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I was going over in my head today the topological proof that a continuous function defined on a closed, bounded interval achieves its maximum and minimum value, and I was trying to prove that a closed, bounded subset $K$ of a metric space $(X,d)$ is compact. The proof that a closed, bounded subset of $\mathbb{R}^n$ is compact relies, in some sense, on the notion of "dividing the subset in half" (and then goes on to use "one half must not be covered by a finite subcollection of the open cover", etc.).

If $K$ cannot be embedded in $\mathbb{R}^n$ for any $n$, I'm not sure what would be an appropriate version of "dividing the space in half". If $K$ has finite Lebesgue Covering Dimension, it embeds in a finite dimensional Euclidean space, and we're done.

So, are there closed, bounded subsets of metric spaces that do not have finite Lebesgue covering dimension? Equivalently (from my point of view), is there some general notion of dividing a closed, bounded subset $K$ of a metric space in two pieces, each with half the diameter, that works generally for all such possible $K$? Is there a counter-example to the statement that a closed, bounded subset of a metric space is compact? We likely went over this in my Ph.D. level general topology course, but I don't remember off the top of my head.

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Yes, there are closed, bounded metric spaces that don't embed in a finite dimensional Euclidean space, and are not compact. For example, a closed unit ball in an infinite-dimensional Banach space.

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  • $\begingroup$ You're right, of course; I don't know what I was thinking. $\endgroup$ – Jeffrey Rolland Mar 14 '19 at 19:26
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Take any metric space which is not compact (say the real line with the usual metric) and consider $X$ with the new metric $d'(x,y)=\frac {d(x,y)} {1+d(x,y)}$. Then the whole space $X$ is closed and bounded but not compact.

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  • $\begingroup$ You're right, of course. That's a really good example, because both the original Euclidean metric and yours induce the same topology, and so are "equivalent metrics", but, of course, the theorem is true for one but not the other. $\endgroup$ – Jeffrey Rolland Mar 14 '19 at 19:27

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