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Thread: 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 $\displaystyle (a-r,a+r)\subset A\subset\mathbb{R} $ can I say that $\displaystyle a+r\in A$?

    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; Oct 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 $\displaystyle (-a,a)\subset A\subset\mathbb{R} $ can I say that $\displaystyle a\in A$?
    Quick reply: no.
    It could be that $\displaystyle (-2a,a)=A$.
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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 $\displaystyle (-2,2)=A$ now clearly $\displaystyle (1-1,1+1)\subset A$ right?
    Does $\displaystyle 1+1\in A~?$
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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 $\displaystyle supA\not\in A$

    proof

    Assume that $\displaystyle \alpha=sup A\in A$.

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

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

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

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

    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 $\displaystyle supA\not\in A$proof
    Assume that $\displaystyle \alpha=sup A\in A$.
    Then as A is open from the definition we have that for $\displaystyle \alpha\in A\ \mbox{and}\ r>0$
    $\displaystyle (\alpha-r,\alpha+r)\subset A$.
    Contradiction- as we assumed that $\displaystyle \alpha=Sup A$ but $\displaystyle \alpha+r>\alpha\ \mbox{and}\ \alpha+r\in A$.
    I was asking if I can always choose an r such that $\displaystyle \alpha+r\in A$?
    Well thank you for finally posting an intelligible problem.
    It is well know that an open set in $\displaystyle \mathbb{R}$ 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 $\displaystyle \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 $\displaystyle \alpha+r\in A $?
    No you cannot say that.
    But you can say $\displaystyle \left( {\exists s} \right)\left[ {\alpha < s < r} \right]$ thus $\displaystyle s\in A~.$
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