Proving a set is open by complement Let $f$ be continuous and defined on $\mathbb{R}$. Prove that for each real number $\alpha$, the set $A_\alpha = \{x \in \mathbb{R}: f(x) > \alpha\}$ is an open set.
I know that I need to prove the complement is closed, but I'm confused as to where I start (and how to finish for that matter).
Help?
 A: Since $\mathbb{R}$ is a metric space you can use: a set $C$ is closed if and only if for every sequence $\{x_n\}\subset C$ converging to a $x$ we must have $x\in C$.
Denote by $C:=A_\alpha^c=\{x\in\mathbb{R}:f(x)\le \alpha\}$. We want to prove that $C$ is closed. Let $\{x_n\}\subset C$ be a sequence converging to $x$. By definition we have $f(x_n)\le \alpha$ for every $n$ and $x_n\to x$. Since $f$ is continuous we know $f(x_n)\to f(x)$, therefore $f(x)=\lim_n f(x_n)\le \alpha$, hence $x\in C$ and you are done. 
A: Well, you know $A_{\alpha} = f^{-1}((\alpha, \infty))$. If you can show $(\alpha, \infty)$ is open, then by continuity of $f$, $A_{\alpha}$ must be open.
A: don't try to prove closeness.. instead  try for some thing like this :
to prove $A_{\alpha}$ is open what you need to do is :
given $x\in A_{\alpha}$... try finding $r>0 $ such that for all $y$ with $|x-y|<r$ implies $y\in A_{\alpha}$
believe me it is nothing more than using $\epsilon-\delta$ definition of continuity...
Sorry for giving more hints...
