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A problem in Real Analysis/Topology

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Let A be a non empty part of ℝ, such that A and A complement are two open subsets of ℝ.



    2. Relevant equations

    1- Prove that A is not bounded above.
    2- Assuming that A complement is non-empty and let x in A complement.
    and let B={xεA such that x≤t}. Prove that B is non empty and has a lower bound m such that m≥x.
    3- Prove that: [tex]m\notin A[/tex] and [tex]m\notin A^{c}[/tex] . Conclude that A=ℝ.
    4- Let F be non-empty set in ℝ such that F and F complementary are closed in R. What can we say about the set F?





    3. The attempt at a solution

    1- A is an open subset of R so we're going to use a proof by contradiction.

    Suppose that a=sup(A) so for all x in A : x≤a . But since A is open then there exists an ε>0 such that: [tex]]a-\epsilon,a+\varepsilon[\cap A\neq\varnothing[/tex] so that means there is another number say a+ε/2 that's also in A so therefore A is unbounded above.

    2- B is a subset of A so there for B is nonempty. Second part I know I have to conduct a proof by contradiction but I don't know how to start . Well I can suppose that B doesn't have a lower bound.

    Any help will be appreciated. Thank you very much.
     
  2. jcsd
  3. Dec 8, 2013 #2

    Dick

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    Part 1 isn't going well. Some of the parts are there but the reasoning is all muddled. The idea is to show that ##a=sup(A)## CANNOT be an element of ##A##. Suppose it was? Then there's a neighborhood of ##a## that is contained in ##A##. Can you see how that would create a problem with ##a## being ##sup(A)##?
     
    Last edited: Dec 8, 2013
  4. Dec 8, 2013 #3
    If there is a neighborhood of a that is contained in A then there is another element other than a that's also in A so therefore a cannot be Sup(A). Is that right?
     
  5. Dec 8, 2013 #4

    HallsofIvy

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    This is only true because R is a connected set. At some point you are going to have to use the connectedness of R.
     
  6. Dec 8, 2013 #5
    Or for number 1 can I do this:

    Let N be any natural number and define V={a: lal>N} a neighborhood of a at infinity.
    Thus, there exists some a in A such that A is in V.
    Hence, there exists some a in A such that |x|>N.
    Therefore, A is unbounded.
     
  7. Dec 8, 2013 #6

    Dick

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    There being another element besides ##a## in A does not contradict ##a## being a sup. How can it? If you could show there is an element ##b## in A such that a<b, that would contradict ##a## being a sup. Can you do that?
     
  8. Dec 8, 2013 #7

    Dick

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    No, stick to the original approach. There's no such thing as "a neighborhood of a at infinity".
     
  9. Dec 8, 2013 #8

    Dick

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    That's a good point. But I think this is a proof of the connectedness of R. Can't really use that R is connected in the proof. I think the important property of R that's not being explicitly stated is that if A is bounded above then there is such a real number as sup(A).
     
  10. Dec 8, 2013 #9
    Okay A is open and we supposed that a=sup(A) so there exist an ε>0 such that:

    [tex]]a-\epsilon,a+\epsilon[\subset A[/tex] so:

    [tex]\exists b\neq a\in]a-\epsilon,a+\epsilon[\subset A[/tex]

    Therefor b is in the neighborhood of a and in the subset A. So for b=a+ε/2 we still have b in the neighborhood of a and in A so there for a is not the supremum of A and finally A is unbounded above.
     
  11. Dec 8, 2013 #10

    Dick

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    There is such an ε because we ASSUME that ##a=sup(A)## is an element of A when A is bounded above. That's an assumption. We may have to reconsider it.

    Yes, a<b=a+ε/2 is in the neighborhood saying ##a## is not the supremum. That's a contradiction, because we defined ##a=sup(A)##. When you reach a contradiction you have to go back and look at the last thing you assumed. That must be wrong. What is it?
     
  12. Dec 8, 2013 #11
    Since we found a b>a in the neighborhood of a, and since we had assumed that a=Sup(A) we have reached a contradiction with a=Sup(A).
     
  13. Dec 8, 2013 #12

    Dick

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    You are finding the wrong contradiction at this point in the proof. I'll give you an example, take the open interval A=(0,1). a=sup(A)=1. So far, there's no contradiction. What assumption really created the contradiction. Look back. It's an assumption you keep failing to mention.
     
  14. Dec 8, 2013 #13

    pasmith

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    You've shown that if [itex]\sup A \in A[/itex] then [itex]A[/itex] cannot be open. But you're given that [itex]A[/itex] is open.

    So either [itex]\sup A \in A^c[/itex] or [itex]A[/itex] is not bounded above, which is what you are asked to prove.

    So how do you exclude the possibility that [itex]\sup A \in A^c[/itex], given that [itex]A^c[/itex] is open?
     
  15. Dec 8, 2013 #14
    We assumed that a is in A so our contradiction is with a in A right?
     
  16. Dec 8, 2013 #15

    Dick

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    Yes, your conclusion is that ##a## is not in ##A##. Therefore ##a## must be in A complement, which is also open by assumption. Now what?
     
  17. Dec 8, 2013 #16
    If a is not in A then if it's in A complement, that means that a is a limit point of A right?
     
  18. Dec 8, 2013 #17

    Dick

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    I'll agree with that if you can give me reason.
     
  19. Dec 8, 2013 #18
    a is a limit point of A because:

    [tex]\exists b\neq a\in]a-\epsilon,a+\epsilon[\cap A[/tex] by the definition of a limit point. b is what found before to be greater than a.
     
  20. Dec 8, 2013 #19

    Dick

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    That doesn't make any sense to me as a reason. ##a## is a limit point of ##A## just because ##a=sup(A)##. How would being the sup imply that it's a limit point?
     
  21. Dec 8, 2013 #20
    A complement is open and we know that if a is not in A then it's in the adherent of A which is always a closed set because it contains all of the limit points of A?
     
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