Example which proves independence of Kuratowski's fourth closure axiom

Question

In his first volume of Topology, Kuratowski mentions an example which shows that his fourth axiom, $\overline{\overline{A}} = \overline{A}$, is independent from the other three of the closure operator. Take $X$ the set of all functions $\mathbb{R} \rightarrow \mathbb{R}$ and for $A \subseteq X$ define $\overline{A}$ as all functions obtained as pointwise limits of sequences of functions in $A$. Then, if $A$ is the continous functions, he claims that the Dirichlet function $f$ with $1$ on rationals and $0$ on irrationals is in $\overline{\overline{A}}$ but not in $\overline{A}$.

To show that $f \in \overline{\overline{A}}$, I noted that for any finite set of rationals, we can find $(g_n)_n$ a sequence of continous functions which converges to $0$ outside that set and $1$ on those rationals, by patching together different pieces of graphs of $a_nx^n$ and $1/(b_nx^n)$; this shows that $\overline{A}$ contains a function which is $1$ on any given finite set of rationals and $0$ outside. So by using a bijection from $\mathbb{Q}$ to $\mathbb{N}$ (like the diagonal one), we can choose these finite sets of rationals to be in a sequence which is $\mathbb{Q}$ in the limit, hence we have a sequence of functions in $\overline{A}$ which converges to $f$. Is this correct?

For $f \not\in \overline{A}$, I'd have to show that the pointwise limit of a continous sequence of functions cannot be everywhere discontinous (my intuition says it can probably only be discontinous on a Lebesgue negligible set), as suggested by this question. I can't think of how to prove this; can anyone provide the proof/a sketch of it/a resource where it is found?

Thank you for your time.

Of course, if all you want is to prove the independence of the fourth axiom, you can do it with a trivial example like $X=\mathbb N$, $\overline A=A\cup{x+1:x\in A}$. But Kuratowski chose to give a natural example which is interesting for its own sake. — bof, Sep 07 '20 at 11:54
In case you're interested about generalized topological spaces satisfying the first three axioms (but not necessarily satisfying the fourth axiom), see the sections titled 6. Čech Closure Operators and 7. Topological Closure Operators in my answers to General topological space with closure operation as in Russian translation of Hausdorff's 1914 and 1927 Mengenlehre. — Dave L. Renfro, Sep 07 '20 at 15:08

score 2 · Accepted Answer · answered Sep 07 '20 at 11:49

Assume for a contradiction that $\{f_n\}$ is a sequence of continuous functions converging pointwise to the Dirichlet function $f$.

For $N\in\mathbb N$ let $S_N=\{x\in\mathbb R:|f_n(x)-f(x)|\le\frac13\text{ for all }n\ge N\}$.

Since $f_n\to f$ pointwise, $\bigcup_{N=1}^\infty S_N=\mathbb R$. By the Baire category theorem, not all of the sets $S_N$ are nowhere dense. Choose $N$ so that the set $S_N$ is dense in some interval $(a,b)$. Choose $c,d$ so that $a\lt c\lt d\lt b$, $f(c)=0$, $f(d)=1$. Choose $n\ge N$ so that $f_n(c)\lt\frac12\lt f_n(d)$.

Since $n\ge N$, we have $f_n(x)\notin(\frac13,\frac23)$ for $x\in S_N$. Since $S_N$ is dense in $(c,d)$, and since $f_n$ is continuous, we have $f_n(x)\notin(\frac13,\frac23)$, and in particular $f_n(x)\ne\frac12$, for $x\in(c,d)$, but this contradicts the intermediate value theorem.

Example which proves independence of Kuratowski's fourth closure axiom

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