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?

  • $\begingroup$ 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

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 $\{x: osc(f,x) <r\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.