Union of Covers is Cover of Union

Theorem

Let $\sequence {S_i}_{i \in I}$ be an indexed family of sets.

For each $i \in I$, let $\CC_i$ be a cover of $S_i$.

Then, $\ds \bigcup_{i \mathop \in I} \CC_i$ is a cover of $\ds \bigcup_{i \mathop \in I} S_i$.

Proof

Let $\ds x \in \bigcup_{i \mathop \in I} S_i$ be arbitrary.

By definition of union, there is some $i \in I$ such that:

$x \in S_i$

Then, by definition of cover, there is some $C \in \CC_i$ such that:

$x \in C$

But, by definition of union:

$\ds C \in \bigcup_{i \mathop \in I} \CC_i$

Hence, there is some $\ds C \in \bigcup_{i \mathop \in I} \CC_i$ such that $x \in C$.

As $x$ was arbitrary, the result follows from the definition of cover.

$\blacksquare$