# Show that the following is an open set

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

Show that $$A= \{ (x,y) : a 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?

• Do you know a property saying that a function $f : A \to B$ is continuous if an only if the preimage $f^{-1}(U)$ of any open subset $U$ of $B$ is open in $A$ ? Oct 20, 2020 at 19:41
• Sorry I've not mentioned this before - I'm not supposed to use that. Need to prove it with what I've wanted to show or show that A is a union of open sets.
– yong
Oct 20, 2020 at 19:45

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 $$p\in A$$.