Union of Arb. Coll. of Open Sets; Int. of a Fin. Coll. of Open Sets

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

Recall from the Open and Closed Sets in Metric Spaces page a set $S \subseteq M$ is said to be open if $\mathrm{int} (S) = S$ and $S$ is said to be closed if $S^c$ is open.

We will now look at some rather important theorems regarding the unions and intersections of open sets.Theorem 1: Let $(M, d)$ be a metric space. If $\{ A_i \}_{i \in I}$ is any arbitrary collection of open sets where $I$ is an indexing set then $\displaystyle{\bigcup_{i \in I} A_i}$ is an open set. |

**Proof:**Let $\{ A_i \}_{i \in I}$ be a collection of open sets for some index set $I$ and let $\displaystyle{A = \bigcup_{i \in I} A_i}$. Let $x \in A$. Then $x \in A_i$ for all $i \in I$. Since $A_i$ is open for all $i \in I$ there exists a positive real number $r > 0$ such that the ball centered at $x$ with radius $r$ is contained in $A_i$, i.e.:

\begin{align} \quad B(x, r) \subseteq A_i \end{align}

- But $A_i \subseteq \bigcup_{i \in I} A_i$. Hence, for all $x \in A$ there exists an $r > 0$ such that $B(x, r) \subseteq A$ and so $x \in \mathrm{int} (A)$. Hence $A \subseteq \mathrm{int} (A)$. Furthermore, if $x \in \mathrm{int} (A)$ then there exists a positive real number $r > 0$ such that $B(x, r) \subseteq A$. Clearly $x \in B(x, r)$ so $x \in A$. Therefore $\mathrm{int} \subseteq A$.

- Since $A \subseteq \mathrm{int} (A)$ and $\mathrm{int} (A) \subseteq A$ we conclude that $A = \mathrm{int} (A)$ so $\displaystyle{A = \bigcup_{i \in I} A_i}$ is an open set. $\blacksquare$

Theorem 2: Let $(M, d)$ be a metric space. If $\{ A_1, A_2, ..., A_n \}$ is a finite collection of open sets then $\displaystyle{\bigcap_{i=1}^{n} A_i}$ is an open set. |

**Proof:**Let $\{ A_1, A_2, ..., A_n \}$ be a finite collection of open sets and let $\displaystyle{A = \bigcap_{i=1}^{n} A_i}$. For all $x \in A$ we have that $x \in A_i$ for all $i \in \{ 1, 2, ..., n \}$. Since each $A_i$ is an open set, we have that there exists positive real numbers $r_i > 0$ such that the ball centered at $x$ with radius $r_i$ is contained in $A_i$, that is, for each $i \in \{ 1, 2, ..., n \}$ we have that:

\begin{align} \quad B(x, r_i) \subseteq A_i \end{align}

- Let $r > 0$ be defined to be:

\begin{align} \quad r = \min \{ r_1, r_2, ..., r_n \} \end{align}

- Then $B(x, r) \subseteq B(x, r_i)$ for all $i \in \{ 1, 2, ..., n \}$. Hence $B(x, r) \subseteq A_i$ for all $i \in I$, i.e., $\displaystyle{B(x, r) \subseteq A = \bigcup_{i=1}^{n} A_i}$. So for all $x \in A$ there exists a positive real number $r > 0$ such that $B(x, r) \subseteq A$ so $x \in \mathrm{int} (A)$. Hence $A \subseteq \mathrm{int}(A)$. If $x \in \mathrm{int} (A)$ then there exists a positive real number $r > 0$ such that $B(x, r) \subseteq A$, so since $x \in B(x, r)$ we have that $x \in A$ so $\mathrm{int} (A) \subseteq A$ again.

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