Connected subsets one of which meets the closure of the other

olivierd

Jul 22, 2017 2 min read

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.

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