Union of Open Subsets

Theorem

Let $M = \left({A, d}\right)$ be a metric space.

The union of any collection of open subsets of $M$ is open in $M$.

Proof

Let $I$ be any indexing set.

Let $U_i$ be open in $M$ for all $i \in I$.

Let $\displaystyle x \in \bigcup_{i \in I} U_i$.

Then $x \in U_k$ for some $k \in I$.

Since $U_k$ is open in $M$, $\displaystyle \exists \epsilon > 0: N_\epsilon \left({x}\right) \subseteq U_k \subseteq \bigcup_{i \in I} U_i$.

The result follows.

$\blacksquare$

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Proposition $2.3.19$