Discrete Metric Compact?

1. Feb 28, 2007

mattmns

Here is the exercise:
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Let $(X,d_{disc})$ be a metric space with the discrete metric.
(a) Show that X is always complete
(b) When is X compact, and when is X not compact? Prove your claim.
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Now (a) is pretty simple, but for (b) I am still not sure.

Here is our definition of compact: A metric space (X,d) is said to be compact iff every sequence in (X,d) has at least one convergent subsequence.

We also have the following Proposition: Let (X,d) be a compact metric space. Then (X,d) is both complete and bounded.

This tells us that we must have X bounded and complete (the latter we already have).

But I seem to be out of ideas. I almost feel as if the answer will be any X, or something of that nature (just from the way the exercise is written).

The entire exercise feels silly, as if I am just missing some silly detail that pulls the whole thing together. Any ideas? Thanks!

edit...

Since X needs to be bounded we know that there must be a ball B(x,r) in X which contains X. This would imply X is open (since X would be a ball), but that ... I am not sure. It just seems as though there is not much to go on here.

I was also maybe thinking about contradiction. Assuming that there is some sequence which has no convergent subsequence. But all of these feel strange without having any idea of what X is, or might be.

Last edited: Feb 28, 2007
2. Mar 1, 2007

dextercioby

What's the discrete metric ? How is it defined ?

3. Mar 1, 2007

gammamcc

What sort of sequences must X contain (or not contain) so that they have converging subsequences?

Last edited: Mar 1, 2007
4. Mar 1, 2007

matt grime

If you just write down what a convergent sequence is in the discrete metric you'll see the answer (and if not then look up one post ot gammamcc).

5. Mar 1, 2007

HallsofIvy

Staff Emeritus
Have you proved that, in the discrete metric, every set is an open set? In particular, every "singleton" set, every set containing exactly one ,is open. Given any set, X, the collection of all singleton sets, containing the points of X, is an open cover for X. Under what conditions does there exist a finite subcover?

Last edited: Mar 1, 2007
6. Mar 1, 2007

matt grime

That isn't the definition the OP has of compactness. I thought of posting it too, but they are using sequential compactness (which is of coiurse equivalent to the proper definition of compactness in a metric space).

7. Mar 1, 2007

mattmns

Thanks for the responses.

So X must contain a finite number of points to be compact.

The book has an exercise that says to prove that if every open cover of X has a finite subcover, then X is compact. Which I am currently messing around with.