Let $f:\mathbb{R}\rightarrow\bar{\mathbb{R}}$. Prove for $C(f)=\{x\in\mathbb{R} :f \text{ is continuous in }x\}$ is a $G_\delta$

Question

Definition: $A\subset X$ is a set $G_\delta$ iff $A=\bigcap_{n=1}^\infty G_n$ where $G_n$ is a open set for all $n$.

My attempt:

We need prove $C(f)=\bigcap_{n=1}^\infty G_n$ where $G_n$ is a open set for all $n$.$

Let $x\in C(f)$ then $f$ is continuous in $x$. This implies for all neighbourhood $ V \in (\bar{\mathbb{R}},\tau_\bar{\mathbb{R}}) $ then $f^{-1}(V)$ is open.

here i'm stuck. can someone help me?

A $G_\delta $ is $\bigcap_{n\in\mathbb N}O_n$ where $O_i$ are open. What you wrote is juste an open set. — user657324, May 14 '19 at 20:36
okay, then $f^{-1}(V)$ is a $G_i$ and how can i prove x in all $G_n$? — rcoder, May 14 '19 at 20:40
Check this one: https://math.stackexchange.com/questions/138072/how-to-show-that-the-set-of-points-of-continuity-is-a-g-delta — Danny Pak-Keung Chan, May 14 '19 at 21:26

Julian Mejia · Answer 1 · 2019-05-14T22:01:01.717

$x$ is point of continuity iff for all $n$ there is a neighborhood $U\ni x$ such that $diam f(U)<1/n$.

So, $$C(f)=\bigcap_{n} \bigcup_{U\in F_n} U$$ where $F_n=\{U open: diam f(U)<1/n\}$.

Since $\bigcup_{U\in F_n} U$ is a union of open sets it is open. Hence, $C(f)$ is $G_\delta$.

Edit To treat the case with image in $\overline{\mathbb{R}}$, we actually need to consider $F_n=\{U open: diam f(U)<1/n\hbox{ or } f(U)\subset (n,\infty)\}\hbox{ or } f(U)\subset (-\infty,-n)\}$.

Let $f:\mathbb{R}\rightarrow\bar{\mathbb{R}}$. Prove for $C(f)=\{x\in\mathbb{R} :f \text{ is continuous in }x\}$ is a $G_\delta$

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