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

    Proof: We first need a lemma

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

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

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

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

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

    \iota_X\oplus\iota_X:X\mapsto X\times X and \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 X to \Delta_X and \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 X is connected Hausdorff and \varphi:X\mapsto Y continuous then is \Gamma_\varphi\subseteq X\times Y connected? ( \Gamma_\varphi is the graph of \varphi). I think the answer is yes.

    Proof: We first need a lemma

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

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

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

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

    Recalling the lemma and that connectedness is invariant under continuous mappings finishes the argument. \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 \rightarrow and the correspondence relation with \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 \rightarrow and the correspondence relation with \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 \phi:G\longrightarrow G' and 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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