Intersection of Closed Sets is Closed/Normed Vector Space

Theorem
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.

Then the intersection of an arbitrary number of closed sets of $M$ (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 $M$ indexed by $I$.

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

From De Morgan's laws: Difference with Intersection:


 * $\displaystyle X \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$

We have that $\displaystyle \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$ is the union of a indexed family of open sets of $M$ indexed by $I$.

By Union of Open Sets of Normed Vector Space is Open, $\displaystyle \bigcup_{i \mathop \in I} \paren {X \setminus V_i} = X \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $M$.

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