A compact => d(x, A) = d(x, a) for some a in A

[radou]

Homework Statement

The title pretty much suggests everything.

Let (X, d) be a metric space, and A a non empty compact subset of X.

If A is compact, then there exists some a in A such that d(x, A) = d(x, a), where d(x, A) = inf{d(x, a) : a is in A}, i.e. the set {d(x, a) : a in A} has a least element.

The Attempt at a Solution

Let x be a fixed point of X. Define the mapping d' : {x} x A -->R with d(x, a). Since the metric d : X x X --> R is continuous, so is d'. Since {x} and A are compact, so is {x} x A, and we can apply the extreme value theorem to d'. So, there exist points x1 = (x, b) and x2 = (x, c) such that d'(x1) <= d'(a) <= d'(x2), for any a in A, i.e. d(x, b) <= d(x, a) <= d(x, c), so for any a in {x} x A, we have d(x, b) <= d(x, a), for any a in A. Hence, b is the element of A we were looking for.

 

[micromass]

This is entirely correct imho.

 

[radou]

Thanks!