Welcome human.

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

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


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.

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