Suppose $\mathscr{A}$ is a collection of connected subsets in some topological space such that for any two $A$, $B$ in $\mathscr{A}$, there are $A=C_0,C_1,\dots,C_n=B$ such that $C_i$ and $C_{i+1}$ are not separated. My question then is, why is the union of the collection, $\bigcup_{A\in\mathscr{A}}A$, connected?
About a finite chain of connected sets.
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0For reference, this is an exercise in Kelley's General Topology p.60-61 – 2018-02-10
1 Answers
I'll assume, as André noted, that the $C_i$ are supposed to be in $\mathscr{A}$, since otherwise the statement is clearly false.
Let $S$ and $T$ be two non-empty separated sets with $S\cup T=\bigcup_{A\in\mathscr{A}}A$. Then for every $A\in\mathscr{A}$, the sets $S\cap A$ and $T\cap A$ are separated. Since $A$ is connected, this means that one of them is empty and the other one is $A$. Thus, all sets in $A\in\mathscr{A}$ are either fully in $S$ or fully in $T$. Since $S$ and $T$ are non-empty, each of them contains at least one of the $C_i$. But then there must be some $C_i$ and $C_{i+1}$ such that one is in $S$ and the other is in $T$. These two are separated, which is false by assumption. The contradiction shows that such sets $S$ and $T$ don't exist, and thus $\bigcup_{A\in\mathscr{A}}A$ is connected.
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0Oh yeah, the $C_i\in\mathscr{A}$. Thanks for answering. – 2011-12-13