If $f:R \to R $ is continuous and $r_0 \in R $. Prove that {$x:f(x) \ne r_0$} is an open set.
Pf:Let $A=${$x:f(x) \ne r_0$}
If f is continuous then $\exists \delta >0 : \lvert x-x_0 \rvert < \delta \implies \lvert f(x)-f(x_0) \rvert < \epsilon$
Let $a \in A$, then $f(a) \ne r_0 $
Let $\epsilon =\frac{1}{2} \lvert f(a)-r_0 \rvert >0$
We have to prove that {$x: \lvert x-a \rvert <\delta $} $\subset A$
Since f is continuous, therefore {$f(x):\lvert f(x)-f(a) \rvert <\frac{\epsilon}{2}$}
Consider that $\lvert f(a)-r_0 \rvert \le \lvert f(x)-r_0 \rvert + \lvert f(x)-f(a) \rvert$
or $2 \epsilon < \lvert f(x) -r_0 \rvert + \frac{\epsilon}{2}$
$\therefore \lvert f(x)-r_0 \rvert > \frac{3 \epsilon}{2}$ and this implies $f(x) \ne r_0$ and {$x:\lvert x-a \rvert< \delta$} $\subset A$ or every point of x in A has a neighborhood inside A, thus A is open