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Thread: prove that every continuous function is separately continuous

  1. #1
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    prove that every continuous function is separately continuous

    Let (M,d),(N,p),(Y,q) be metric spaces. A function f:MxN-->Y is called separately continuous if for every $\displaystyle a\in M $ and $\displaystyle b \in M$, the function $\displaystyle g:N-->Y$ and $\displaystyle h:M-->Y$, given by $\displaystyle g(y)=f(a,y)$ and $\displaystyle h(x)=f(x,b)$, are continuous.

    Prove that every continuous function f:MxN-->Y is separately continuous.

    the deinition given is confusing, I have not done any separately continuous question before, need some help here.
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    Re: prove that every continuous function is separately continuous

    Quote Originally Posted by wopashui View Post
    Let (M,d),(N,p),(Y,q) be metric spaces. A function f:MxN-->Y is called separately continuous if for every $\displaystyle a\in M $ and $\displaystyle b \in M$, the function $\displaystyle g:N-->Y$ and $\displaystyle h:M-->Y$, given by $\displaystyle g(y)=f(a,y)$ and $\displaystyle h(x)=f(x,b)$, are continuous.

    Prove that every continuous function f:MxN-->Y is separately continuous.

    the deinition given is confusing, I have not done any separately continuous question before, need some help here.
    Prove that you have the obvious embedding $\displaystyle \iota:N\to M\times N:h\mapsto (a,y)$ is continuous and note then that $\displaystyle h=f\circ\iota$, etc.
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    Re: prove that every continuous function is separately continuous

    Quote Originally Posted by Drexel28 View Post
    Prove that you have the obvious embedding $\displaystyle \iota:N\to M\times N:h\mapsto (a,y)$ is continuous and note then that $\displaystyle h=f\circ\iota$, etc.
    sorry, what is embedding? can you explain a bit more about the proof?
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    MHF Contributor Drexel28's Avatar
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    Re: prove that every continuous function is separately continuous

    Quote Originally Posted by wopashui View Post
    sorry, what is embedding? can you explain a bit more about the proof?
    There are two ways you can think about it. Perhaps the easiest is, despite what I said, to merely note that $\displaystyle h$ is just $\displaystyle f$ restricted to $\displaystyle \{a\}\times N$ and restrictions of continuous functions are continuous.
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    Re: prove that every continuous function is separately continuous

    Quote Originally Posted by Drexel28 View Post
    There are two ways you can think about it. Perhaps the easiest is, despite what I said, to merely note that $\displaystyle h$ is just $\displaystyle f$ restricted to $\displaystyle \{a\}\times N$ and restrictions of continuous functions are continuous.
    hmm, what about g(x), is g the same as f restricted to Mx{b}?
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    Re: prove that every continuous function is separately continuous

    Quote Originally Posted by wopashui View Post
    hmm, what about g(x), is g the same as f restricted to Mx{b}?
    Indeed. Of course, technically we are talking about the restrictions with the obvious identifications $\displaystyle M\approx M\times\{b\}$ and $\displaystyle N\approx \{a\}\times M$, of course this identification is via the embedding I previously mentioned.
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    Re: prove that every continuous function is separately continuous

    ic, but how does this help for the proof?
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