Union of Fin. Coll. of Closed Sets; Int. of Arb. Coll. of Closed Sets

# The Union of a Finite Collection of Closed Sets and The Intersection of an Arbitrary Collection of Closed Sets

Recall from The Union of an Arbitrary Collection of Open Sets and The Intersection of a Finite Collection of Open Sets page that if $\{ A_i \}_{i \in I}$ is an arbitrary collection of open sets for some indexing set $I$ then $\displaystyle{\bigcup_{i \in I} A_i}$ is an open set and if $\{ A_1, A_2, ..., A_n \}$ is a finite collection of open sets then $\displaystyle{\bigcup_{i=1}^{n} A_i}$ is an open set.

We will now look at two similar theorems regarding closed sets. Note that subtle (BUT VERY IMPORTANT) differences between the theorems on this page and the theorems on the page linked above.

Theorem 1: Let $(M, d)$ be a metric space. If $\{ A_1, A_2, ..., A_n \}$ is a finite collection of closed sets then $\displaystyle{\bigcup_{i=1}^{n} A_i}$ is an closed set. |

**Proof:**Let $\{ A_1, A_2, ..., A_n \}$ be a finite collection of closed sets and let $\displaystyle{A = \bigcap_{i=1}^{n} A_i}$. Consider the complement of $A$:

\begin{align} \quad A^c = M \setminus A = M \setminus \bigcup_{i=1}^{n} A_i \end{align}

- By De Morgan's Laws we see that:

\begin{align} \quad = \bigcap_{i=1}^{n} (M\setminus A_i) \end{align}

- Since each $A_i$ is a closed set, we must have that $M \setminus A_i$ is an open set. So $A^c$ is the intersection of a finite collection of open sets which must be an open set. Hence $\displaystyle{A = \bigcup_{i=1}^{n} A_i}$ must be a closed set. $\blacksquare$

Theorem 2: Let $(M, d)$ be a metric space. If $\{ A_i \}_{i \in I}$ is any arbitrary collection of closed sets where $I$ is an indexing set then $\displaystyle{\bigcap_{i \in I} A_i}$ is a closed set. |

**Proof:**Let $\{ A_i \}_{i \in I}$ be an arbitrary collection of closed sets for some indexing set $I$ and let $\displaystyle{A = \bigcap_{i \in I} A_i}$. Consider the complement of $A$:

\begin{align} \quad A^c = M \setminus A = M \setminus \bigcap_{i \in I} A_i \end{align}

- By De Morgan's laws we see that:

\begin{align} \quad = \bigcup_{i \in I} (M \setminus A_i) \end{align}

- Since each $A_i$ is a closed set, we must have that $M \setminus A_i$ is an open set. So $A^c$ is the union of an arbitrary collection of open sets which must be open. Hence $\displaystyle{A = \bigcap_{i \in I} A_i}$ must be a closed set. $\blacksquare$