Union of Open Sets of Metric Space is Open

Theorem

Let $M = \struct {A, d}$ be a metric space.

The union of a set of open sets 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 $\ds x \in \bigcup_{i \mathop \in I} U_i$.

Then by definition of set union, $x \in U_k$ for some $k \in I$.

Since $U_k$ is open in $M$:

$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k$

where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.

By Set is Subset of Union:

$\ds U_k \subseteq \bigcup_{i \mathop \in I} U_i$

Hence by Subset Relation is Transitive:

$\ds \map {B_\epsilon} x \subseteq \bigcup_{i \mathop \in I} U_i$

and the result follows by definition of open set in metric space.

$\blacksquare$

Sources

• 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous) ... (next): $2.24 a$
• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Proposition $2.3.19$