The Union and Intersection of Collections of Closed Sets

The Union and Intersection of Collections of Closed Sets

Recall from The Union and Intersection of Collections of Open Sets page that if $\mathcal F$ is an arbitrary collection of open sets then $\displaystyle{\bigcup_{A \in \mathcal F} A}$ is an open set, and if $\mathcal F = \{ A_1, A_2, ..., A_n \}$ is a finite collection of open sets then $\displaystyle{\bigcap_{i=1}^{n} A_i}$ is an open set. We will now prove two analogous theorems regarding the union and intersection of collections of closed sets.

 Theorem 1: If $\mathcal F = \{ A_1, A_2, ..., A_n \}$ is a finite collection of closed sets then $\displaystyle{\bigcup_{i=1}^{n} A_i}$ is a closed set.
• Proof: Let $\mathcal F = \{ A_1, A_2, ..., A_n \}$ be a finite collection of closed sets and let:
(1)
\begin{align} \quad S = \bigcup_{i=1}^{n} A_i \end{align}
• By applying the generalized De Morgan's Law, we see that the complement $S^c$ is:
(2)
\begin{align} \quad S^c = \left ( \bigcup_{i=1}^{n} A_i \right )^c = \bigcap_{i=1}^{n} A_i^c \end{align}
• For each $A_i$ for $i \in \{1, 2, ..., n \}$ we have that $A_i$ is closed, so $A_i^c$ is open. The intersection of a finite collection of open sets is open, so $S^c$ is open and hence $(S^c)^c = S$ is closed. Therefore $\displaystyle{\bigcup_{i=1}^{n} A_i}$ is closed. $\blacksquare$
 Theorem 2: If $\mathcal F$ is an arbitrary collection of closed sets then $\displaystyle{\bigcap_{A \in \mathcal F} A}$ is a closed set.
• Proof: Let $\mathcal F$ be any arbitrary collection of closed sets and let:
(3)
\begin{align} \quad S = \bigcap_{A \in \mathcal F} A \end{align}
• By applying the generalized De Morgan's Law, we see that the complement $S^c$ is:
(4)
\begin{align} \quad S^c = \left( \bigcap_{A \in \mathcal F} A \right )^c = \bigcup_{A \in \mathcal F} A^c \end{align}
• For all $A \in \mathcal F$ we have that $A$ is closed, so $A^c$ is open. The union of an arbitrary collection of open sets is open, so $S^c$ is open. Therefore $(S^c)^c = S$ is closed. $\blacksquare$