Is the Distance to a Closed Subset in a Metric Space Always Finite?

[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) = inf_{a 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?

 

[micromass]

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??

 

[foxjwill]

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) = inf_{a in A}{d(x,a)}

Show that d(x,A) < infinity.

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&lt;d(x,A)&lt;\infty for all x\notin A?

 

[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?

 

[micromass]

Try to prove the following:

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