Closed intervals in open sets

Let be a collection of open sets such that

Prove a finite number of sets such that

Need some help understanding the question first.

Clarification

The statement means an open set containing number , another (or the same set) containing number and for every such that , so this implies ?

Re: Closed intervals in open sets

Re: Closed intervals in open sets

There is one little change needed.

The statement says that the cover is in open sets, not open intervals, so doesn't guarantee that [0, s] is covered by that same finite subcover. That's because s could be in a separate open component of than the component of containing c.

This is easy enough to account for. You could declare that this was a theorem about covers by open intervals, and then have the general open cover case as an immediate corollary. You could also say that the cover used in this proof was actually a refinement of the original cover that used the open intervals that comprised the open sets of the original cover - and then easily deduce that there's a finite subcover of the original cover, because you've shown that there's a finite subcover of the open intervals that comprised the open sets of the original cover.