Intersection of Closed Sets is Closed

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Then the intersection of an arbitrary number of closed sets of $T$ (either finitely or infinitely many) is itself closed.

Proof
Let $I$ be an indexing set (either finite or infinite).

Let $\displaystyle \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $T$ indexed by $I$.

Then from De Morgan's laws: Difference with Intersection:


 * $\displaystyle S \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \left({S \setminus V_i}\right)$

By definition of closed set, each of $S \setminus V_i$ are by definition open in $T$.

We have that $\displaystyle \bigcup_{i \mathop \in I} \left({S \setminus V_i}\right)$ is the union of a indexed family of open sets of $T$ indexed by $I$.

Therefore, by definition of a topology, $\displaystyle \bigcup_{i \mathop \in I} \left({S \setminus V_i}\right) = S \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $T$.

Then by definition of closed set, $\displaystyle \bigcap_{i \mathop \in I} V_i$ is closed in $T$.