Union of Open Sets of Metric Space is Open

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Theorem

Let $M = \left({A, d}\right)$ 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 $\displaystyle x \in \bigcup_{i \mathop \in I} U_i$.

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


Since $U_k$ is open in $M$:

$\displaystyle \exists \epsilon > 0: B_\epsilon \left({x}\right) \subseteq U_k \subseteq \bigcup_{i \mathop \in I} U_i$

where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$ in $M$.

The result follows.

$\blacksquare$


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