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First, I found two other posts of people who asked this question.

How to show that the set of points of continuity is a $G_{\delta}$

Set of points of continuity are $G_{\delta}$

However, because of the way that I'm attempting to solve it, I didn't find these threads to be helpful.

The question is to show that

Let $f$ be a real valued function on $\mathbb{R}$, the set of points at which $f$ is continuous is a $G_\delta$ set.

Note that a $G_{\delta}$ set is a the intersection of a countable collection of open sets.

My solution idea:

We know that for a set $E$ of real numbers, $f$ is continuous on $E$ if and only if for each open set $O$ in the range of the function, $f^{-1}(O) = E \cap U$ where $U$ is an open set$.

So denote the set of points at which $f$ is continuous by $E$. Then for each open set $O_i$ in the range of $f$, we have $f^{-1}(O_i) = E \cap U_i$.

Now maybe I could use the fact that the $O_i$ in the range must be countable?

I feel like I'm close, but I'm stuck at this point. Can anyone point me in the right direction?

Thanks.

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  • What do you mean by "the $O_i$ in the range must be countable"? – Jonas Meyer Feb 03 '17 at 22:20
  • Note that having the restriction of $f$ to $E$ being continuous is a weaker condition than having $f$ with domain $\mathbb R$ be continuous at each point in $E$, so it doesn't seem like this approach will work. This is because being continuous at $x\in\mathbb R$ requires good behavior of the function near $x$, including at points that might not be in $E$. – Jonas Meyer Feb 03 '17 at 22:22
  • Each open set $O_i$ in the range must contain a rational number and we can put the rational numbers in one to one correspondence with $\mathbb{N}$ so we can put the open intervals in the range in one to one correspondence with $\mathbb{N}$. Maybe I should have said there is a countable number of disjoint open intervals in the range? – 1233211 Feb 03 '17 at 22:23
  • There are an uncountable number of open intervals in $\mathbb R$. What are these disjoint intervals? Perhaps you are thinking of the fact that an open set in $\mathbb R$ is a countable disjoint union of open intervals, but you are not restricting to one open set, but taking an arbitrary open set, of which there are uncountably many. – Jonas Meyer Feb 03 '17 at 22:24
  • Ahh, I see what you're saying. Thank you. I will try another approach. – 1233211 Feb 03 '17 at 22:27

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