in exercise 4.16 here: Jech - Set Theory, we are asked to prove that:
Given a function $f: \mathbb{R} \rightarrow \mathbb{R}$, the set of all points $x$ in which $f$ is continuous is $G_{\delta}$.
Something bothers me here since it seems that more of that is true. Given $f$ is open on $x$, there is an open neighborhood $U_x=(a_x,b_x)$ of $x$ where $f$ is continuous on every point of $U_x$. So the set of all points $x$ in which $f$ is continuous equals $\bigcup U_x$ and therefore open..
Am i missing something in here?
Thank you!! Shir
It is quite possible for a function to be continuous at a point without being continuous in a neighborhood of that point, as user73985 aptly demonstrates in the comments above.
As a hint, note that if $f:\Bbb R\to\Bbb R,$ then for all positive integers $n$ and all $x$ at which $f$ is continuous, there exists $\delta_{n,x}>0$ such that $$f(U_{n,x})\subseteq\left(f(x)-\frac1n,f(x)+\frac1n\right) $$ where $$U_{n,x}=\left(x-\delta_{n,x},x+\delta_{n,x}\right).$$ Let $U_n$ be the union of all $U_{n,x}$ where $f$ is continuous at $x,$ and consider $$\bigcap_n U_n.$$