# Thread: The collection of continuous point of a real-valued function

1. ## The collection of continuous point of a real-valued function

Let f be a real-valued function defined for all real numbers, C be the set of points at which f is continuous, Show that C is a $G_\delta$ set.
Any help is appreciate !

2. Originally Posted by Shanks
Let f be a real-valued function defined for all real numbers, C be the set of points at which f is continuous, Show that C is a $G_\delta$ set.
Write $B(x,r)$ for the open interval (x–r,x+r). For n=1,2,3,..., define $G_n = \{x\in\mathbb{R}:\exists\delta>0\text{ such that }f(B(x,\delta)\subseteq B(f(x),1/n)\}$. Then $G_n$ is open, so $G = \textstyle\bigcap_{n=1}^\infty G_n$ is a $G_\delta$-set. Check that $G$ is the set of points at which f is continuous.

3. ## counterexample

I did the same things as you, but I can't proof that each $G_n$ is open.
Here I give a counterexample that $G_1$ is not open :
let f be defined as :
$f(0)=0$;
$f(x)=\frac{3}{4}$ if x is in the set of rational numbers except 0;
$f(x)=-\frac{3}{4}$ if x is in the irrational numbers set.
Then $G_1=\{0\}$ is not open.
Any way , thank you all the same !
I've also got a new idea!
I made a little change to $G_n$: By let
$G_n = \{x\in\mathbb{R}:\exists\delta>0\text{ such that }|f(x_1)-f(x_2)|\leq 1/n \text{ for any } x _1, x_2\in B(x,\delta)\}$

4. Originally Posted by Shanks
I made a little change to $G_n$: By let
$G_n = \{x\in\mathbb{R}:\exists\delta>0\text{ such that }|f(x_1)-f(x_2)|\leq 1/n \text{ for any } x _1, x_2\in B(x,\delta)\}$
You're quite right. I realised overnight that my sets $G_n$ need not be open. Your modification above seems to repair that error.