## 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 $x^{-0.99\dots 9}$ 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

$\displaystyle \int_{0}^{1} x^{-0.99999999999999999999999999999999999999999999}\,dx < \infty$

while

$\displaystyle \int_{0}^{1}x^{-1}dx = \infty$

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 "$\{x_{n}\}$ is Cauchy" is.  Also, it’s also a slick proof, so I couldn’t resist posting it.

Theorem: Let $X$ be a metric space with metric $d$.  Let $\{x_{n}\}$ be Cauchy.  Then $\{x_{n}\}$ is bounded.

Proof.  Pick any $\epsilon > 0$; say, $\epsilon = 1$.  Then by assumption, there exists some $N_{1}\in {\mathbb N}$ such that for all $n,m\geq N_{1}$ we have that $d(x_{n},x_{m}) < 1$.  Now let $k\in {\mathbb N}$ and $k \geq N_{1}$.  Note that $d(x_{k}, x_{m}) < 1$, so there are infinitely many terms of this sequence inside the ball of radius 1 centered at $x_{k}$; thus, these terms are bounded.

But now we only have to consider finitely many terms which are potentially not inside this ball, namely $\{x_{1}, x_{2}, \dots, x_{k-1}\}$.  But since there are only finitely many of these, there exists one which is a maximum distance away from $x_{k}$.  Let’s call this maximum distance $M$.  Then all of the points in the sequence lie inside the ball of radius $M + 1$ away from $x_{k}$.  Hence the sequence is bounded.  $\Box$

### 6 Responses to “Cute Proofs: Every Cauchy Sequence in a Metric Space is Bounded.”

1. thank you said

I was stuck on this proof for a while now using my book. Seeing the little picture and last paragraph it all cleared up. Thanks a lot!

2. Izuddeen Kabir said

Thanks for the briefness. it really helps!

thank you sir

4. Anonymous said

what is ball

5. Abhishek said

beautiful.the unit ball arguement is very nice and intuitive. Normally in other texts,people would break the Cauchy sequence in a finite part and infinite part. We can show that every finite sequence is bounded and for the infinite part we say it is smaller than a term + the boundedness criterion(say epsilon). Hence the whole sequence is bounded. Instead of that,we can use the unit ball arguement of yours.

6. Anonymous said

awsume proof