Metric Space defined by Closed Sets

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Then:

Proof
From Metric Space is Closed in Itself, $C1$ holds.

From Empty Set is Closed in Metric Space, $C2$ holds.

Let $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ be the union of a finite number of closed sets of $M$.

Then from De Morgan's laws:


 * $\displaystyle A \setminus \bigcup_{i \mathop = 1}^n V_i = \bigcap_{i \mathop = 1}^n \left({A \setminus V_i}\right)$

By definition of closed set, each of the $A \setminus V_i$ is by definition open in $M$.

We have that $\displaystyle \bigcap_{i \mathop = 1}^n \left({A \setminus V_i}\right)$ is the intersection of a finite number of open sets of $M$.

Therefore, by Finite Intersection of Open Sets of Metric Space is Open, $\displaystyle \bigcap_{i \mathop = 1}^n \left({A \setminus V_i}\right) = A \setminus \bigcup_{i \mathop = 1}^n V_i$ is likewise open in $M$.

Then by definition of closed set, $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ is closed in $M$.

Thus $C3$ holds.

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

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

Then from De Morgan's laws:


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

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

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

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

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

Thus $C4$ holds.