Finite intersections of a countable family of sets is countable?

Question

Let $\mathcal{A} = \{A_n\}_{n \in \mathbb{N}}$ a countable family of sets. Is it true that the following set $$\mathcal{F}:=\{ B : B=\bigcap_{j=1}^n A_{a_j} \text{ for a finite choice of } A_{a_j} \text{ in } \mathcal{A} \}$$ is countable?

Answer 1

There is only one element of $\mathcal F$ for each $n$

Hint: For the revised question, how many finite subsets of a countable set are there?

Answer 2

$\mathcal{F}$ is countable.

There is a surjective map $$\bigcup_n \mathcal{A}^n \to \mathcal{F},$$ in fact the correspondence $\{A_{a_1}, \dots, A_{a_n} \}\mapsto \bigcap_1^n A_{a_j}$ is surjective; thus $$| \mathcal{F}| \leq |\bigcup_n \mathcal{A}^n| \leq | \sum_{n} \aleph_0 | \stackrel{1}{ =} | \aleph_0 \times \aleph_0| = \aleph_0. $$

[1] A chessboard is the union of its rows.

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Notice that $\bigcup_n \mathcal{A}^n$ is a proper subset of $\mathcal{A}^{\mathbb{N}}! $