Let $\displaystyle S = (0,1) $ and $\displaystyle \mathcal{O} = \{(1/n,1): n \in \bold{N} \} $. Show that $\displaystyle \mathcal{O} $ is an open cover of $\displaystyle S $.

So choose $\displaystyle x \in (0,1) $. Pick $\displaystyle n_0 \in \bold{N} $ such that $\displaystyle 1/n_0 < x $. This implies that $\displaystyle x \in (1/n_0, 1) \subset \bigcup_{n \in \bold{N}} (1/n,1) = \bigcup_{G \in \mathcal{O}} G $. Is this correct? Could I pick $\displaystyle n_0 $ in a different way?

What if we had a finite subset $\displaystyle \mathcal{O'} \subset \mathcal{O} $?