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Math Help - Prove continuous -> graph is closed

  1. #1
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    Prove continuous -> graph is closed

    Prove that if f: X \rightarrow Y is a continuous map of a space X to a Hausdorff space Y then its graph

    G_{f} = ({ (x,f(x)) \in X \times Y : x \in X})

    is a closed subset of X x Y.
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  2. #2
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    Let (x,y) be a point in the closure of G_f, and let \bigl((x_n,f(x_n))\bigr) be a sequence of points in G_f with limit point (x,y). Then x_n\to x and f(x_n)\to y. But f is continuous, and therefore f(x_n)\to f(x). Therefore y=f(x) and hence (x,y) = (x,f(x))\in G_f. Thus every limit point of G_f is in G_f, and so G_f is closed.

    (If the space is "too big" for closures to be detected by sequences then you will have to replace "sequence" by "directed net" in that proof.)
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    Wow, nice answer. I have a feeling though that it's a bit "too" nice - for example we haven't even used that Y is Hausdorff. It seems strange to specify something irrelevant in the question...But still, thanks anyway.
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  4. #4
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    Quote Originally Posted by Amanda1990 View Post
    Wow, nice answer. I have a feeling though that it's a bit "too" nice - for example we haven't even used that Y is Hausdorff. It seems strange to specify something irrelevant in the question...But still, thanks anyway.
    The proof does use the information that Y is Hausdorff (though I didn't actually point that out). It is used to deduce that if f(x_n)\to f(x) and f(x_n)\to y then f(x) = y. (In a Hausdorff space, a sequence has a unique limit. In a non-Hausdorff space, it's possible for a sequence to have more than one limit.)
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    Good point. I had completely forgotten about this.
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