- #1

mattmns

- 1,128

- 6

Here is the exercise:

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Let [itex](X,d_{disc})[/itex] 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.

---------

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.

----------

Let [itex](X,d_{disc})[/itex] 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.

---------

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.

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