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Thread: 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:
    $\displaystyle I_1=\{(x,y) \in R^2|y=0, 0\leq x\leq 1\}$
    $\displaystyle I_2=\{(x,y)\in R^2|x=1, 0\leq y\leq 1\}$
    $\displaystyle I_3=\{(x,y)\in R^2|y=x; x,y \in [0 \;1]\}$
    $\displaystyle Q=I_1 \cup I_2 \cup I_3$

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

    Solution:

    $\displaystyle Q$ is connected and compact so even $\displaystyle Q/\sim$ must be connected and compacted. At this point we have to prove that $\displaystyle Q/\sim$ is $\displaystyle T_2$ (Hausdorff). That is to prove that the set
    $\displaystyle F= \{ (x;y),(u;v): \pi(x;y)=\pi(u;v) \}=\{ (x;y),(u;v): (x;y)\sim(u;v) \}$ is closed on $\displaystyle Q\times Q$.
    $\displaystyle C_1=\{ (x;y),(x;y)\in Q\times Q| (x;y)\in Q}$, $\displaystyle C_2=\{ (x;0),(x;0)\in Q\times Q| x\in[0\;1]\}$
    $\displaystyle C_3=\{ (1;y),(1;y)\in Q\times Q| y\in[0\;1]\}$.
    $\displaystyle C_1$ is closed because $\displaystyle Q$ is $\displaystyle T_2$. $\displaystyle C_2$, $\displaystyle C_3$ are closed being subspaces of space $\displaystyle T_2$ and immages of the compact $\displaystyle [0\;1]$ on $\displaystyle R$ through continuous applications. Because $\displaystyle F=C_1 \cup C_2 \cup C_3 $ then even $\displaystyle F$ will be closed, this means that $\displaystyle Q/\sim$ is $\displaystyle T_2$. I conclude that $\displaystyle \pi$ is closed.
    Last edited by mameas; Jul 1st 2011 at 06:00 AM.
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  2. #2
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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 $\displaystyle Q \times Q$, then I use the theorem that say: If F is closed in $\displaystyle Q\times Q$ then $\displaystyle Q/\sim$ is $\displaystyle T_2$ and this means that $\displaystyle \pi$ is closed.
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  4. #4
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    Re: Prove that the application is closed

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