# The set of continuity points is a $G_{\delta}$ set

I know there are other proofs to this question on this site, but my question is about this specific proof:

The set of continuity points of a function $$f$$ is a $$G_{\delta}$$ set.

A proof I read online went like this:

Let $$U$$ be an open neighborhood of a point $$x.$$ Then,

$$\text{Continuity points of f = }\left \{ x \in \mathbb{R} : \text{osc} (f,x) = 0 \right \} = \bigcap_{n=1}^{\infty}\left \{ x \in \mathbb{R} : \text{osc} (f,x) < 1/n \right \} = \bigcap_{n=1}^{\infty}\bigcup_{U}^{}U_i.$$

Now, I understand the the conclusion is an intersection of open sets, but I am not understanding this proof. Can someone break this down for me a little? Or perhaps provide a little more detail in what is going on?

• what are the sets $U_i$, is it a countable basis? Sep 14 at 22:16
• You have to be more specific. What exactly is your doubt? Sep 14 at 23:25
• @KaviRamaMurthy I am specifically not seeing the last equality Sep 14 at 23:26

What you need is the fact that $$\{x: osc(f,x) is an open set for each $$r>0$$. Let $$x$$ be a point in this set. Then $$\sup \{|f(y)-f(x)| :|y-x| \leq \delta\} for some $$\delta >0$$ (by definition of $$osc(f,x)$$). If $$z$$ is any point in $$(x-\delta, x+\delta)$$ then there exist $$s>0$$ such that $$[z-s,z+s] \subseteq (x-\delta, x+\delta)$$. Now $$osc (f,z) because $$\sup \{|f(y)-f(z)| :|y-z| \leq s\} \leq \sup \{|f(y)-f(x)| :|y-x| \leq \delta\}. We have proved that $$x$$ is an interior point of $$\{x: osc(f,x) .