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The graph is connected

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f:X→Y we a continuous map. Let
    Γ(x)={(x,f(x))∈X×Y}

    a) Show that if X is connected, then Γ is connected.
    b) Show that if Y is hausdorrf then Γ is a closed subset of X×Y.

    2. Relevant equations
    thm: if X is connected the image of a continuous map f:X→Y is connected.
    thm: The product space of two connected spaces is connected.
    thm: X is hausdorrf iff the diagonal is closed (?)
    def: A space is hausdorrf if each pair of distinct points have neighborhoods that are disjoint.


    3. The attempt at a solution

    a) Since f:X→Y is continuous, the image is going to be connected. Therefore image(f(X))[itex]\subset[/itex] Y is connected. Also given two connected spaces the product space is also connected, therefore X×image(f(X)) is connected. Therefore Γ is connected? This seemed too easy so im not sure if I left a huge hole here...

    b) Y is hausdorrf. Therefore given any points y1≠y2 we can find a disjoint neighborhoods U and V around y1 and y2 respectfully. I also know that the diagonal, D, is closed, {(y,y)|y is in Y}. Not sure what to do next. Maybe consider the complement and try to do a proof similar to the diagonal proof? The complement of D would be open. I dont think Y being hausdorrf implies that f(x1) and f(x2) have disjoint neighborhoods so I'm not sure where to go next.

    Edit: *Being Hausdorrf implies that sequences converge to at most one point. *Does this mean that f(x_n) converges to f(x) in this case? *

    I don't want any flat out answers, I just want to know what I should be looking for.
    Also new here to the physics forums, if my write up is awful please tell me :)
     
    Last edited: Oct 3, 2012
  2. jcsd
  3. Oct 3, 2012 #2
    I might have it. consider y≠f(x) then (y,f(x)) is in the complement of Γ. Take y in U and f(x) in V. Since Y is hausdorrf x and f(x) have disjoint nighborhoods in Y. Suppose U[itex]\cap[/itex] V ≠∅ then there exists x=f(x), which is a contradiction, therefore given any ordered pair (y,f(x)) the neighborhood UxV is open in the complement of the graph? Therefore the graph is closed?
     
  4. Oct 3, 2012 #3

    jbunniii

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    How did you conclude that? It's certainly not true that an arbitrary subset of a connected set is connected.

    Try working with the definition. Assume [itex]\Gamma(X)[/itex] is disconnected. Then [itex]\Gamma(X)[/itex] is the disjoint union of two open sets [itex]U[/itex] and [itex]V[/itex]. What can you say about the inverse images, [itex]\Gamma^{-1}(U)[/itex] and [itex]\Gamma^{-1}(V)[/itex], of these sets?
     
  5. Oct 3, 2012 #4
    I concluded this with a theorem from my book. "The image of a connected space under a continuous map is connected." So I thought maybe given X is connected the image of f was connected, and I just threw in the subset of Y because every point in the graph is an element in XxY.

    And to answer your question.
    Γ−1(U) and Γ−1(V) are disjoint and open in X?
     
    Last edited: Oct 3, 2012
  6. Oct 3, 2012 #5
    Actually since this is the inverse image of f, then wouldn't that mean that Γ−1(U) and Γ−1(V) are disjoint and open in X and their union is the whole space X? Contradicting the fact that X was connected?
     
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