# Prove continuous -> graph is closed

• Feb 10th 2009, 11:57 AM
Amanda1990
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.
• Feb 11th 2009, 12:32 PM
Opalg
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.)
• Feb 11th 2009, 12:42 PM
Amanda1990
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 PM
Opalg
Quote:

Originally Posted by Amanda1990
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.)
• Feb 11th 2009, 11:40 PM
Amanda1990