Problem in a Metric Space!

1. Sep 2, 2011

aodesky

Suppose (X,d) is a metric space and A, a subset of X, is closed and nonempty. For x in X, define d(x,A) = infa in A{d(x,a)}

Show that d(x,A) < infinity.

I really don't have much of an idea on how to show it must be finite. An obvious thought comes to mind, namely that a metric is real-valued by definition, so it must be a real number and hence finite, but I don't feel that that reasoning captures the gist of the inherent problem.

Does anyone have any ideas?

2. Sep 2, 2011

micromass

Staff Emeritus
So you must show that the set

$$\{d(x,a)~\vert~a\in A\}$$

has a finite infimum. The only thing you need to check here is that the set is nonempty. Do you agree??

3. Sep 3, 2011

foxjwill

Are you sure you wrote down the problem correctly? As micromass pointed out, whether or not $A$ is closed is irrelevant. Maybe you have to show that $0<d(x,A)<\infty$ for all $x\notin A$?

4. Sep 3, 2011

aodesky

Thanks for the responses. To micromass: I don't quite see how the set's being nonempty necessarily implies that its infimum is finite. And to foxjwill: the set's being closed has no pertinence (at least I don't think it does) to the part of the question I asked here, but there are two other parts of the question to which it does play a role; however, I knew how to answer those so I didn't post them here, and I didn't omit the fact that the set was closed because I wasn't positive that it played no role whatsoever in the question I asked here. The question is quoted here correctly, I just still cannot see a direct implication toward a finite infimum based off of the information here. Can you prove it micromass?

5. Sep 3, 2011

micromass

Staff Emeritus
Try to prove the following:

if a set of real numbers is nonempty and bounded below, then its infimum is finite.