# 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.