Countable set - $\aleph_0$ or finite?

Question

I have 100 sets $A_1,\ldots,A_{100}$. They are all subsets of $\Bbb R$. For each $A_i$ the complement of $A_i$ in $\Bbb R$ is a countable set.

$A= A_1 \cap A_2 \cap \ldots \cap A_{100}$.

$B$ is the complement of $A$. What is the cardinal number of $B$? There are 4 options as an answer:

1. $0$
2. finite number but not $0$
3. $\aleph_0$
4. c

I think that is can be $2$ or $3$ because we don't know if the 'complement of $A_i$ in $\Bbb R$ is a countable set' is all finite or one of them is $\aleph_0$.

Am I right?

The formal answer for this question is $3$.

Answer 1

The answer depends only on what your definition of "countable" is. As you point out, if a countable set can be finite, then option $2$ is indeed a possibility. See this Wikipedia article: "Some authors use countable set to mean countably infinite alone."

Edited to add: As Keen-amateur points out, option $1$ is also possible if a countable set can be empty.

Answer 2

If you take countable to mean exactly countable, i.e. not finite, and not empty then $3$, is your only answer since

$$B = A^c = (\bigcap(A_i))^c = \bigcup(A_i)^c$$

which is a finite union of exactly countable sets. Otherwise, both $1$ and $2$ could be answers.