Show graph is closed if f is continuous

Question

I need help figuring out how to complete this proof:

Show that the graph of $f$ is closed if $f$ is continuous.

Answer 1

Suppose $(x,y) \notin \text{Graph}(f)$, then $f(x)\neq y$, so $\exists V_1, V_2$ open neighbourhoods of $f(x)$ and $y$ respectively such that $V_1\cap V_2 = \emptyset$. Now note that $$ U := f^{-1}(V_1)\times V_2 \subset M\times \mathbb{R} $$ is an open neighbourhood of $(x,y)$ in $M\times \mathbb{R}$ and, for any $(x',y') \in U$, it follows that $f(x') \neq y'$. Hence $$ U \subset M\times \mathbb{R}\setminus \text{Graph}(f) $$ whence $M\times \mathbb{R}\setminus\text{Graph}(f)$ is open, which is what you want.

Answer 2

A direct proof is also possible in a metric setting. Let $\{(x_n,f(x_n))\}_n$ be a sequence from the graph of $f$. Assume that $x_n \to x_0$ and $f(x_n) \to y_0$. Since $f$ is continuous (at $x_0$), $\lim_n f(x_n)=f(x_0)$, and by uniqueness of the limit, $y_0=f(x_0)$. We conclude that $(x_0,y_0)=(x_0,f(x_0))$, a point on the graph of $f$. This implies that the graph is closed.