Metric Space defined by Closed Sets

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Theorem

Let $M = \struct {A, d}$ be a metric space.

Then:

\((C1)\)   $:$   $A$ is closed in $M$             
\((C2)\)   $:$   $\O$ is closed in $M$             
\((C3)\)   $:$   The union of a finite number of closed sets of $M$ is a closed set of $M$             
\((C4)\)   $:$   The intersection of arbitrarily many closed sets of $M$ is a closed set of $M$             


Proof

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

$\Box$


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

$\Box$


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:

$\displaystyle A \setminus \bigcup_{i \mathop = 1}^n V_i = \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i}$

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

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

Therefore, by Finite Intersection of Open Sets of Metric Space is Open, $\displaystyle \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i} = A \setminus \bigcup_{i \mathop = 1}^n V_i$ 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.

$\Box$


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 $I$.

Then from De Morgan's laws:

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

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

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

Therefore, by definition of a topology, $\displaystyle \bigcup_{i \mathop \in I} \paren {A \setminus V_i} = A \setminus \bigcap_{i \mathop \in I} V_i$ 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.

$\blacksquare$


Sources