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Open set is a collection of regions

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Let U be a nonempty open set. The U is the union of a collection of regions.


    2. The attempt at a solution

    Let x be an element of U. There exists a region R such that x is in R and R is in U. So x is also in the union of a collection of regions Rx. Then U is a subset of Rx → This is the part I don't understand even though the class agrees upon it.

    Also, since for all x, x is in R and is in U, so Rx is a subset of U. → I don't understand this part either!

    Therefore U = Rx
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 7, 2011 #2
    Can someone please help? A new proof is good too. Thanks!
     
  4. Dec 7, 2011 #3

    HallsofIvy

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    What is your definition of "region"?
     
  5. Dec 7, 2011 #4
    If a,b \in C and a<b, then the set of points between a and b is the region ab.

    Thanks!
     
  6. Dec 7, 2011 #5

    Dick

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    Every x in U is in a region contained in U. Doesn't that make the union of all of those regions U?
     
  7. Dec 7, 2011 #6
    Yeah. Then why is U a subset of the union of regions?

    Since I'm showing U=union of regions, I want to show that U is a subset of the union of regions, and the union of regions is the subset of U.

    Thanks a lot!
     
  8. Dec 7, 2011 #7

    Dick

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    Because every element of U is in one of the regions. So U is a subset of the union of the regions. Since every region is a subset of U, then the union of the regions is a subset of U.
     
  9. Dec 7, 2011 #8
    This TOTALLY makes my day. WOOHOO!

    THANKS :)
     
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