Exercise 225 (Chapter 32, “Connectedness”):

Prove the following generalization of theorem 40. Let ( in ) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by if have just one equivalence class. Then is connected.

For clarity, rather than a family I will consider a set of connected subsets of . The condition is that the equivalence relation in generated by the above relation has only one class.

Let .

Let and be eventual open subsets of such that and . To show that is connected, we must show that either or is empty.

First, let be an eventual element of . Since , the two subsets of , and are disjoint, and their union is . Furthermore, they are open in . Since is connected, this implies that one or the other is empty. If is empty, then ; if instead is empty, then . Hence in all cases an element of is entirely in or entirely in .

Let us consider the relation on defined by ; in other words, if and only if both are subsets of , or both are subsets of . Clearly, this is an equivalence relation.

Furthermore, if , that is, , we cannot have and , for and are disjoint. Hence and are either both in or both in ; that is, . Hence is an equivalence relation greater than the equivalence relation generated by .

If had two or more classes, there would be some element of in one class and another in another; these two elements would not be equivalent following . Since is greater than the equivalence relation generated by , they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence too can have only one class, which means that for all and in , either both are in or both are in . But this implies that all are in or all are in , which in turn implies that their union is a subset of either or , hence that either or is empty.