Intersection of Closed Sets is Closed

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \struct {S, \tau}$ 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:

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

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

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

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

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

$\blacksquare$


Sources