Intersection of two algebras on different spaces not an algebra Reviewing my lecture notes, I have trouble understanding why the intersection of the following two algebras behaves as it does:
$$ \mathcal{A} = \left\{A \subseteq \mathbb{N}: |A| < \infty \text{ or } |A^c| < \infty  \right\}$$
$$ \mathcal{B} = \left\{B \subseteq \mathbb{Z}: |B| < \infty \text{ or } |B^c| < \infty  \right\},$$
where $A^c$ and $B^c$ denote the complements of $A$ and $B$, respectively.
I was told that the intersection of $\mathcal{A}$ and $\mathcal{B}$ is equal to $\left\{A \subseteq \mathbb{N}: |A| < \infty \right\}$.
Why isn't $\mathcal{A} \cap \mathcal{B}$ equal to $\mathcal{A}$ itself? Why isn't it an algebra?
Thanks for enlightening me! Much obliged for any input!
 A: Think about a cofinite set in $\mathbb N$. It contains all natural numbers except finitely many.
But if you consider it as a subset of $\mathbb Z$, there are infinitely many negative numbers that it doesn't contain, so its complement is not finite. Thus the cofinite sets in $\mathbb N$ are not cofinite in $\mathbb Z$.
(But the finite sets in $\mathbb N$ are automatically finite sets in $\mathbb Z$!)
A: It's important to note that, when taking complements, the "universal" set matters. In the first set, $A^c$ means $\mathbb{N}\setminus A$; whereas in the second, $B^c$ means $\mathbb{Z}\setminus B$.

It is indeed true that $\mathcal{A}\cap\mathcal{B}=\{A\subseteq\mathbb{N}: |A|<\infty\}$.

Proof:
Suppose $A\in \mathcal{A}\cap\mathcal{B}$. Then $A\subseteq\mathbb{N}$, so $\mathbb{Z}\setminus A$ is infinite. Since $A\in\mathcal{B}$, we conclude that $A$ is finite.
The converse is easy. $\square$
For an explicit example why $\mathcal{A}\cap\mathcal{B}\not\supseteq\mathcal{A}$, consider the set $A:=\mathbb{N}$.
