finite intersection of a family of sets same as countable intersection?

Question

I am reading definition of semi-algebra and one of its properties is that it is closed under finite intersection. In that case can't it be proved that it is also closed under countably infinite intersections using induction? Could someone please help me understand why induction proof fails?

Answer 1

Induction, in general, only allows you to prove that for every finite number of sets, their intersection is still in the (semi)-algebra.

But it is not necessarily the case that you can "pull" that induction to the transfinite realm. For example, you can prove by induction that adding $1$ to itself finitely many times is still a natural number; but can you do that countably many times? No, you cannot.

Consider the set $S=\{A\subseteq\Bbb N\mid A\text{ is finite, or }\Bbb N\setminus A\text{ is finite}\}$. This is an algebra on $\Bbb N$, and in particular a semi-algebra. But it is not closed under countable intersections, since $\Bbb N\setminus\{2k\}$ is in the algebra for each $k\in\Bbb N$, but $\bigcap_{k\in\Bbb N}\Bbb N\setminus\{2k\}=\Bbb N\setminus\{2k\mid k\in\Bbb N\}\notin S$ (since it is the set of odd integers, and it is neither finite nor co-finite).