Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The title is a claim my classmate made during our summer vacation :D

He showed me a TeX file describing a proof of his claim, and it contains a fairly short but elegant proof. He says that the motivation was to see if the converse of the statement '$(X, d)$ is compact $\Rightarrow$ All metrics homeomorphic to $X$ is bounded.

I was surprised by his work, and wanted to know if it's already discovered and/or published.

So the question is,

  1. Is the claim my classmate made is already discovered? If it is, any references?

  2. How about more general one? like, can we remove/reduce the condition of the claim more weaker?

  3. Are there any underlying theorems or theories in it?

  4. If it is not yet discovered, where should I recommend him to publish his work? (LOL) I think it is worth sharing the proof.

share|cite|improve this question
@johnny: The open disk is homeomorphic to the unbounded metric space $\mathbb{R}^2$. – Nate Eldredge Aug 21 '11 at 14:30
@John M but the usual metric on it is unbounded. – Olivier Bégassat Aug 21 '11 at 15:12
I'm not sure what you're looking for, but one characterization of compact metric spaces is: A metrizable space is compact if and only if it is complete and totally bounded with respect to some metric inducing the topology. If that is the case then all compatible metrics are complete and totally bounded. – t.b. Aug 21 '11 at 15:31
In Joshi, Introduction to General Topology, pg. 287, Exercise 1.13: "Let X be a metrisable space. Prove X is compact iff every metric which induces the given topology is bounded." Book gives hints on how to solve. – John M Aug 21 '11 at 15:59
@Olivier: If $(X,d)$ is non-compact, it contains a closed subspace $Y$ homeomorphic to $\mathbb{N}$ (take a sequence $(y_n)$ without convergent subsequence). Put $f(y_n) = n$. Use Tietze to extend $f:Y \to \mathbb{R}$ to a continuous function $F: X \to \mathbb{R}$. Check that $d$ and $d_{F} (x,y) = d(x,y) + |F(x) - F(y)|$ have the same convergent sequences, hence the metric spaces $(X,d)$ and $(X,d_F)$ have the same closed sets. Here's the link to Joshi (I can't see it, but I guess it must be a similar argument). – t.b. Aug 21 '11 at 17:18
up vote 17 down vote accepted

Since this is a nice result and people asked how to prove this in the comments, let me provide a proof of a somewhat stronger result.

Note: A locally compact space is metrizable if and only if it is second countable (hence separable) by the Urysohn metrization theorem. However, there are compact separable non-metrizable spaces (e.g. the Stone-Čech compactification $\beta\mathbb N$ of $\mathbb{N}$).

A metrizable space $X$ is compact if and only if every compatible metric is bounded.

If $X$ is compact then every compatible metric is bounded since it is a continuous function on the compact space $X \times X$.

If $X$ is non-compact, then it contains a closed subspace $Y$ homeomorphic to $\mathbb{N}$ (choose a sequence of points without convergent subsequence, hence without accumulation point). Choose an enumeration $\{y_n\}_{n \in \mathbb{N}}$ of $Y$ and define $f: Y \to \mathbb{R}$ by $f(y_n) = n$. Then $f$ is continuous. Use the Tietze extension theorem to extend $f$ to a continuous function $F: X \to \mathbb{R}$. Given any metric $d$ on $X$ compatible with the topology, the metric $d_F(x,y) = d(x,y) + |F(x)-F(y)|$ is compatible as well.

Why? A $d_{F}$-convergent sequence is also $d$-convergent since $d \leq d_F$. On the other hand, a $d$-convergent sequence converges with respect to $d_F$, as $d_F$ is continuous with respect to $d$. This means that $(X,d)$ and $(X,d_F)$ have the same closed sets, hence the metrics $d$ and $d_F$ induce the same topology. Since $d_F(y_n, y_0) = d(y_n,y_0) + n \geq n$ the metric $d_F$ is unbounded.

I don't see any easier way how to prove this. One can make this argument a little more explicit by replacing $F$ by an explicit function depending on the metric, but little is gained except that the proof becomes a bit more explicit and basic.

Because Google doesn't like me, I can't see the relevant page 287 of John M.'s reference to Exercise 1.13 in K.D. Joshi, Introduction to general topology, New Age International, 1983, but as Pierre-Yves kindly transcribed the exercises in the comments, I see no need of reproducing them here.

You asked about applications. Here I must say I can't think of anything interesting (I can't come up with any example of a space where I can check that it has the property that every compatible metric is bounded without proving that it is compact before).

Maybe it is worth mentioning the important property of total boundedness. A metric space is compact if and only if it is complete and totally bounded. This can be used for many compactness results involving function spaces, e.g. the all-important Arzelà-Ascoli theorem.

Finally, let me commend your friend for finding this. It is a beautiful observation, and it seems to have impressed several people (including me). It is an extremely important reflex to ask yourself what about the converse? Having the perseverance to answer that question nicely and convincingly certainly is a sign of talent. So: no need to feel bad that it was already known! It is impressive if somebody can come up with such a result with only little background in topology.

Concerning publication, I would show such work to one of your professors who could maybe point you to a college math journal and, more importantly in my opinion, could be interested in supporting your friend further by guiding and advising him. You might also want to look at amWhy's answer here for some related thoughts and links that might be interesting and helpful.

share|cite|improve this answer
Could somebody please confirm that the Google Books link to Joshi's book works and points to the correct page? I can't see it. – t.b. Aug 21 '11 at 19:52
Yes, the Google Books link works great here (and is to the correct page). – Jonas Meyer Aug 21 '11 at 20:00
Great, thanks! ${}$ – t.b. Aug 21 '11 at 20:01
Thanks for all who considered my question! – Jineon Baek Aug 23 '11 at 5:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.