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Math Help - Continuity on M

  1. #1
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    Continuity on M

    Fix a point a in a metric space (M,d), and define f:M->R by f(x)=d(x,a). Show f is continuous on M.

    I took a sequence Xn->x in (M,d)
    So f(Xn)=d(Xn,a)
    taking the limit, f(Xn)->d(x,a)=f(x)
    Since we have Xn->x => f(Xn)->f(x), f is continuous on M.

    But I feel like that's wrong, because how do we know f(x) is in R?
    I need to show if Xn->x in (M,d), then f(Xn)->f(x) in R.....but I don't know whether or not f(x) is in R do I? Am i missing something obvious?
    Thanks guys.
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  2. #2
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    f(x) is in R because f(x) is a distance, whowever I think you nedd to prove d(xn,a)->d(x,a)
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    Fix a point a in a metric space (M,d), and define f:M->R by f(x)=d(x,a). Show f is continuous on M.

    I took a sequence Xn->x in (M,d)
    So f(Xn)=d(Xn,a)
    taking the limit, f(Xn)->d(x,a)=f(x)
    Since we have Xn->x => f(Xn)->f(x), f is continuous on M.

    But I feel like that's wrong, because how do we know f(x) is in R?
    I need to show if Xn->x in (M,d), then f(Xn)->f(x) in R.....but I don't know whether or not f(x) is in R do I? Am i missing something obvious?
    Thanks guys.
    Why not use the normal definition of continuity? It's easier.

    Then, d(x,a)-d(x_0,a)\leqslant d(x,x_0)+d(x_0,a)-d(x_0,a)=d(x,x_0) and d(x_0,a)-d(x,a)\leqslant d(x_0,x)+d(x,a)-d(x,a)=d(x_0,x). Combining these two we see that \left|d(x,a)-d(x_0,a)\right|\leqslant d(x,x_0) and thus f is Lipschitz and thus trivially continuous.
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  4. #4
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    I'm not familiar with Lipschitz What is that?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    I'm not familiar with Lipschitz What is that?
    If \varphi:M\to M' is a function (where M,M' are metric spaces) then \varphi is said to be Lipschitz with Lipschitz constant C>0 if d_{M'}(\varphi(x),\varphi(y))\leqslant C d_M(x,y). I actually didn't prove Lipschitz per se (since I fixed x_0) but it can easily be proved using the EXACT same idea that f is Lipschitz.

    Every Lipschitz function is uniformly continuous and thus continuous. It's clear for if \varepsilon>0 is given then choosing \delta=\frac{\varepsilon}{C} we see that d_M(x,y)<\delta\implies d_{M'}(\varphi(x),\varphi(y))\leqslant C d_M(x,y)\leqslant C\frac{\varepsilon}{C}=\varepsilon
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