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Real analysis

  1. Sep 29, 2010 #1
    Hey guys, doin another rudin-related question. Here Goes:

    Show that if E [tex]\subseteq[/tex] [tex]\Re[/tex] is open, then E can be written as an at most countable union of disjoint open intervals, i.e., E=[tex]\bigcup[/tex]n(an,bn). (It's possible that an=-[tex]\infty[/tex] bn=+[tex]\infty[/tex] for some n.)

    My attempt:
    Take the set of all Neighborhoods of all of the rationals of a rational radius in R to be A. Now all members of E intersect A make up E. Take the union of all of the neighborhoods in this set E intersect A and this is a countable union of disjoint sets.

    Is there a problem with this?
     
    Last edited: Sep 30, 2010
  2. jcsd
  3. Sep 30, 2010 #2
    I now realize this doesnt create a disjoint union. I had the idea of taking an arbitrarily large neighborhood of a point in E that is still a proper subset of E and then filling up the gaps on either side with more neighborhoods, always still contained in E, until I have a countably dense set of neighborhoods contained in E...

    Would this do?
     
  4. Oct 2, 2010 #3
    Here is my suggestion:
    First let E be a union of disjoint open sets F. Then E is an open set. We know that by Lindelof there is a countable subset F' of F. However, since F is disjoint, F=F'. So F is countable.

    Accurately, F is a collection of sets and E is the union of sets taken from F.

    I guess that your first suggestion proves the lindelof that countable subets exist. Although they are "not disjoint", you have proven a lindelof and you can apply this property to the open set which is a union "disjoint" open sets.
     
  5. Oct 3, 2010 #4

    Landau

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    This is a familiar exercise/result. See e.g. here, here, and your own (!!) thread here.
     
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