Given two points \(a\) and \(b\) of a space \(S\), a collection of sets \(A_1,\ldots,A_n\) in \(S\) is called a simple chain from \(a\) to \(b\) if \(a\in A_1\), \(b\in A_n\), and \(A_i\cap A_j\neq\varnothing\) if and only if \(|i-j|\leq 1\).
Prove that if \(\{U_\alpha\}\) is a collection of open sets covering \(S\) and \(S\) is a connected space, then there is a simple chain of elements in \(\{U_\alpha\}\) joining \(a\) and \(b\) for any \(a\) and \(b\). Hint: Consider the set \(C_a\) of all \(b\) so that there is a simple chain of elements joining \(a\) and \(b\). Prove that \(C_a\) is open and closed.
Solution
Fix \(a\in S\) and let \(C_a\) be the set of all points \(b\in S\) for which there is a simple chain of members of the cover from \(a\) to \(b\).
1. The set \(C_a\) is nonempty.
Because the family \(\{U_\alpha\}\) covers \(S\), choose a member \(U\) containing \(a\). The one-term chain \[ U \] is a simple chain from \(a\) to \(a\). Thus \(a\in C_a\).
2. The set \(C_a\) is open.
Let \(b\in C_a\) and choose a simple chain \[ A_1,\ldots,A_n \] from \(a\) to \(b\). Since \(b\in A_n\) and \(A_n\) is a member of the open cover, \(A_n\) is an open neighborhood of \(b\).
For every \(c\in A_n\), the same sequence \(A_1,\ldots,A_n\) is a simple chain from \(a\) to \(c\): the intersection conditions among the sets are unchanged, while \(c\in A_n\). Hence \[ A_n\subseteq C_a. \] Therefore \(C_a\) is open.
3. If a member \(U\) of the cover meets \(C_a\), then every point of \(U\) belongs to \(C_a\).
Suppose \[ c\in U\cap C_a. \] Choose a simple chain \[ A_1,\ldots,A_n \] from \(a\) to \(c\). Then \(U\cap A_n\ne\varnothing\).
Let \(j\) be the least index such that \[ U\cap A_j\ne\varnothing. \] If \(U=A_j\), then \[ A_1,\ldots,A_j \] is already a simple chain from \(a\) to every point of \(U\).
Suppose instead that \(U\ne A_j\). By minimality of \(j\), \[ U\cap A_i=\varnothing \qquad (i<j), \] while \(U\cap A_j\ne\varnothing\). Consequently \[ A_1,\ldots,A_j,U \] is a simple chain: the original prefix is simple, \(U\) meets its immediate predecessor \(A_j\), and \(U\) is disjoint from every earlier member of the prefix. Thus this chain joins \(a\) to every point of \(U\).
In either case, \[ U\subseteq C_a. \]
4. The complement \(S\setminus C_a\) is open; equivalently, \(C_a\) is closed.
Let \(b\in S\setminus C_a\). Choose a member \(U\) of the cover with \(b\in U\). If \(U\) met \(C_a\), then 3 would give \[ U\subseteq C_a, \] contradicting \(b\notin C_a\). Hence \[ U\cap C_a=\varnothing, \] so \(U\) is an open neighborhood of \(b\) contained in \(S\setminus C_a\). Thus the complement is open.
5. One has \(C_a=S\).
By 1, \(C_a\) is nonempty. By 2 and 4, it is both open and closed in the connected space \(S\). The only nonempty clopen subset of a connected space is the whole space, so \[ C_a=S. \]
6. Any two points \(a,b\in S\) can therefore be joined by a simple chain of members of the given open cover.
Fix \(a\in S\). By 5 every \(b\in S\) lies in \(C_a\), which is exactly the asserted conclusion.