Connected subsets one of which meets the closure of the other

Exercise 224 (Chapter 32, “Connectedness”):

Let A and B be connected subsets of topological space X such that A intersects \Cl(B). Prove that A \cup B is connected. Find an example to show that it is not enough to assume that \Cl(A) intersects \Cl(B).

Let us assume that V and W are open subsets of X such that V \cap W \cap (A \cup B) is empty and that A \cup B \subseteq V \cup W. To show that A \cup B is connected, we need to show that either V or W does not intersect A \cup B.

Since A is connected, and V \cap W \cap A is empty and A \subseteq V \cup W, necessarily A is entirely in V or in W. Similarly, B is entirely in V or entirely in W.

Let us suppose that they are not both in V or both in W; for instance, that A \subseteq V and B \subseteq W.

If a \in A, then a \in V with V open in X. Since V does not intersect B, a cannot be in \Cl(B). Hence A cannot intersect \Cl(B). This contradicts our assumption.

Hence it is impossible that A \subseteq V and B \subseteq W. Similarly, it is impossible that A \subseteq W and B \subseteq V. Thus, A and B are either both in V or both in W, which implies that either V or W does not intersect A \cup B. \blacksquare

If we only assume that \Cl(A) and Cl(B) intersect, it does not always follow that A \cup B is connected. For instance, let X be the topological plane and A and B 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 A and B is then not connected.

Union of transitively overlapping connected spaces

Exercise 225 (Chapter 32, “Connectedness”):

Prove the following generalization of theorem 40. Let A_\lambda (\lambda in \Lambda \neq \emptyset) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by A_\lambda \approx A_{\lambda'} if A_\lambda \cap A_{\lambda'} \neq \emptyset have just one equivalence class. Then \bigcup_\lambda A_\lambda is connected.

For clarity, rather than a family (A_\lambda)_{\lambda \in \Lambda} I will consider a set \mathcal A of connected subsets of X. The condition is that the equivalence relation in \mathcal A generated by the above relation \approx has only one class.

Let K = \bigcup_{A \in \mathcal A} A.

Let V and W be eventual open subsets of X such that K \subseteq V \cup W and V \cap W \cap K = \emptyset. To show that K is connected, we must show that either V \cap K or W \cap K is empty.

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

Let us consider the relation \mathcal R on \mathcal A defined by A_1 \mathcal R A_2 \iff (A_1 \subseteq V \wedge A_2 \subseteq V) \vee (A_1 \subseteq W \wedge A_2 \subseteq W); in other words, A_1 \mathcal R A_2 if and only if both are subsets of V, or both are subsets of W. Clearly, this is an equivalence relation.

Furthermore, if A_1 \approx A_2, that is, A_1 \cap A_2 \neq \emptyset, we cannot have A_1 \subseteq V and A_2 \subseteq W, for V and W are disjoint. Hence A_1 and A_2 are either both in V or both in W; that is, A_1 \mathcal R A_2. Hence \mathcal R is an equivalence relation greater than the equivalence relation generated by \approx.

If \mathcal R had two or more classes, there would be some element of \mathcal A in one class and another in another; these two elements would not be equivalent following \mathcal R. Since \mathcal R is greater than the equivalence relation generated by \approx, they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence \mathcal R too can have only one class, which means that for all A_1 and A_2 in \mathcal A, either both are in V or both are in W. But this implies that all are in V or all are in W, which in turn implies that their union K is a subset of either V or W, hence that eitherV \cap K or W \cap K is empty. \blacksquare