Winter 1995 #1

Problem Statement: Using only the definition of a Cauchy sequence show that every Cauchy sequence \{a_n\} in \mathbb{R} is bounded.

Proof: Let \{a_n\} be a Cauchy sequence in \mathbb{R}. Let \varepsilon >0, then there exists a N\in\mathbb{N} such that for every n,m\geq N it follows that |a_n-a_m|<\varepsilon.

Let M=max\{|a_1|,|a_2|,\dots,|a_N|,|a_N|+\varepsilon\}. I claim that latex M$ is a bound on \{a_n\}.

Let n\in\mathbb{N}. If n\leq N then |a_n|\leq M by construction. If n>N then it follows that |a_n-a_N|<\varepsilon since \{a_n\} is Cauchy. The reverse triangle inequality implies that |a_n|-|a_N|<\varepsilon, thus, |a_n|<|a_N|+\varepsilon\leq M by construction.

Therefore |a_n|\leq M for all n\in\mathbb{N} and thus it follows that any Cauchy sequence is bounded.


Reflection: This one wasn’t too bad. When I first approached it I wasn’t exactly sure how it was going to play out but I drew a sketch of how a Cauchy sequence behaves. Then I realized that there were finitely many points that could be acting differently than the rest of the sequence and this allowed me to come up with my bound. The main idea behind this is that a Cauchy sequence could be acting crazy at first, but it has to converge and so eventually things have to be close together, within \varepsilon to be precise.

This entry was posted in Analysis, Compact, Math, Sequence. Bookmark the permalink.

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