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B Can open sets be described in-terms of closed sets?

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  1. Apr 9, 2016 #1
    Let A be an open set and A=(a,b). Can A be described, as closed set as
    "or every x>0, all the elements of closed set [a+x,b-x] are elements of A"?
     
  2. jcsd
  3. Apr 9, 2016 #2

    pwsnafu

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    The statement in quotes means
    ##\forall x>0##, ##[a+x, b-x] \subset A##,
    which is a true statement.

    But
    ##\forall x>0##, ##[a+x, b-x] \subset [a,b]##
    is also a true statement, so you can't use it to define A.
     
  4. Apr 9, 2016 #3
    Yes you are right....It just becomes a subset. But is there any way to describe it?
     
  5. Apr 9, 2016 #4

    pwsnafu

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    Sure. For example, let "A be the union of all ##[a+x,b-x]## with ##x>0##." Or "A is the smallest set that contain ##[a+x,b-x]## with ##x>0##".
     
  6. Apr 9, 2016 #5
    Yes,,,,the union part....If there were no union, every set (a+x,b-x) is a subset of both (a,b) and [a,b]
     
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