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Problem.Let X be a topological space, and Y a T2-space (i.e. a Haussdorf topological space). Let f : X --> Y be a continuous function. One needs to show that the graph of , i.e. the set G = {(x, f(x)) : x is in X} is closed in X x Y.

Attempt of proof. To show what we need to show, we have to prove that the complement of G is open. Now, since Y is a Haussdorf space, one can show that it is a T1-space too, i.e. that every set containing a single point is closed. So, for every x, {f(x)} is closed in Y. Further on, since f is continuous, the preimage of every closed set is a closed set too, so, for any f(x), {x} is closed in X. I feel I'm getting very warm, but I got stuck anyway. The complement of G in X x Y is X x Y \ G = X x Y \ U (x, f(x)) (where the union goes through all x in X) = [itex]\cap[/itex] [(X x Y) \ (x, f(x))].Now,if X and Y are T1-spaces, can one show that X x Y is a T1 space, too? But then we'd in general have an infinite intersection of open sets, which doesn't need to be open.

Thanks in advance for a push.

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# Showing a graph is closed

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