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

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Show that the points at which $f$ is continuous is a $G_{\delta}$ set.

$$A_n = \{ x \in \mathbb{R} | x \in B(x,r) \text{ open }, f(x'')-f(x')<\frac{1}{n}, \forall x',x'' \in B(x)\}$$

I saw that this proof was already on here, but I wanted to confirm and flesh out more details.

"$\Rightarrow$" If f is continuous at $x$, then $f(x'')-f(x')<\frac{1}{n}$ for $x'',x' \in B(x, r_{n})$. That is, there is a ball of radius $r$ where $r$ depends on $n$. Then $x \in A_n$ and thus $x \in \cap A_n$.

"$\Leftarrow$" If $x \in \cap A_n$, then there is an $\epsilon > 0$ and a $\delta > 0$ such that $x' , x'' \in B(x, \delta_n)$ for all $n$ and $$|f(x'')-f(x')|<\epsilon.$$ Take $\epsilon = \frac{1}{n}$.

Answer 1

The definition of $A_n$ is a bit confusing. I think that what you want here is to let

$$A_n=\left\{x\in\Bbb R:\exists r_n(x)>0\,\forall x',x''\in B\big(x,r_n(x)\big)\left(\left|f(x'')-f(x')\right|<\frac1n\right)\right\}\;.$$

You definitely want the absolute values, and you need to say that it’s the points $x$ for which such a neighborhood $B\big(x,r_n(x)\big)$ exists. You don’t have to indicate explicitly the dependence of $r$ on $n$ and $x$ as I did here, but it doesn’t hurt, especially when you’re learning.

Now let $G=\bigcap_{n\in\Bbb Z^+}A_n$, and let $C=\{x\in\Bbb R:f\text{ is continuous at }x\}$. You need to show three things:

1. Each $A_n$ is open.
2. $C\subseteq G$. This is your ‘$\Rightarrow$’.
3. $G\subseteq C$. This is your ‘$\Leftarrow$’.

You omitted (1) altogether, but it’s not hard: just show that if $x\in A_n$, then $B\big(x,r_n(x)\big)\subseteq A_n$, and conclude that $A_n=\bigcup_{x\in A_n}B\big(x,r_n(x)\big)$ and hence is open.

You’ve essentially got (2), but it could be stated much more clearly. Suppose that $x\in C$ and $n\in\Bbb Z^+$. Then there is an $r_n(x)>0$ such that $|f(x')-f(x)|<\frac1{2n}$ for all $x'\in B\big(x,r_n(x)\big)$. But then by the triangle inequality $$|f(x'')-f(x')|\le|f(x'')-f(x)|+|f(x)-f(x')|<\frac1n$$ for all $x',x''\in B\big(x,r_n(x)\big)$, so $x\in A_n$. And since $n\in\Bbb Z^+$ was arbitrary, $x\in G$.

Much the same applies to (3). Suppose that $x\in G$, and let $\epsilon>0$ be arbitrary. There is an $n\in\Bbb Z^+$ such that $\frac1n\le\epsilon$, and $x\in A_n$, so $|f(x')-f(x)|<\frac1n\le\epsilon$ for all $x'\in B\big(x,r_n(x)\big)$, i.e., for all $x'$ such that $|x'-x|<r_n(x)$, and it follows immediately that $x\in C$.

Answer 2

Here's a slightly different approach.

Let $G$ be the set of points where $f$ is continuous, $A_{n,x} = (x-\frac{1}{n}, x+ \frac{1}{n})$ is an open set where $f$ is continuous, and $A_n = \bigcup_{x \in G} A_{n,x}$.

Since $A_n$ is union of open sets which is open, $\bigcap_{n \in \mathbb{N}} A_n $ is a $G_\delta$ set. We want to show $\bigcap_{n \in \mathbb{N}} A_n = G$.

$\forall n \in \mathbb{N}, A_{n+1, x} \subset A_{n,x} \\ \Rightarrow \bigcup_{x \in G}A_{n+1,x} \subset \bigcup_{x \in G}A_{n,x} \\ \Rightarrow A_{n+1} \subset A_n \\ \Rightarrow \{A_n\} \text{ is a decreasing sequence of nested sets}$

Furthermore, $\bigcap_{n \in \mathbb{N}} A_n = \{x\} $. (i.e. intersection is non-empty by Nested Set Theorem / Cantor Theorem). Therefore,

$\begin{align}\bigcap_{n \in \mathbb{N}} A_n &= \lim_{n \to \infty} A_n \\ &= \lim_{n \to \infty} \bigcup_{x \in G} A_{n,x} \\ &= \bigcup_{x \in G} (\lim_{n \to \infty} A_{n,x} ) \\ &= \bigcup_{x \in G} \{x\} \\ &= G\end{align}$

So $G$ is the intersection of open sets and is a $G_\delta$ set. Can someone see why interchanging the limit and union is legal?