Show that the following is an open set

Question

Assume $f: (a,b) \to (0, \infty)$ is a continuous function.

Show that $ A= \{ (x,y) : a<x<b, 0<y<f(x) \} $ is open.

I want to show that for every point $(x_0,y_0) \in A$, there is $r>0$ such that $B((x_0,y_0), r)\subset A$.

I tried to set $r = \min ( x_0 - a, \,b - x_0,\, y_0, \,\inf \{||(x,f(x)) - (x_0,y_0)|| \}) $ but couldn't complete the proof.

I'm supposed to prove it with what I've wanted to show or to show that A is a union of open sets.

Am I on the right direction? Someone has a more elegant idea?

Answer 1

You are mostly on the right track, but you need to use the definition of $f$ being continuous somewhere in your attempt, as this is not true when $f$ isn't required to be continuous.

For any $p_0=(x_0,y_0)\in A$, let $\epsilon={\max(y_0,\frac {f(x_0)-y_0}2)}$. By definition of $f$ being continuous at $x_0$, there exist $\delta$ such that $|x-x_0|<\delta$ imples $|f(x)-f(x_0)|<\epsilon$.

Let $r=\min(\epsilon,\,\delta,\, x_0-a,\,b-x_0)$, then for any $p=(x,y)\in B_r(\,p_0)$, $|x-x_0|<\delta$, so $f(x)>f(x_0)-\epsilon$. We also have $y<\epsilon+y_0$, then $f(x)-y>f(x_0)-y_0-2\epsilon>0$, so $f(x)>y$. Obviously $y>y_0-\epsilon>0$, $x\in (a,b)$, so $p\in A$.