The Union and Intersection of Collections of Open Sets

# The Union and Intersection of Collections of Open Sets

Recall from the Open and Closed Sets in Euclidean Space page that a set $S \subseteq \mathbb{R}^n$ is said to be an open set if $S = \mathrm{int} (S)$ and is said to be a closed set if $S =\mathrm{int} (S) \cup \mathrm{bdry} (S)$.

We will now look at some very important theorems regarding the union of an arbitrary collection of open sets and the intersection of a finite collection of open sets.

Theorem 1: If $\mathcal F$ is an arbitrary collection of open sets then $\displaystyle{\bigcup_{A \in \mathcal F} A}$ is an open set. |

*By "arbitrary" we mean that $\mathcal F$ can be a finite, countably infinite, or uncountably infinite collection of sets.*

**Proof:**Let $\mathcal F$ be an arbitrary collection of open sets and let:

\begin{align} \quad S = \bigcup_{A \in \mathcal F} A \end{align}

- We want to show that $S = \mathrm{int} (S)$.

- First suppose that $x \in S$. Then $x \in A$ for some set $A \in \mathcal F$. Since $A$ is an open set, there exists an $r > 0$ such that $B(x, r) \subseteq A$. But $A \subseteq S$, so by extension, there exists an $r > 0$ such that $B(x, r) \subseteq S$, so $x \in \mathrm{int} (S)$ and hence $S \subseteq \mathrm{int} (S)$.

- Now suppose that $x \in \mathrm{int} (S)$. Then for some $r > 0$ there exists a $B(x, r) \subseteq S$. Since $x \in B(x, r)$ we have that by extension, $x \in S$, so $\mathrm{int} (S) \subseteq S$.

- Since $S \subseteq \mathrm{int} (S)$ and $S \supseteq \mathrm{int} (S)$ we conclude that $S = \mathrm{int} (S)$. Therefore $\displaystyle{\bigcup_{A \in \mathcal F} A}$ is an open set. $\blacksquare$

Theorem 2: 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. |

**Proof:**Let $\mathcal F = \{ A_1, A_2, ..., A_n \}$ be a finite collection of open sets and let:

\begin{align} \quad S = \bigcap_{i=1}^{n} A_i \end{align}

- Once again, we want to show that $S = \mathrm{int} (S)$.

- Let $x \in S$. Then $x \in A_i$ for all $i \in \{1, 2, ..., n \}$ and so for each $i$ there exists some $r_i > 0$ such that:

\begin{align} \quad B(x, r_i) \subseteq A_i \: \mathrm{for \: all \:} i = 1, 2, ..., n \end{align}

- Let $r = \mathrm{min} \{ r_1, r_2, ..., r_n \}$. Then we have that $B(x, r) \subseteq A_i$ for all $i \in I$, so $B(x, r) \subseteq S$. Hence there exists an $r > 0$ such that $B(x, r) \subseteq S$ so $x \in S$ and $S \subseteq \mathrm{int} (S)$.

- Now suppose that $x \in \mathrm{int} (S)$. Then once again there exists an $r > 0$ such that $B(x, r) \subseteq S$. Since $x \in B(x, r)$, by extension we have that $x \in S$ so $\mathrm{int} (S) \subseteq S$.

- Since $S \subseteq \mathrm{int} (S)$ and $S \supseteq \mathrm{int} (S)$ we concluded that $S = \mathrm{int} (S)$ so $\displaystyle{\bigcap_{i=1}^{n} A_i}$ is an open set. $\blacksquare$