# Proving completeness of a metri space X

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## Homework Statement

I feel there's something wrong with my solution, so I'd like to check it tout.

Let X be a metric space and suppose that there exists some ε > 0 such that every ε-ball in X has a compact closure. Show that X is complete.

(btw, it would be enough, in the formulation of the problem to assume that every ε-ball in X is compact, since X is Hausdorff, and compact subspaces are closed, right?)

## The Attempt at a Solution

So, let xn be a Cauchy sequence in X. Choose ε > 0 such that every ε-ball has a compact closure. Choose N such that for any n, m >= N, we have |xn - xm| < ε. Hence, there exists an ε-ball containing all but finitely many members of the sequence xn. For this finite number of members of the sequence, choose for each an ε- ball containing it. Now, a finite union of such compact ε-balls is compact, and hence sequentially compact. So, xn has a convergent subsequence, so X is complete. (actually, more precisely, the union of the ε-balls is complete, but since they are a subspace of X, and we have chosen a Cauchy sequence in X, we arrived at a convergent subsequence in X again, hence I concluded that X is complete=

Thanks in advance, but for some reason, after dealing with more abstract issues, I feel a but unsure about metric spaces.

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Fredrik
Staff Emeritus
Gold Member
Let X be a metric space and suppose that there exists some ε > 0 such that every ε-ball in X has a compact closure. Show that X is complete.

(btw, it would be enough, in the formulation of the problem to assume that every ε-ball in X is compact, since X is Hausdorff, and compact subspaces are closed, right?)
This would be a stronger assumption, since "is compact" implies "has compact closure".

Choose N such that for any n, m >= N, we have |xn - xm| < ε. Hence, there exists an ε-ball containing all but finitely many members of the sequence xn.
I think you need to choose |xn - xm| < ε/2, and use the triangle inequality to prove that your conclusion holds.

For this finite number of members of the sequence, choose for each an ε- ball containing it. Now, a finite union of such compact ε-balls is compact, and hence sequentially compact. So, xn has a convergent subsequence, so X is complete. (actually, more precisely, the union of the ε-balls is complete, but since they are a subspace of X, and we have chosen a Cauchy sequence in X, we arrived at a convergent subsequence in X again, hence I concluded that X is complete=
I would add a few words to explain why "has a convergent subsequence" and "is a Cauchy sequence" implies "is convergent". Also, you didn't prove that the union of the ε-balls is complete. You proved that the union of their closures is.

So your proof contains a couple of inaccurate (but easily fixed) statements, and a couple that should be clarified, but as far as I can tell it will be a correct proof when you have fixed those issues.

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Fredrik, thanks for the reply. I'll go step by step.

This would be a stronger assumption, since "is compact" implies "has compact closure".
Of course, it would be stronger, but it still implies that the closure would need to be compact, since a compact subspace of a Hausdorff space is necessarily closed.

I think you need to choose |xn - xm| < ε/2, and use the triangle inequality to prove that your conclusion holds.
OK, I'm interested if this statement holds:

If xn is a Cauchy sequence, choose N such that whenever m, n >= N, |xm - xn| < ε holds. Now, choose an ε-ball around xN. Then all but finitely many members of the sequence lie in that ε-ball, right?

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So, if this is correct, then for every member of the sequence (the number of such members is finite) which doesn't lie is the ball B(xN, ε), choose an ε-ball containing it. The union of the closures of these ε-balls, along with B(xN, ε), is compact. By Theorem 28.1. (Munkres), compactness of this set implies its sequential compactness, hence our sequence xn has a convergent subsequence. By Lemma 43.1., X is complete.

Fredrik
Staff Emeritus
Gold Member
Of course, it would be stronger, but it still implies that the closure would need to be compact, since a compact subspace of a Hausdorff space is necessarily closed.
Now that I think about it (more than before), what I should have said is that the alternative assumption doesn't really make sense. An ε-ball is an open ball with radius ε, right? So we can't assume that it's compact, because that would imply that it's also closed. A non-empty proper subset of a metric space can be both open and closed, but only if the space is disconnected. For example, if our space is the union of the open balls with radius 1 around (1,0) and (-1,0) in ℝ2 with the metric inherited from the standard metric on ℝ2, then those two open balls would be both open and closed. However, I don't know a way to make sense of the assumption that all open balls (or even all ε-balls with a specific value of ε) are compact.

OK, I'm interested if this statement holds:

If xn is a Cauchy sequence, choose N such that whenever m, n >= N, |xm - xn| < ε holds. Now, choose an ε-ball around xN. Then all but finitely many members of the sequence lie in that ε-ball, right?
Ah, yes, that was a mistake on my part.

So, if this is correct, then for every member of the sequence (the number of such members is finite) which doesn't lie is the ball B(xN, ε), choose an ε-ball containing it. The union of the closures of these ε-balls, along with B(xN, ε), is compact. By Theorem 28.1. (Munkres), compactness of this set implies its sequential compactness, hence our sequence xn has a convergent subsequence. By Lemma 43.1., X is complete.
Looks good.

Homework Helper
Now that I think about it (more than before), what I should have said is that the alternative assumption doesn't really make sense. An ε-ball is an open ball with radius ε, right? So we can't assume that it's compact, because that would imply that it's also closed. A non-empty proper subset of a metric space can be both open and closed, but only if the space is disconnected. For example, if our space is the union of the open balls with radius 1 around (1,0) and (-1,0) in ℝ2 with the metric inherited from the standard metric on ℝ2, then those two open balls would be both open and closed. However, I don't know a way to make sense of the assumption that all open balls (or even all ε-balls with a specific value of ε) are compact.
Yes, I agree with you... It seems it wouldn't make sense.

Ah, yes, that was a mistake on my part.

Looks good.
OK, thanks!