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Math Help - show that a set is closed

  1. #1
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    show that a set is closed

    I have a set A:=\{(x,x)|x\in[0,1]\}\subset [0,1]\times [0,1] i.e the diagonal of the unit square. How to show that A is closed??

    By definition a set is closed if it's complement is open, so I need to show that A^{c} is an open set. What A^{c} will be in this case?

    A^{c}=\{(x,x)|x\not\in[0,1]\}?
    is A^{c}=\{(x,x)|0>x>1|y=x\}=\{(x,x)|x\in(-\infty,0)\cup(0,\infty)|y=x\}?
    Last edited by rayman; September 23rd 2013 at 03:45 AM.
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    Re: show that a set is closed

    Quote Originally Posted by rayman View Post
    I have a set A:=\{(x,x)|x\in[0,1]\}\subset [0,1]\times [0,1] i.e the diagonal of the unit square. How to show that A is closed??

    By definition a set is closed if it's complement is open, so I need to show that A^{c} is an open set. What A^{c} will be in this case?

    A^{c}=\{(x,x)|x\not\in[0,1]\}?
    Consider that U=[0,1]\times [0,1] is the unit square.
    If P: (p,q)\in(U\setminus A) then the distance that P
    is from the line x-y=0 is \frac{|p-q|}{\sqrt2}.

    So let \delta=\frac{|p-q|}{2\sqrt2}.

    The ball \mathcal{B}(P;\delta) contains P and no point of the line x-y=0.

    That shows that A^c is open.

    BTW. Your notation A^{c}=\{(x,x)|x\not\in[0,1]\} is incorrect. You want A^{c}=\{(x,y)\in[0,1]\times [0,1]:x\ne y\}
    Last edited by Plato; September 23rd 2013 at 03:46 AM.
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    Re: show that a set is closed

    thank you for the help. I have a question.
    Just to clarify
    is the radius of the open ball \delta=\frac{|p-q|}{\sqrt{2}}? (from the formula I get \sqrt{2} in the denominator)

    What about the end points of A? Don't we have to find open balls in the neighborhood of these two with the proper radius?
    Last edited by rayman; September 23rd 2013 at 04:42 AM.
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    Re: show that a set is closed

    Quote Originally Posted by rayman View Post
    thank you for the help. I have a question.
    Just to clarify
    is the radius of the open ball \delta=\frac{|p-q|}{\sqrt{2}}? (from the formula I get \sqrt{2} in the denominator)
    I used \delta=\frac{|p-q|}{2\sqrt{2}} because that makes the closure of the ball disjoint from A.

    That is merely a matter of choice.

    Quote Originally Posted by rayman View Post
    What about the end points of A? Don't we have to find open balls in the neighborhood of these two with the proper radius?
    I don't know what you mean by "What about the end points of A?"

    The proof I gave is in fact the way we show that any line in \mathbb{R}^2 is a closed set.
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    Re: show that a set is closed

    Thanks again for helping.
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