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:

\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:

\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:

\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:

\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. $\blacksquare$