Let $X$ be a set and let $\mathcal F (X)$ dentoe the finite subsets of $X$ and let there be given a function
$$\tag 1 F: X \to \mathcal F (X)$$
Let $K$ be a finite subset of $X$.
Definition: A subset $Y$ of $X$ is said to be an inductive accommodation of $K$ and $F$ if
$$\tag 2 K \subset Y$$ $$\tag 3 \text{For every } y \in Y, \; F(y) \subset Y$$
The intersection of any family of subsets of $X$ that are inductive accommodations of $K$ and $F$ is also an inductive accommodation.
Prove that the minimal inductive accommodation of $K$ and $F$ is a countable set.
My work
I was looking at
and
Tree of primitive Pythagorean triples
and figured that this abstraction is true but not sure how to proceed.
Following bof's suggestions...
For any $Y \subset X$ we can form $\Phi (Y)$,
$$\tag 1 \Phi (Y) = \bigcup_{y \in Y} F(y)$$
If $Y$ is countable so is $\Phi (Y)$,
Let $Y_0 = K$ and for $n \ge 0$ define
$$\tag 2 Y_{n+1} = \Phi (Y_n)$$
By induction. each $Y_n$ is countable, so
$$\tag 3 \hat K = \bigcup_{n \ge 0} Y_n$$
is also countable.
Claim: $\hat K$ is an inductive accommodation of $K$ and $F$:
Clearly $K \subset \hat K$.
If $y \in \hat K$ then there is a $k$ with $y \in Y_k$, and since
$$\tag 4 F(y) \subset \Phi (Y_k) = Y_{k+1} \subset \hat K$$
the claim is established.
It can be demonstrated by induction that $\hat K$ is the minimal inductive accommodation .
Note that we need the axiom of choice in this setting.