Let $X$ be a Hausdorff space. Let $f:X\rightarrow \mathbb{R}$ be such that $\{(x,f(x)):x\in X\}$ is a compact subset of $X\times\mathbb{R}$. Show that $f$ is continuous.
What i have done so far is :
$X$ is Hausdorff and $\mathbb{R}$ is Hausdorff. So, $X\times \mathbb{R}$ is Hausdorff.
Compact subset of Hausdorff space is closed.
Given that $\{(x,f(x)):x\in X\}$ is a compact subset of $X\times\mathbb{R}$.
So, $\{(x,f(x)):x\in X\}$ is a closed subset of $X\times\mathbb{R}$.
Now, i am trying to prove that Graph is closed implies function is continuous..
I have checked some questions in this site but they have mentioned just for metric spaces...
Please give only hints..
We want to show for any open set $U\subset \mathbb{R}$, $f^{-1}(U)$ is open in $X$. Since every open set in $\mathbb{R}$ is a countable union of disjoint open intervals and a countable union of open sets is open, we just need to show $f^{-1}((a, b))$ is open in $X$.
To show this , by definition $f^{-1}((a, b))=\{x\in X: f(x)\in (a, b)\}$, the complement of it is $\{x: f(x)\geq b\}\cup\{x: f(x)\leq a\}$, let's show the complement is closed in $X$.
Notice that $X\times[ b,\infty)$ is closed in $X\times \mathbb{R}$, hence the intersection $(X\times [b, \infty))\cap\{(x, f(x)): x\in X\}=\{(x, f(x)): x\in X, f(x)\geq b\}$ is compact in $X\times \mathbb{R}$. Since the projection map is continuous, we have $\{x: f(x)\geq b\}$ is closed, so is $\{\{x: f(x)\leq a\}\}$.