## Connected subsets one of which meets the closure of the other

Exercise 224 (Chapter 32, “Connectedness”):

Let and be connected subsets of topological space such that intersects . Prove that is connected. Find an example to show that it is not enough to assume that intersects .

Let us assume that and are open subsets of such that is empty and that . To show that is connected, we need to show that either or does not intersect .

Since is connected, and  is empty and , necessarily is entirely in or in . Similarly, is entirely in or entirely in .

Let us suppose that they are not both in or both in ; for instance, that and .

If , then with open in . Since does not intersect , cannot be in . Hence cannot intersect . This contradicts our assumption.

Hence it is impossible that  and . Similarly, it is impossible that and . Thus, and are either both in or both in , which implies that either or does not intersect .

If we only assume that and intersect, it does not always follow that is connected. For instance, let be the topological plane and and two open disks that “almost touch”, that is such that the corresponding closed disks ‑ which are their closures ‑ intersect in one point. The union of and is then not connected.

## Union of transitively overlapping connected spaces

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.