How to prove that there does not exist a sequence of continuous functions that converge pointwise to $\chi_{\mathbb{Q}}$ (definition only)

Question

A fellow member of the community asked: "there isn't a sequence of continuous function on $[0,1]$ that converges pointwise to the function $f$ on $[0,1]$ defined by $f(x)=0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational." This is the reverse of $\chi_{\mathbb{Q}}$, but the proof is obviously the same.

I'm wondering about a solution that only uses the definition of continuity and point-wise convergence. I answered the OP with the following attempt:

Attempt 1: " Suppose that $\{f_{n}\}_{n=1}^{\infty}$ is a point-wise convergent sequence that converges to $f$. Let $x \in [0,1]$ and WLOG, suppose that $x$ is irrational.

Let $\epsilon>0$. WLOG, suppose that $\epsilon<1$.

1) Then there exists $N_{1} \in \mathbb{N}$ such that $b \in \mathbb{N}$ and $b>N_{1}$ imply that $|f_{b}(x)-f(x)|=|f_{b}(x)-1|<\frac{\epsilon}{2}$.

2) Since $\{f_{n}\}_{n=1}^{\infty}$ is a sequence of continuous functions, there exists some $\delta>0$ such that $|x-y|<\delta$ implies that $|f_{b}(y)-f_{b}(x)|<\epsilon$ for each $y \in [0,1]$. Let $y \in [0,1]$ such that $|x-y|<\delta$. Then choose $p$ such that $p \in \mathbb{Q}$, where $x<p<y$.

4) Since $\{f_{n}\}_{n=1}^{\infty}$ is pointwise convergent, there exists $N_{2} \in \mathbb{N}$ such that $a \in \mathbb{N}$ and $a>N_{2}$ imply that $|f_{a}(p)-f(p)|=|f_{a}(p)|<\frac{\epsilon}{2}$.

5) Let $N=\max\{N_{1},N_{2}\}$. Suppose that $n>N$. By hypothesis, $|f_{n}(p)|<\frac{\epsilon}{2}$ and $|f_{n}(x)-1|<\frac{\epsilon}{2}$. However, by (3), we know that $|f_{n}(x)-f_{n}(p)|<\epsilon$. But this is clearly a contradiction."

However, this doesn't work, since it assumes that the same $\delta$ will characterize continuity for any $f_{n}$ where $n>N$.

**as a side-remark, is there any literature on the idea of a single $\delta$ working to describe continuity in a sequence of point-wise convergent functions $f_{n}$ for sufficiently large $n$?

Attempt 3: sigh, here is another proof that ultimately rests on baire category for contradiction:

Let $\{f_n(x)\}$ be a sequence of continuous functions that converge pointwise.

We show that $A=\{x \in \mathbb{R} \mid f_n(x) \textrm{converges}\}$ is a Borel set [and more specifically, $G_\delta$.

If $x \in A$, then there exists $N \in \mathbb{N}$ so that $n,m \geq N$ implies that $V_{x,n,j}=|f_n(x)-f_m(x)|<\frac{1}{j}$ for each $j \in \mathbb{N}$. Then Let $V_{n,j}=\bigcup V_{x,n,j}$. Since $V_{n,j}$ is open, let the open set $U_{n,j}$ be defined by $U_{N,J}=F^{-1}(V_{n,j})$. But then $$\bigcap_{j \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n,m \geq N} U_{n,j}=A$$ is $G_\delta$

Then use Baire category to show that $\mathbb{Q}$ is not $G_\delta$, and so the result follows.

I'm still interested in a proof that does not use this, and also in a possible proof verification for my proof given in the answers section.

Answer 1

A way to show that $ Q$ is not $G_{\delta}$ in $R$ without the Baire Category theorem:

(1).A non-empty completely metrizable space with no isolated points is an uncountable space.(In fact it has a subspace homeomorphic to the Cantor set.)

(2).A subspace $Y$ of a completely metrizable space $X$ is completely metrizable iff $Y$ is $G_{\delta}$ in $X.$

Therefore, since the space $Q$ has no isolated points, it cannot be completely metrizable, else by (1) it would be uncountable. So by (2), $Q$ is not $G_{\delta}$ in $R.$

Both (1) and (2) are provable by elementary, but not brief, methods, without any reference to the Baire Category theorem.

Answer 2

I think I have an answer, I received some assistance from a friend of mine. Slight question: does it depend implictly on any non-stated theorems (I mean, so called "big theorems.")

Suppose that $\{f_{n}\}_{n=1}^{\infty}$ is a point-wise convergent sequence that converges to $f$. Let $a,b \in [0,1]$ such that $0<a<b<1$. We will show that for each non-degenerate segment $A \in [0,1]$, there exists some $B \subseteq A$ and arbitrarily large $N \in \mathbb{N}$ such that $f_{N}(B) = [a,b]$.

Let $A$ be a non-degenerate segment of $[0,1]$. Let $y \in (\mathbb{R}-\mathbb{Q})\cap A$ and $x \in \mathbb{Q} \cap A$ such that $x < y$. Since $\{f_{n}\}_{n=1}^{\infty}$ converges pointwise to $f$, we have that there exists $N \in \mathbb{N}$ such that $f_{N}(x) \geq b$ and $f_{N}(y) \leq a$. But since $f_{N}$ is continuous,by the Intermediate Value Theorem, there exists some $ B \subseteq [x,y] \subseteq A \subseteq [0,1]$ such that $f_{N}(B)=[a,b]$.

Then consider the embedded sequence of $\{A_{n}\}$, where $A_{n+1} \subseteq A_{n}$. Clearly for each $A_{n}$ there exist respective $K_{n} \in \mathbb{N}$ such that $f_{K_{n}}(B_{n})=[a,b]$ for $B_{n} \subseteq A_{n}$. By the nested interval theorem, the intersection of $A_{n}$ is non-empty. Then let $x \in \bigcap_{n \in \mathbb{N}}A_{n}$. But then $f(x)$ is the limit of $f_{n}(x) \in [a,b]$, which is a contradiction since $f(x)=0$ or $f(x)=1$ and $0<a<b<1$.

**edit: i think this proof is incorrect since it makes little reference to the rationals (Although I suppose it assumes denseness.)