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Function continuous, then a subset is closed

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Let M, N be two metric spaces. For f: M --> N, define the function on M,

    graph(f) = {(x,f(x)) [itex]\in[/itex]MxN: x[itex]\in[/itex]M}

    show f continuous => graph(f) is closed in MxN

    2. Relevant equations



    3. The attempt at a solution

    I can't figure out what method to use.

    I have written out many equivalent defintions of continuity and closed sets, so far I know I want to show that.. basically since f is continuous, we have convergent sequences mapped to convergent sequences, ie xn -> x, fn -> f

    And I need to show that (x,f(x)) [itex]\in[/itex] graph(f)

    It's either extremely trivial, because from the definition of continuity we see that all the limits belong to that subset..

    But anyways, I don't think I can say that or make that a proof by just stating the obvious..

    So I tried creating a neighbourhood and showing that M\graph(f) is open.

    Assume f is continuous. Then for all x in M, for all ε, there exists δ=θ/2 s/t dM(x,y)<δ => dN(f(x),f(y))<ε

    So consider B(y,ε'), choose ε' = δ = θ/2.
    For any z in B(y,ε')

    d(z,x) ≤ d(z,y) + d(x,y) < ε' + δ = θ

    Then d(z,x) < θ and z is in M\graph(f).....

    This is all probably very wrong.. I didn't use anywhere the definition of graph(f) which worries me.

    eek.. Help?
     
  2. jcsd
  3. Oct 6, 2011 #2
    Have you consider seeing what happens to points that _are not_ in the graph, i.e., to see what happens to the complement of the graph in MxN? Seeing closed sets as complements of open ones. of course.
     
  4. Oct 6, 2011 #3
    hm, im not sure I understand.. so if I take points in MxN, that are not in the graph, then we have (x,f(x)) but x is not in M, which wouldn't make sense, would it?
     
  5. Oct 6, 2011 #4
    No, I mean, MxN, and then take (m,n), with n not in f(M), i.e., n is in N-f(M); since f is a function, every argument m in M can have--has--exactly one output, so that if (m,f(m)) is in the graph, then (m,n') , with n'≠ f(m) _cannot_ be in the graph of f.
     
  6. Oct 7, 2011 #5
    Ok.. So to clarify, we have points in MxN,

    (m,n) , n not in f(M)
    (m,n') not in graph(f)

    Then do I use the defn of continuity on these two sets of points?
     
  7. Oct 7, 2011 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Which definition of "continuous" are you given? One is "f is continuous if and only if, for every open set U, [itex]f^{-1}(U)[/itex] is open". That may be what Bacle was thinking when he suggested that you look at complements which will be open if the given sets are closed.
     
  8. Oct 7, 2011 #7
    I'm just give that f is continuous... I'm not told that any set U is open though, so I don't know if I can use that definition.

    Is there a way to do this using a delta epsilon definition? Or neighbourhoods?
     
  9. Oct 8, 2011 #8
    What I was suggesting was using continuity to construct an open set UxV; U open in M, V open in N (so that UxV is open in the product MxN) , as an open set containing a point (m,n), with n≠ f(m), i.e., (m,n) is not in the graph of f , so that UxV misses the point (m,f(m)), then UxV is an open set containing a generic point in the complement, so that the complement of the graph is open, and so the graph is closed. Specifically, the continuity of f allows you to contain the image of the graph of f within an ε-ball of f(x)--where you can , again by continuity, make this ε as small as you want. Consider, e.g., the graph of f(x)=x2, and a point not in the graph, e.g., a point p=(x,x2+a) ; a≠ 0. Can you see how to construct an open set containing p, that misses the graph? Here M=N=the reals.
     
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