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Math Help - subset question

  1. #1
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    subset question

    If I have that A is an open interval the from the definition I have that  (a-r,a+r)\subset A\subset\mathbb{R} can I say that  a+r\in A?

    thanks for any help

    ps sorry about the first post I sent it from my phone and clearly something went wrong sorry
    Last edited by hmmmm; October 3rd 2011 at 09:02 AM.
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  2. #2
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    Re: subset question

    Quote Originally Posted by hmmmm View Post
    If I have the following  (-a,a)\subset A\subset\mathbb{R} can I say that  a\in A?
    Quick reply: no.
    It could be that (-2a,a)=A.
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  3. #3
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    Re: subset question

    sorry about the mistake I made above edited now
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  4. #4
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    Re: subset question

    Quote Originally Posted by hmmmm View Post
    sorry about the mistake I made above edited now
    No again.
    Consider (-2,2)=A now clearly (1-1,1+1)\subset A right?
    Does 1+1\in A~?
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  5. #5
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    Re: subset question

    yeah it isn't, however we can always choose an r so that it is?
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  6. #6
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    Re: subset question

    Quote Originally Posted by hmmmm View Post
    yeah it isn't, however we can always choose an r so that it is?
    Look this is a waste of time.
    We have no idea where your going with this thread.
    Post a problem. A complete well formed problem.
    Stop making us guess as to what you are trying to ask.
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  7. #7
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    Re: subset question

    If A is an open subset then supA\not\in A

    proof

    Assume that \alpha=sup A\in A.

    Then as A is open from the definition we have that for \alpha\in A\ \mbox{and}\ r>0

    (\alpha-r,\alpha+r)\subset A.

    Contradiction- as we assumed that \alpha=Sup A but \alpha+r>\alpha\ \mbox{and}\ \alpha+r\in A.

    I was asking if I can always choose an r such that \alpha+r\in A?

    Thanks for any help
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  8. #8
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    Re: subset question

    Quote Originally Posted by hmmmm View Post
    If A is an open subset then supA\not\in Aproof
    Assume that \alpha=sup A\in A.
    Then as A is open from the definition we have that for \alpha\in A\ \mbox{and}\ r>0
    (\alpha-r,\alpha+r)\subset A.
    Contradiction- as we assumed that \alpha=Sup A but \alpha+r>\alpha\ \mbox{and}\ \alpha+r\in A.
    I was asking if I can always choose an r such that \alpha+r\in A?
    Well thank you for finally posting an intelligible problem.
    It is well know that an open set in \mathbb{R} cannot contain a maximal element.
    Recall that between any two real numbers there is a real number.
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  9. #9
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    Re: subset question

    Is my proof of this statement then correct, specifically where I derived the contradiction and said  \alpha+r\in A ?
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  10. #10
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    Re: subset question

    Quote Originally Posted by hmmmm View Post
    Is my proof of this statement then correct, specifically where I derived the contradiction and said  \alpha+r\in A ?
    No you cannot say that.
    But you can say \left( {\exists s} \right)\left[ {\alpha < s < r} \right] thus s\in A~.
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