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If I is open interval, prove I is an open set

  1. Sep 11, 2011 #1
    Is this a good-enough proof? I could have used neighborhoods to show this, but it seems like this way is a bit easier.

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110911_113143.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 11, 2011 #2
    You have really only changed the problem from showing that one set is open to showing that two are closed; you'll need to use neighborhoods either way. Much simpler to just take [itex]x \in (a,b)[/itex] (assuming [itex]a < b[/itex]) and take [itex]r[/itex] to be the minimum of [itex]b-x[/itex] and [itex]x-a[/itex] so that [itex]x - r >= x - (x-a) = a[/itex] and [itex]x + r <= x + (b-x) = b[/itex]. Then [itex](x-r, x+r) \subset (a,b)[/itex] holds so that [itex]x[/itex] is an interior point of [itex](a,b)[/itex].
     
  4. Sep 11, 2011 #3
    Well, I also used the fact that if one set is open, its complement is closed. I suppose that's not enough, though.

    The professor suggested the method you just did.
     
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