Union of Open Sets of Normed Vector Space is Open
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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.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.3$: Normed and Banach spaces. Topology of normed spaces