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Math Help - open and closed sets

  1. #1
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    open and closed sets

    In this problem you will prove that the only subsets of R that are both open and closed are R and the empty set. Let E be a nonempty subset of R, and assume that E is both open and closed. Since E is nonempty there is an element a in E. Denote the set
    Na(E) = {x > 0|(a-x, a + x) contained in E}
    (a) Explain why Na(E) is nonempty.
    (b) Prove that if x in Na(E) then [a-x, a+x] is contained in E.
    (c) Prove that if [a-x, a+x] is contained in E, then there is a y in Na(E) satisfying y > x.
    (d) Show that Na(E) is not bounded above (argue by contradiction).
    (e) Prove that E = R.


    This is what I tried:
    a) Well I know that a is both open and closed, so if its closed, it contains its limit points. Since x is a limit point, and a is in E, then by properties of sets that would mean that a+x and a-x would be contained in E, but I really don't know how to show how the interval is contained in E.

    b)Think I need to use part a for this one

    c)well if [a-x,a+x] is contained in E, and E is also open, that means that there must be an open interval that is also contained in E. So there must be a y in Na(E) so that y>x that would yield the interval (y-a,y+a)

    d)Not too sure how to do this

    e) Not sure either

    Any help would be great!!!
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  2. #2
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    Because E is closed and x+a is a limit point of (a-x,a+x) then x+a \in E
    Same can be said for a-x.

    Because E is open \left( {\exists z > 0} \right)\left[ {\left[ {a + x - z,a + x + z} \right] \subset E} \right] Why?
    So let y=x+z then y>x and y\in E
    The point of all of that is that for any number in E there is some number greater in E.
    Thus E can't be bounded above nor below.

    Now for part e). If t>s~\&~t\in E must it follow that s\in E?
    If so how is it possible for E\ne R to be true?
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  3. #3
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    Thanks for the help, but do you think you can point me in the right direction toward showing how to prove that Na(E) is nonempty. It seems like it would be so easy yet I can't figure out how to do it. I've been trying to use the definiton of limit points but it isn't working.
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  4. #4
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    Quote Originally Posted by mathgirlie View Post
    Thanks for the help, but do you think you can point me in the right direction toward showing how to prove that Na(E) is nonempty. It seems like it would be so easy yet I can't figure out how to do it. I've been trying to use the definiton of limit points but it isn't working.
    Good grief.
    \emptyset  \ne E \subseteq R so \left( {\exists a \in E} \right)
    But we know that E is open so \left( {\exists x > 0} \right)\left[ {a \in (a - x,a + x) \subset E} \right].
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  5. #5
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    thanks, I knew it was easy... must be tired or something
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