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Problem in a Metric Space!

  1. Sep 2, 2011 #1
    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. jcsd
  3. Sep 2, 2011 #2

    micromass

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    So you must show that the set

    [tex]\{d(x,a)~\vert~a\in A\}[/tex]

    has a finite infimum. The only thing you need to check here is that the set is nonempty. Do you agree??
     
  4. Sep 3, 2011 #3
    Are you sure you wrote down the problem correctly? As micromass pointed out, whether or not [itex]A[/itex] is closed is irrelevant. Maybe you have to show that [itex]0<d(x,A)<\infty[/itex] for all [itex]x\notin A[/itex]?
     
  5. Sep 3, 2011 #4
    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?
     
  6. Sep 3, 2011 #5

    micromass

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    Try to prove the following:

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