For every connected space X and an open cover U, every two points has a simple chain containing them

Question

I am trying to prove this theorem saying: " A space X is connected, if and only if for an open cover U of X, every two points in X has a chain between them". I cant prove only if part (a connected space has a simple chain between every two points). I know to show that for every two points in X , say a,b , i can find a family of open groups Uα in U such that the union of Uα is connected and contains a and b. But I couldn't understand why there is always a finite group of groups as above. Any idea? thank alot

Answer 1

Given a connected space $X$ and an open cover $\mathcal{C}$, define an equivalence relation $\sim_{\mathcal{C}}$ by $x \sim_{\mathcal{C}} y$ if there is a chain from $x$ to $y$. Show that each equivalence class is open. (Of course equivalence classes are nonempty by definition.) Equivalence classes are disjoint; if there were more than one equivalence class, you could easily contradict connectedness.