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

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

is a closed subset of X x Y.

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- Feb 10th 2009, 11:57 AMAmanda1990Prove continuous -> graph is closed
Prove that if $\displaystyle f: X \rightarrow Y$ is a continuous map of a space X to a Hausdorff space Y then its graph

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

is a closed subset of X x Y. - Feb 11th 2009, 12:32 PMOpalg
Let (x,y) be a point in the closure of $\displaystyle G_f$, and let $\displaystyle \bigl((x_n,f(x_n))\bigr)$ be a sequence of points in $\displaystyle G_f$ with limit point (x,y). Then $\displaystyle x_n\to x$ and $\displaystyle f(x_n)\to y$. But f is continuous, and therefore $\displaystyle f(x_n)\to f(x)$. Therefore y=f(x) and hence $\displaystyle (x,y) = (x,f(x))\in G_f$. Thus every limit point of $\displaystyle G_f$ is in $\displaystyle G_f$, and so $\displaystyle G_f$ 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.) - Feb 11th 2009, 12:42 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 11th 2009, 11:32 PMOpalg
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 $\displaystyle f(x_n)\to f(x)$ and $\displaystyle f(x_n)\to y$ then $\displaystyle f(x) = y$. (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.)

- Feb 11th 2009, 11:40 PMAmanda1990
Good point. I had completely forgotten about this.