Show that the set of points where a real-valued continuous sequence of functions converges is $F_{\sigma\delta}$

Question

By $F_{\sigma\delta}$, I mean that the set can be expressed as a countable intersection of $F_\sigma$ sets.

Let this sequence of functions be $f_n$, and the set of points where $f_n$ converges be $C$. Since $f_n$ must be Cauchy, I can define $C$ as

$$C = \{x \in \mathbb{R}\:|\: \forall\epsilon\in\mathbb{N} \; \exists N\in\mathbb{N} \::\:|f_m(x) - f_n(x)| < \frac{1}{\epsilon} \: \forall m,n > N\}$$.

Looking at this set, I'm trying to prove $F_{\sigma\delta}$ by firstly trying to find some countable intersection. I try to rewrite $C$ as

$$C = \bigcap\limits_{\epsilon \in \mathbb{N}}\{x \in \mathbb{R}\:|\: \exists N\in\mathbb{N} \::\:|f_m(x) - f_n(x)| < \frac{1}{\epsilon} \: \forall m,n > N\}$$.

Now I need to express each set being intersected as a countable union of closed sets to prove that it is $F_\sigma$. Then the countable intersection of these must be $F_{\sigma\delta}$. I tried to pick $N$ to describe this union, and rewrote $C$ as

$$C = \bigcap\limits_{\epsilon \in \mathbb{N}}\bigg[\;\bigcup\limits_{N \in \mathbb{N}}\{x \in \mathbb{R}\:|\:|f_m(x) - f_n(x)| < \frac{1}{\epsilon} \: \forall m,n > N\}\;\bigg]$$.

But this option confuses me because $N$ depends on $\epsilon$, so I don't know whether it would make sense to let both vary freely like that. Am I going in the wrong direction?

Answer 1

Actually I feel a hint is not enough here. I needed to think a bit more as well so here is an attempt at a solution.

The first identity to notice is this:

$$C = \bigcap_{k \in \Bbb N} \; \underbrace {\{ x \in \Bbb R\ : \exists N \in \Bbb N \; \text{s.t.} \; m, n \gt N \implies |f_m(x) - f_n(x)| \le \frac 1 k \}}_{B_k} $$

The second is:

$$B_k = \bigcup_{n \in \Bbb N} \; \underbrace{\{ \ x \in \Bbb R : |f_m(x) - f_n(x)| \le \frac 1 k\ \; \text{for each} \;m\ge n \}}_{A_{n, k}}$$

The third is:

$$A_{n, k} = \bigcap_{t \ge n} \; \underbrace{\{x \in \Bbb R : |f_t(x) - f_n(x)| \le \frac 1 k \}}_{F_{t, n,k}}$$

Now each $F_{t, n,k}$ is a closed set since it is the pre-image of a closed set under a continuous function, $g = |f_t - f_n|$. Since any intersection of closed sets is closed, $A_{n,k}$ is closed. Hence, $B_k$ is an $F_\sigma$ set for each $k \in \Bbb N$.