

Only when you’ve had time to think about the question, and try it...

Try the sequence \(1,2,3,3.9,3.99,3.999,...\) and set \(\epsilon=0.1\). Then \(N=3\) works, with \(max\{1,2,3\}+\epsilon=3+0.1\) which is not an upper bound. Daaaaammmmmmnnnnn.