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Closed set as infinite intersection of open sets

  1. Mar 9, 2012 #1
    This is not a homework problem, just something I was thinking about. In a general metric space, is it true that every closed set can be expressed as the intersection of an infinite collection of open sets?

    I don't really know where to begin. Since the finite intersection of open sets is open, and the infinite intersection of open sets may or may not be open or closed, this suggests to me that the statement may not be true; however, I'm not sure how I would use this fact to construct a counter-example. Any help is much appreciated!
  2. jcsd
  3. Mar 9, 2012 #2
    The answer is yes. Every closed set can be written as a countable intersection of open sets in a metric space. (this is not true anymore in an arbitrary topological space)

    The reason is this: take a closed set F. We have a continuous function [itex]f(x)=d(x,F)[/itex]. Notice that [itex]f^{-1}(0)=F[/itex].
    Of course, we have that [itex]\{0\}=\bigcap_{n>0} ]-1/n,1/n[[/itex], thus

    [tex]F=\bigcap_{n>0} f^{-1}(]-1/n,1/n[)[/tex]
  4. Mar 9, 2012 #3
    Thanks for the reply. I hope you can clarify a couple of points in your answer, however. When you say "We have a continuous function [itex]f(x)=d(x,F)[/itex]," am I correct in saying that you mean d(x,F) = inf{d(x,f): f lying in F}?

    Also, what do you mean by [itex]]-1/n,1/n[[/itex] ? Is it something like an open neighbourhood of radius 1/n around the zero element, and if so, is there any reason why you write it that way as opposed to (-1/n, 1/n)?
  5. Mar 9, 2012 #4

    I meant (-1/n,1/n). But I always write ]-1/n,1/n[. It's an accepted notation.
  6. Mar 9, 2012 #5
    Ok, thanks very much for the help and the quick reply!
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