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Math Help - Continuous functions in metric spaces

  1. #1
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    Continuous functions in metric spaces

    Let M, N be metric spaces with metrics dm and dn respectively.
    Let f be a function from M to N.

    Suppose that if (an) is any convergent sequence in M, then (f(an)) is a convergent sequence in N.

    Show that f is a continuous function.



    Note that the question does not say that (an) --> a implies (f(an)) --> f(a). If it did then I could solve the question. But, as it is currently stated, I do not know the answer.
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    You know that f is continuous if and only if  a_n \rightarrow a \Rightarrow f(a_n) \rightarrow f(a) , right?

    I would assume that  f(a_n) \rightarrow c and then show that  c = a . Or assume  f(a_n) approaches a limit distinct from  a and then derive a contradiction.
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  3. #3
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    what happened to plato post??
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    Hint: Suppose that a_n\to a and that f(a_n)\to b. What can you say about the limit of the sequence a_1,a,a_2,a,a_3,a,\ldots?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Let M, N be metric spaces with metrics dm and dn respectively.
    Let f be a function from M to N.

    Suppose that if (an) is any convergent sequence in M, then (f(an)) is a convergent sequence in N.

    Show that f is a continuous function.



    Note that the question does not say that (an) --> a implies (f(an)) --> f(a). If it did then I could solve the question. But, as it is currently stated, I do not know the answer.
    Alternatively to Opalg's (this is true in any topological space) but let N be an arbitrary neighborhood of f(a) then f^{-1}(N) is a neighborhood of a (by f's continuity) and so we see that f^{-1}(N) is an open set containing a and since a_n\to a it contains all but finitely many of \{a_n\}_{n\in\mathbb{N}} so then ff^{-1}(N)\subseteq N contains all but finitely many values of f(a_n).
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    Quote Originally Posted by Drexel28 View Post
    f^{-1}(N) is a neighborhood of a (by f's continuity) .
    But the point is to show that f is continuous.
    So we cannot assume that it is, can you?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    But the point is to show that f is continuous.
    So we cannot assume that it is, can you?
    Oops! Misread the question. And well...I probably shouldn't give the full solution...but I do want to mention that this is true whenever the domain is a first countable space.
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    Quote Originally Posted by Opalg View Post
    Hint: Suppose that a_n\to a and that f(a_n)\to b. What can you say about the limit of the sequence a_1,a,a_2,a,a_3,a,\ldots?
    Thanks for hint, I see how it implies that (f(an)) converges to f(a).

    Just out of curiosity, is this really the best way to go about it? I would think there's a more "direct" route to take. Maybe I'm wrong.

    Either way, thanks a lot!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Thanks for hint, I see how it implies that (f(an)) converges to f(a).

    Just out of curiosity, is this really the best way to go about it? I would think there's a more "direct" route to take. Maybe I'm wrong.

    Either way, thanks a lot!
    There is, a way dealing with open balls around f(x) and x. Try writign out what continuity means and see if something doesn't "strike" you
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    Quote Originally Posted by southprkfan1 View Post
    I see how it implies that (f(an)) converges to f(a). Just out of curiosity, is this really the best way to go about it? I would think there's a more "direct" route to take.
    That is the very defintion of continuity.
    How can that be inproved upon?
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    Quote Originally Posted by Drexel28 View Post
    There is, a way dealing with open balls around f(x) and x. Try writign out what continuity means and see if something doesn't "strike" you
    Well, I'll tell you off the bat I have know idea what "strike" represents

    But, here's what I was working on before, and you can feel free to give me your thoughts.


    Suppose f is not continuous:

    Then, there exists e>0 and a xn (n=1,2,3,...) in M where

    dM(xn, a) < 1/n but dN(f(xn), f(a)) > e

    clearly (xn) converges to a, so by assumption (f(xn)) is a convergent sequence. Say it converges to f(x)

    Thus, for n large enough, we have dN(f(xn), f(x))< e + d(f(x),f(a)) and dM(an, a)<1/n

    but d<(xn, a) < 1/n --> e < dN(f(xn), f(a)) <= d(f(xn), f(x)) + d(f(x),f(a))

    --> dN(f(xn), f(a)) > e - dN(f(x),f(a))

    --> e + d(f(x),f(a)) > e + dN(f(x),f(a))

    --> d(f(x),f(a)) > 0

    --> f(x) /= f(a)

    And I'm stuck
    (well, not really, I could use the last posters hint to say f(x) = f(a), but in that case all this crap above would be unnecessary)
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Well, I'll tell you off the bat I have know idea what "strike" represents

    But, here's what I was working on before, and you can feel free to give me your thoughts.


    Suppose f is not continuous:

    Then, there exists e>0 and a xn (n=1,2,3,...) in M where

    dM(xn, a) < 1/n but dN(f(xn), f(a)) > e

    clearly (xn) converges to a, so by assumption (f(xn)) is a convergent sequence. Say it converges to f(x)

    Thus, for n large enough, we have dN(f(xn), f(x))< e + d(f(x),f(a)) and dM(an, a)<1/n

    but d<(xn, a) < 1/n --> e < dN(f(xn), f(a)) <= d(f(xn), f(x)) + d(f(x),f(a))

    --> dN(f(xn), f(a)) > e - dN(f(x),f(a))

    --> e + d(f(x),f(a)) > e + dN(f(x),f(a))

    --> d(f(x),f(a)) > 0

    --> f(x) /= f(a)

    And I'm stuck
    (well, not really, I could use the last posters hint to say f(x) = f(a), but in that case all this crap above would be unnecessary)
    How about this. I am going to assume a lot of stuff and let you generalize! Mainly that you call a function f continuous at x if \lim_{\xi\to x}f(\xi)=f(x).

    So, now suppose that x_n\to x\implies f(x_n)\to f(x) but \lim_{\xi\to x}f(\xi)=y\ne f(x). Then, since any metric space is Hausdorff we may find open sets O_y,O_{f(x)} such that y\in O_y,f(x)\in O_{f(x)} but O_y\cap O_{f(x)}=\varnothing, and since f(x_n)\to f(x) we have that O_{f(x)} contains all but finitely many points of f(x_n). Now the definition for \lim_{\xi\to x}f(x)=y means that every neighborhood E of y there exists a neighborhood N of x such that f\left(N-\{x\}\right)\subseteq E. But, given any neighborhood N of x we have that there exists all but finitely many points of \{x_n\} and so it follows that f\left(E-\{x\}\right)\cap O_{f(x)}\ne\varnothing\implies f\left(E-\{x\}\right)\nsubseteq O_y. This is clearly contradicts that \lim_{\xi\to x}f(\xi)=y. It follows that \lim_{\xi\to x}f(\xi)=f(x).

    Note: It is really late here and I may have a) made a mistake b) missed a much easier method. Also, I assumed that the limit exists...and it is easy to prove that it must.
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