Let $(f_1,\cdots,f_n)$ be a finite sequence of natural numbers where $f_i>0$ for all $i\le n$.
Let $A$ be a set and $F_i$ be an $f_i-$array operation on $A$ for all $i\le n$.
$B\subseteq A$ is closed if and only if $F_i(a_1,\cdots,a_{f_i})\in B$ for all $i\le n$ and $a_1,\cdots,a_{f_i}\in B$ provided that $F_i(a_1,\cdots,a_{f_i})$ is defined.
The closure of $B$, to be denoted by $\overline B$, is defined by $$\overline B=\bigcap\{C\in\mathcal P(A)\mid B\subseteq C \text{ and }C \text{ is closed}\}$$
If $B$ is countable, then $\overline B$ is countable
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!
My attempt:
Lemma 1: $\overline B=\bigcup\limits_{j\in\Bbb N}B_j$ where $(B_j\mid j\in\Bbb N)$ is defined recursively by $$B_0=B \text{ and } B_{j+1}=B_j\bigcup \bigg(\bigcup_{i=1}^{n}F_i[B_j^{f_i}]\bigg)$$ (I presented a proof here)
Lemma 2: Countable union of countable sets is countable (I presented a proof here)
Lemma 3: The Cartesian product of a finite number of countable sets is countable (I presented a proof here)
By Lemma 1, $\overline B=\bigcup\limits_{j\in\Bbb N}B_j$ where $(B_j\mid j\in\Bbb N)$ where $$B_0=B \text{ and } B_{j+1}=B_j\bigcup \bigg(\bigcup_{i=1}^{n}F_i[B_j^{f_i}]\bigg)$$
First, we prove that $B_j$ is countable for all $j\in\Bbb N$ by induction on $j$. Since $B_0=B$ and $B$ is countable, $B_0$ is countable. Assume $B_k$ is countable. Thus $B_k^{f_i}$ is countable for all $i\le n$ by Lemma 3. Hence $B_k\bigcup \bigg(\bigcup_{i=1}^{n}F_i[B_k^{f_i}]\bigg)$ is countable by Lemma 2. It follows that $B_{k+1}$ is countable. By principle of induction, $B_j$ is countable for all $j\in\Bbb N$. Hence $\overline B=\bigcup\limits_{j\in\Bbb N}B_j$ is countable by Lemma 2.
