# Union of Open Sets of Normed Vector Space is Open

## Theorem

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector 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 $x \in U_k$ for some $k \in I$.

Since $U_k$ is open in $M$:

$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k \subseteq \bigcup_{i \mathop \in I} U_i$

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

The result follows.

$\blacksquare$