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A set is closed iff it equals an intersection of closed sets

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Let M be a metric space, A a subset of M, x a point in M.

    Define the metric of x to A by

    d(x,A) = inf d(x,y), y in A

    For [itex]\epsilon[/itex]>0, define the sets

    D(A,[itex]\epsilon[/itex]) = {x in M : d(x,A)<[itex]\epsilon[/itex]}

    N(A,[itex]\epsilon[/itex]) = {x in M: d(x,A)[itex]\leq[/itex][itex]\epsilon[/itex]}

    Show that A is closed iff A = [itex]\bigcap[/itex]N(A,[itex]\epsilon[/itex]) for [itex]\epsilon[/itex]>0

    2. Relevant equations



    3. The attempt at a solution

    I was able to do the <= implication.
    I'm having trouble doing the => one...

    If I assume A is closed, all I know is that its compliment is open, which is a union of open sets... then should I just use that fact and DeMorgans laws to show A is in fact an intersection of those closed N(A,[itex]\epsilon[/itex])?

    OR should I take an arbitrary point in A, and show it belongs to the intersection..??

    Just need some help on what method I should try to use.

    Thanks!
     
  2. jcsd
  3. Sep 28, 2011 #2
    So, for [itex]\Rightarrow[/itex], you must prove that

    [tex]A=\bigcap{N(A,\varepsilon)}[/tex]

    There are 2 things to prove now: [itex]\subseteq[/itex] and [itex]\supseteq[/itex]. Only [itex]\supseteq[/itex] is nontrivial. For this: take a point in the intersection, and prove that it is in A.
     
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