Function is continuous if graph is compact.

Question

Let $X$ be a Hausdorff space and let $f:X\to \mathbb{R}$. If grapph of $f$ is compact we have to show that $f$ is continuous.

Since every closed subset of a Hausdorff space is closed, therefore grapph of $f$ is closed. WE know that if $f:X\to Y$ and $Y$ is compact, then graph of $f$ is clsed implies $f$ is continuous. But here $\mathbb R$ is not compact. Please help!

Answer 1

Proof without nets:

Let $G$ be the graph of $f$.

Suppose $G$ is compact.

Let $A$ be closed in $\mathbb R$. We show $f^{-1} [A]$ is closed.

Note that $f^{-1}[A]=\pi_X [G\cap (X\times A)]$.

Since $G\cap (X\times A)$ is compact and the projection $\pi _X$ is continuous, $f^{-1}[A]$ is compact.

$X$ is Hausdorff, so every compact subset of $X$ is closed. DONE!

Answer 2

Fix $x_0 \in X$ and let $x_a$ be a net converging to $x_0$ in $X$. You need to show $f(x_a)\rightarrow f(x_0)$ in $\mathbb{R}$. Let $G_f$ denote the graph of $f$, which is a subset of $X \times \mathbb{R}$. Then $(x_a,f(x_a))$ is a net in $G_f$, which is compact, so we can extract a convergent subnet $(x_b,f(x_b)) \rightarrow (x,y) \in G_f$. But then $x_b \rightarrow x$ and since $x_a \rightarrow x_0$, we have $x=x_0$ by uniqueness of limits (Hausdorff). Then $y=f(x_0)$ since $(x,y) \in G_f$. This means $f(x_b) \rightarrow f(x_0)$.

Now, do this argument for ANY subnet of $x_a$ to conclude that the whole net $f(x_a)$ converges to $f(x_0)$.