Product topology, closed subset, Hausdorff

[complexnumber]

Homework Statement

Let $$(X,\tau_X)$$ and $$(Y,\tau_Y)$$ be topological spaces, and let $$f : X \to Y$$ be continuous. Let $$Y$$ be Hausdorff, and prove that the graph of $$f$$ i.e. $$\graph(f) := \{ (x,f(x)) | x \in X \}$$ is a closed subset of $$X \times Y$$.

Homework Equations

The Attempt at a Solution

Which property of closed set should I use to prove this? Should I assume a sequence inside the graph set converging to some $$(x,y) \in X$$ and then somehow show that this limit point belongs to the graph? Or should I prove that the complement of the graph set is not open? I don't know how to finish the proof with either approach. Please give me some hint.

 

[VeeEight]

The idea here is to find (x,y) not on the graph such that every neighbourhood misses the graph. To do this, pick (x,y) not on the graph, so it is different than (x,f(x)) and separate them by neighbourhoods. You must also use continuity here.