Every finitely generated group is at most countable

Question

Context. We are are working in $\mathsf{ZF}$, but my professor told we can also use (if needed) that if $X$ is at most countable, then $|\mathcal{P}^{<\aleph_0}(X)|\leq \aleph_0$ (where $\mathcal{P}^{<\aleph_0}(X)$ is the set of all finite subsets of $X$).

Problem. Given a set $X$, a function $f : X \times X \to X$ and $A \subseteq X$ such that $|A| \leq \aleph_0$, prove that there exists a subset $\overline A \subseteq X$, such that $A \subseteq \overline A$, $|\overline A| \leq \aleph_0$ and $f[\overline A \times \overline A] \subseteq \overline A$. From this conclude that every finitely generated group is at most countable.

I cannot find neither an idea to prove the first statement, nor I can see how it is related to the second question, does anyone has a suggestion?

Edit 1.: there was actually a mistake in the text, my sincere apologies to everyone who was trying to help in the comments, my bad :'-(

Edit 2.(Almost solution): Thanks everyone for the helpful suggestions! I think I've almost put all the pieces together, but there's still something left. I defined (as suggested by @ArturoMagidin) the following recursion: $$A_n := \begin{cases} A_0 = A \\ A_{n + 1} = f[A_n \times A_n] \cup A_n \end{cases}$$ and then $\overline{A} = \bigcup_{n \in \omega} A_n$. Now it's easy to check that $\overline A \subseteq X$ and $f[\overline A \times \overline A] \subseteq \overline A$, the only thing that it's causing me troubles is to prove $|\overline A| \leq \aleph_0$. An idea I've come up with is to use that the at most countable union of at most countable sets is at most countable (which is easly follow from $\mathsf{AC}$), but if I can provide an explicit sequence $(\alpha_n)_{n \in \omega}$ of enumerations for $\{A_n\}_{n \in \omega}$ (i.e. for each $n \in \omega$ $\alpha_n : \omega \to A_n$ is surjective) I can use the fact above without relying the axiom of choice.

So the main idea woud be to define the map $n \mapsto \alpha_n$ by recursion and then prove by induction that is surjective. From the hypothesis that $|A| \leq \aleph_0$ I can easily define $\alpha_0$ as a fixed enumeration, but I cannot find a good expression for $\alpha_{n + 1}$ from $\alpha_n$.

I've not found a simpler manner to prove $|\overline A| \leq \aleph_0$ so far.

Answer 1

Assuming you mean to require $\overline{A}$ to contain $A$, the way to use the hint to prove the result is to consider the function $(x,y)\mapsto xy^{-1}$. Then the fact that $\overline{A}$ satisfies $f(\overline{A},\overline{A})\subseteq \overline{A}$ tells you that, if nonempty, it is a subgroup. Then take $A$ to be a generating set.

For the first part, I would suggest constructing $\overline{A}$ as a countable union of countable sets constructed recursively from $A$ via $f$.

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Can we prove that $\overline{A}$ defined in your most recent edit (2024-04-04, 18:04:17Z) is countable without invoking $\mathsf{AC}$?

I don't think the naive idea you have wouldn't quite work if $A$ is infinite, since your initial $\alpha_0$ may already be a bijection, not sure how you would "extend it".

But we can certainly inject $\overline{A}$ to $\omega\times\omega$, which can be proven to be countable in $\mathsf{ZF}$ without invoking $\mathsf{AC}$, since $\omega$ is already well-ordered. Give $\omega\times\omega$ the "zig-zag" order: $$(a,b)\lt(r,s)\iff a+b\lt r+s\text{ or }(a+b=r+s\text{ and }b\gt s).$$ This gives $\omega\times\omega$ the order $$(0,0)\lt (0,1)\lt (1,0) \lt (0,2)\lt (1,1) \lt (2,0) \lt\cdots$$ which is of type $\omega$, so we get an explicit injection $\omega\times\omega\to \omega$.

Proceed as follows: let $\varepsilon\colon A\to\omega$ be a one-to-one function. This exists because $A$ is countable.

We can define $\alpha_0\colon A_0\to \omega\times\{0\}$ by $\alpha_0(a) = (\varepsilon(a),0)$.

Fix $n\in\omega$, and assume that for all $k\leq n$ we have a one to one function $\alpha_k\colon A_k\to \omega\times\{0,\ldots,k\}$ with $k\leq\ell\leq n\implies \alpha_{\ell}|_{A_k} = \alpha_k$. We wish to define $\alpha_{n+1}\colon A_{n+1}\to \omega\times\{0,\ldots,n+1\}$ with $\alpha_{n+1}|_{A_k} = \alpha_k$ for all $k\leq n$.

Using $\alpha_n$, we can define a well-ordering on $A_n\times A_n$ by $(a,b)\leq(c,d)\iff (\alpha_n(a),\alpha_n(b))\leq_z(\alpha_n(c),\alpha_n(d))$, where $\leq_z$ is the zig-zag order on $\omega\times \omega$.

Let $B=f(A_{n}\times A_n)\setminus A_{n}$. Then we well-order $B$ by letting $b_1\leq b_2\iff \min(f^{-1}(b_1))\leq \min(f^{-1}(b_2))$, where the order on the right is the well-ordering of $A_n\times A_n$ defined using $\alpha_n$.

This ordering defines an injection $\varepsilon_{n+1}\colon B\to \omega$.

Now define $\alpha_{n+1}\colon A_{n+1}\to\omega\times\{0,\ldots,n+1\}$ as follows: $$\alpha_{n+1}(a) = \left\{\begin{array}{ll} \alpha_n(a)&\text{if }a\in A_n,\\ (\varepsilon_n(a),n+1)&\text{otherwise.} \end{array}\right.$$ Then $\alpha_{n+1}$ is one-to-one, $\alpha_{n+1}|_{A_n}=\alpha_n$ (and therefore, $\alpha_{n+1}|_{A_k} = \alpha_k$ for every $k\leq n$), and maps $A_{n+1}$ into $\omega\times\{0,\ldots,n+1\}$.

Now let $\alpha\colon\overline{A}\to\omega\times\omega$ be given by $\alpha=\cup_{n\in\omega}\alpha_n$.

Note that once we have selected $\varepsilon$, there are no more choices left to be made. So the existence of $\alpha$ does not require the use of $\mathsf{AC}$.