Does Compactness Depend on Metric Choice in Metric Spaces?

[JG89]

Suppose that A is a subset of a metric space X. Does compactness of A depend on which metric is given to X? For example, if d(x,y) and p(x,y) are two possible metrics for X, is A compact with respect to the metric d(x,y) if and only if A is compact with respect to the metric p(x,y)?

 

[mynameisfunk]

I am no expert on this forum but I believe that if there is an open cover of A in one metric, the other will follow

 

[HallsofIvy]

Yes, "compactness" does depend strongly on the metric. A simple example is the metric d(x,y)= 1 if $x\ne y$, 0 otherwise. That is called the "discrete metric" because using that metric on any set gives the "discrete topology" in which all sets are open. To see that, recognize that the "neighborhood of point p with radius 1/2" is just the singleton set {p} itself- since the distance from p to any other point is 1, the only point, x, for which it is true that d(p, x)< 1/2 is p itself. Every point, in every set, is an interior point because that singleton set is a subset neighborhood contained in the set.

If X is an infinite set, with the discrete metric, then the compact sets are exactly the finite sets. That is because we could always use the individual singleton sets, {a} where a is any member of A, as our open cover. Since every point of A is in only one of those, we cannot remove any of them, much less reduce to a finite cover.


So, for example, take X to be the set of all real numbers.

With the "usual metric", d(x,y)= |x- y|, the interval [0, 1] is compact because it is closed and bounded. With the discrete metric, it is not compact because it is not finite.

Another example: X is the set of real numbers formed by the sequence {0, 1, 1/2, 1/3, ..., 1/n, ...}, again with the usual metric on the real numbers. Let $\{U_n\}$ be any open cover. Since 0 is in A, there exist some $U_0$ which contains 0. Since $U_0$ is open, 0 is an interior point- there exist $\delta$ such that $\{x | |x|< \delta\}$ is a subset of $U_0$. But the sequence 1, 1/2, 1/3, ... converges to 0. There exist some N such that if n> N, $|1/n|< \delta$ and so all 1/n, for n> N, is in $U_o$. Pick a single $U_n$ that contains 1/n for n< N. That is a finite collection and it, together with $U_0$, makes a finite subcollection that covers A. A is compact.

But that same set of real numbers, with the discrete metric, is infinite and so is not compact.

 

[JG89]

Great reply Halls. That helped a lot!