The set of all points $x$ in which $f$ is continuous is $G_{\delta}$

Question

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

$f(x) = x$ if $x \in \mathbb{Q}$, $f(x) = 0$ otherwise is continuous only at $0$. — Christopher, Dec 19 '13 at 14:19
What does it mean "$f$ open on x"? Do you mean continuous? Notice that a continuous function is not necessarily open. — Emanuele Paolini, Dec 19 '13 at 14:21
See here it's duplicate. http://math.stackexchange.com/questions/138072/how-to-show-that-the-set-of-points-of-continuity-is-a-g-delta — Haha, Dec 19 '13 at 14:29
@Mitsos: I must disagree. This particular question is related, but is asking about a specific misconception. — Cameron Buie, Dec 19 '13 at 14:58

score 4 · Accepted Answer · answered Dec 19 '13 at 14:37

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 $$\bigcap_n U_n.$$

The set of all points $x$ in which $f$ is continuous is $G_{\delta}$

1 Answers1