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Intro Topology: boundry Q

  1. May 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that every nonempty proper subset of Rn has a nonempty boundry.

    3. The attempt at a solution

    First of all, I let S be an nonempty subset of Rn and S does not equal Rn.

    I tried to go about this in 2 different ways:

    1) let x be in S and show that B(r,x) ∩ S ≠ ø and B(r,x) ∩ Sc≠ ø. I figured this wouldn't work with just one x in S. Or perhaps, I thought I should use induction on the number of elements in S?
    2) Assume that bdS is empty and find a contradiction. However, I wasn't able to figure out a contradiction here. Unless, this implies that S equals Rn, then that's a contradiction. But I'm not quite sure it implies that. I think that this is the proof you use to show that Rn and the empty set are the only 2 that are both open and closed.

    Thanks for your help! =)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: May 24, 2007
  2. jcsd
  3. May 24, 2007 #2

    NateTG

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    What's the definition of boundry?
     
  4. May 24, 2007 #3
    bd(S) = {x in Rn s.t. B(r,x)∩ S ≠ ø and B(r,x) ∩ Sc≠ ø for every r>0}
     
  5. May 24, 2007 #4

    Dick

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    Take a point x in S and a point y in S^C and consider the line t*x+(1-t)*y for t in [0,1].
     
  6. May 24, 2007 #5
    Well you could assume the contrary and first prove that S must be closed and similarly that S must be open. Now which are the only sets in a connected space with that property?
     
  7. May 25, 2007 #6

    NateTG

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    Redyelloworange is aware of that, he mentions that he thinks the goal is to prove that the n-dimensional reals are connected under the usual topolgy:
     
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