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Thread: Connected Hausdorff spaces and graphs

  1. #1
    MHF Contributor Drexel28's Avatar
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    Connected Hausdorff spaces and graphs

    I came up with the following question, and would appreciate if I could get some validation/invalidation.

    If $\displaystyle X$ is connected Hausdorff and $\displaystyle \varphi:X\mapsto Y$ continuous then is $\displaystyle \Gamma_\varphi\subseteq X\times Y$ connected? ($\displaystyle \Gamma_\varphi$ is the graph of $\displaystyle \varphi$). I think the answer is yes.

    Proof: We first need a lemma

    Lemma: Let $\displaystyle X$ be connected Hausdorff, then $\displaystyle \Delta_X\subseteq X\times X$ is connected ($\displaystyle \Delta_X$ is the diagonal).
    Proof: Clearly the map $\displaystyle \iota_X\oplus\iota_X:X\mapsto X\times X$ given by $\displaystyle x\mapsto (x,x)$ is continuous. But, it is not hard to see that $\displaystyle \Delta_X=\left(\iota_X\oplus \iota_X\right)\left(X\right)$. But, connectedness is invariant under continuous mapping. So, the conclusion follows $\displaystyle \blacksquare$

    Now, it is easy to prove that the product of two continuous maps is continuous. In particular, since $\displaystyle \varphi$ and the identity map $\displaystyle \iota_X$ are continuous we have that $\displaystyle \iota_X\times\varphi:X\times X\mapsto X\times Y$ is continuous. I now claim that $\displaystyle \Gamma_\varphi=\left(\iota_X\times\varphi\right)\l eft(\Delta_X\right)$.

    To see this, let $\displaystyle (y,y')\in\Gamma_\varphi$ then $\displaystyle (y,y')=(x,\varphi(x))$ for some $\displaystyle x\in X$. Clearly then $\displaystyle \left(\iota_X\times\varphi\right)\left((x,x)\right )=\left(x,\varphi(x)\right)=(y,y')$ and so $\displaystyle (y,y')\in \left(\iota_X\times\varphi\right)\left(\Delta_x\ri ght)$

    Conversely, let $\displaystyle (y,y')\in\left(\varphi\times\iota_X\right)\left(\D elta_X\right)$. Then, $\displaystyle (y,y')=\left(\varphi\times \iota_X\right)\left((x,x)\right)$ for some $\displaystyle x\in X$ but that means that $\displaystyle (y,y')=(x,\varphi(x))$ and so $\displaystyle (y,y')\in\Gamma_{\varphi}$.

    Recalling the lemma and that connectedness is invariant under continuous mappings finishes the argument. $\displaystyle \blacksquare$



    Is that right? I feel as though I am making a stupid mistake...especially because I realized that I didn't use Hausdorffness anywhere.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Idiot moment. Of course it's true.

    $\displaystyle \iota_X\oplus\iota_X:X\mapsto X\times X$ and $\displaystyle \iota_X\times\varphi:X\times X\mapsto X\times Y$ are both continuous. So, of course any topological property invariant under continuous mappings transfers from $\displaystyle X$ to $\displaystyle \Delta_X$ and $\displaystyle \Gamma_\varphi$

    Sorry for the hiccup
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    I came up with the following question, and would appreciate if I could get some validation/invalidation.

    If $\displaystyle X$ is connected Hausdorff and $\displaystyle \varphi:X\mapsto Y$ continuous then is $\displaystyle \Gamma_\varphi\subseteq X\times Y$ connected? ($\displaystyle \Gamma_\varphi$ is the graph of $\displaystyle \varphi$). I think the answer is yes.

    Proof: We first need a lemma

    Lemma: Let $\displaystyle X$ be connected Hausdorff, then $\displaystyle \Delta_X\subseteq X\times X$ is connected ($\displaystyle \Delta_X$ is the diagonal).
    Proof: Clearly the map $\displaystyle \iota_X\oplus\iota_X:X\mapsto X\times X$ given by $\displaystyle x\mapsto (x,x)$ is continuous. But, it is not hard to see that $\displaystyle \Delta_X=\left(\iota_X\oplus \iota_X\right)\left(X\right)$. But, connectedness is invariant under continuous mapping. So, the conclusion follows $\displaystyle \blacksquare$

    Now, it is easy to prove that the product of two continuous maps is continuous. In particular, since $\displaystyle \varphi$ and the identity map $\displaystyle \iota_X$ are continuous we have that $\displaystyle \iota_X\times\varphi:X\times X\mapsto X\times Y$ is continuous. I now claim that $\displaystyle \Gamma_\varphi=\left(\iota_X\times\varphi\right)\l eft(\Delta_X\right)$.

    To see this, let $\displaystyle (y,y')\in\Gamma_\varphi$ then $\displaystyle (y,y')=(x,\varphi(x))$ for some $\displaystyle x\in X$. Clearly then $\displaystyle \left(\iota_X\times\varphi\right)\left((x,x)\right )=\left(x,\varphi(x)\right)=(y,y')$ and so $\displaystyle (y,y')\in \left(\iota_X\times\varphi\right)\left(\Delta_x\ri ght)$

    Conversely, let $\displaystyle (y,y')\in\left(\varphi\times\iota_X\right)\left(\D elta_X\right)$. Then, $\displaystyle (y,y')=\left(\varphi\times \iota_X\right)\left((x,x)\right)$ for some $\displaystyle x\in X$ but that means that $\displaystyle (y,y')=(x,\varphi(x))$ and so $\displaystyle (y,y')\in\Gamma_{\varphi}$.

    Recalling the lemma and that connectedness is invariant under continuous mappings finishes the argument. $\displaystyle \blacksquare$



    Is that right? I feel as though I am making a stupid mistake...especially because I realized that I didn't use Hausdorffness anywhere.
    Please, maps to is not equal to a function...please...you're driving me crazy...

    (I'm an analyst...please...quit that sh...whatever...)
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jose27 View Post
    Please, maps to is not equal to a function...please...you're driving me crazy...

    (I'm an analyst...please...quit that sh...whatever...)
    Take that!
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Okay, I guess it's my bad since I didn't specify what I was ranting about...(sorry it came off a bit insulting)

    What I'm talking about is that usually one writes a function with $\displaystyle \rightarrow$ and the correspondence relation with $\displaystyle \mapsto$ (not trying to impose, but it really bugs me out...I really need to relax a bit with notation...)
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jose27 View Post
    Okay, I guess it's my bad since I didn't specify what I was ranting about...(sorry it came off a bit insulting)

    What I'm talking about is that usually one writes a function with $\displaystyle \rightarrow$ and the correspondence relation with $\displaystyle \mapsto$ (not trying to impose, but it really bugs me out...I really need to relax a bit with notation...)
    It's ok. I appreciate that you'd let met know.

    I understand that one usually puts $\displaystyle \phi:G\longrightarrow G'$ and $\displaystyle g\mapsto g'$. But in my naieve ignorance I got used to doing it the wrong way. It's just kind of persisted. I'll eventually wise up
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