# Show a neighborhood exists

Let $p$ such that $r=f(p)>0$ and $f$ is continous at $p$ then For $\epsilon =r$ there exist a $\delta >0$ such that $\vert f(p)-f(x) \vert < r$ whenever $\vert x-p \vert < \delta$. And so with this $\delta$ we have $f(x)>0$