Union of transitively overlapping connected spaces

olivierd

Jul 22, 2017 3 min read

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 either $V \cap K$ or $W \cap K$ is empty. $\blacksquare$

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