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Math Help - Prove that the application is closed

  1. #1
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    Prove that the application is closed

    Hi, everyone,
    I resolved the following exercise. I need help reviewing it and giving the correct answer (if mine isn't correct of course).


    Let's have:
    I_1=\{(x,y) \in R^2|y=0, 0\leq x\leq 1\}
    I_2=\{(x,y)\in R^2|x=1, 0\leq y\leq 1\}
    I_3=\{(x,y)\in R^2|y=x;   x,y \in [0 \;1]\}
    Q=I_1 \cup I_2 \cup I_3

    where the following equivalence relation is defined
    \A (x,y), (x',y') \in Q, (x,y)\sim (x',y') if (x,y)=(x',y') or (x, y), (x', y')\in I_1,
    (x,y),(x',y')\in I_2
    Prove that \pi : Q\rightarrow Q/\sim is closed.

    Solution:

    Q is connected and compact so even Q/\sim must be connected and compacted. At this point we have to prove that Q/\sim is T_2 (Hausdorff). That is to prove that the set
    F= \{ (x;y),(u;v): \pi(x;y)=\pi(u;v) \}=\{ (x;y),(u;v): (x;y)\sim(u;v) \} is closed on Q\times Q.
    C_1=\{ (x;y),(x;y)\in Q\times Q| (x;y)\in Q}, C_2=\{ (x;0),(x;0)\in Q\times Q| x\in[0\;1]\}
    C_3=\{ (1;y),(1;y)\in Q\times Q| y\in[0\;1]\}.
    C_1 is closed because Q is T_2. C_2, C_3 are closed being subspaces of space T_2 and immages of the compact [0\;1] on R through continuous applications. Because F=C_1 \cup C_2 \cup C_3 then even F will be closed, this means that Q/\sim is T_2. I conclude that \pi is closed.
    Last edited by mameas; July 1st 2011 at 06:00 AM.
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  2. #2
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    Feb 2010
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    Re: Prove that the application is closed

    There are things I don't understant in your explanation. F, as defined, seems to be en element of Q/~ X Q/~ while C1,C2 and C3 are in Q X Q
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  3. #3
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    Re: Prove that the application is closed

    Sorry, but F is defined in Q \times Q, then I use the theorem that say: If F is closed in Q\times Q then Q/\sim is T_2 and this means that \pi is closed.
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  4. #4
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    Re: Prove that the application is closed

    I think that C_2\subset C_1 and C_3\subset C_1 and threre are elements of F that are not in C1
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