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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?

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  • $\begingroup$ what are the sets $U_i$, is it a countable basis? $\endgroup$ Sep 14 at 22:16
  • $\begingroup$ You have to be more specific. What exactly is your doubt? $\endgroup$ Sep 14 at 23:25
  • $\begingroup$ @KaviRamaMurthy I am specifically not seeing the last equality $\endgroup$ Sep 14 at 23:26
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What you need is the fact that $\{x: osc(f,x) <r\}$ is an open set for each $r>0$. Let $x$ be a point in this set. Then $\sup \{|f(y)-f(x)| :|y-x| \leq \delta\} <r$ 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 $osc (f,z) <r$ because $\sup \{|f(y)-f(z)| :|y-z| \leq s\} \leq \sup \{|f(y)-f(x)| :|y-x| \leq \delta\}<r$. We have proved that $x$ is an interior point of $\{x: osc(f,x) <r\}$.

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