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

is a closed subset of X x Y.

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- Feb 10th 2009, 12:57 PMAmanda1990Prove continuous -> graph is closed
Prove that if is a continuous map of a space X to a Hausdorff space Y then its graph

is a closed subset of X x Y. - Feb 11th 2009, 01:32 PMOpalg
Let (x,y) be a point in the closure of , and let be a sequence of points in with limit point (x,y). Then and . But f is continuous, and therefore . Therefore y=f(x) and hence . Thus every limit point of is in , and so 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.) - Feb 11th 2009, 01:42 PMAmanda1990
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.

- Feb 12th 2009, 12:32 AMOpalg
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 and then . (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.)

- Feb 12th 2009, 12:40 AMAmanda1990
Good point. I had completely forgotten about this.