Cute Proofs: Every Cauchy Sequence in a Metric Space is Bounded.
February 25, 2011
For the seasoned mathematician, this is a relatively obvious statement. In a complete metric space, the sequence will even have a limit, and thus trivially be bounded by just taking a ball around the limit and noting only finitely many things lie outside. But when a student asked about this question and the question did not specify the space was complete, I thought that I might be able to construct a counterexample — you know, the points are close together but just far enough apart such that they go off to infinity.
This sort of thing happens all the time in mathematics: for example, the integral of from 0 to 1 is integrable so long as there are finitely many 9’s in the decimal expansion (you can actually prove this directly by integrating). This means that even something like
Even though we can make the former arbitrary close to the latter, the latter is just big enough that it does not have a finite integral.
Of course, this just doesn’t happen with Cauchy sequences metric spaces. After sketching a few pictures, I had an "oh, duh." moment and was able to sketch a quick proof. The proof of this theorem was actually an enlightening experience for me, since it demonstrated to me just how strict the criteria " is Cauchy" is. Also, it’s also a slick proof, so I couldn’t resist posting it.
Theorem: Let be a metric space with metric . Let be Cauchy. Then is bounded.
Proof. Pick any ; say, . Then by assumption, there exists some such that for all we have that . Now let and . Note that , so there are infinitely many terms of this sequence inside the ball of radius 1 centered at ; thus, these terms are bounded.
But now we only have to consider finitely many terms which are potentially not inside this ball, namely . But since there are only finitely many of these, there exists one which is a maximum distance away from . Let’s call this maximum distance . Then all of the points in the sequence lie inside the ball of radius away from . Hence the sequence is bounded.