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Math Help - Real Analysis/Topology problem

  1. #1
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    Real Analysis/Topology problem

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

    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 m is not in A nor in A complement.
    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?

    My attempt:

    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:
    ]a-
    ε,a+ε[nA is nonempty so therefore there exist another number a+ε/2>a so therefore A is unbounded.

    Thank you for any help before hand.

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  2. #2
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    Re: Real Analysis/Topology problem

    Quote Originally Posted by SMAD View Post
    Let A be a non empty part of ℝ, such that A and A complement are two open subsets of ℝ.
    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 m is not in A nor in A complement.
    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?
    This is a trick question (or else whoever wrote it knows really little).
    The only non-empty open subset of \mathbb{R} having its complement open is \mathbb{R} itself.

    This easily seen by noting that \mathbb{R} is a connected set.
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  3. #3
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    Re: Real Analysis/Topology problem

    Yes I believe the goal of this exercise is to get to A=R. I forgot to add to the third question that you have to conclude that A=R.
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