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I have seen two different definitions of disconnected sets. Here (X,T) is a topological space.

A subset $A$ of X is disconnected if there are two open sets $U,V$ of X such that $U \cap A \neq \emptyset $ , $V \cap A \neq \emptyset $, $U \cap A \cap V = \emptyset $ and $A\subset U\cup V$.


Two subsets $A,B$ of X are separated if no point of A lies in the closure of B and vice versa. A subset E of X is disconnected if E is a union of two nonempty separated sets.

I am having trouble understanding the connection between these two definitons. Why do they mean the same thing?

Thank you kindly.

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How can they mean the same thing if the first is about a single set $A$, and the second is about a pair of sets $A,B$? –  Thomas Andrews Dec 12 '11 at 14:49
First is the definition of A is disconnected and expressed in terms of two open sets U,V. Second is the definiton of E is disconnected and expressed in terms of A,B two seperated sets. –  marvinthemartian Dec 12 '11 at 14:54
Whoops, reading comprehension error. –  Thomas Andrews Dec 12 '11 at 15:13
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1 Answer 1

up vote 3 down vote accepted

In your second definition of separated, suppose that $E$ is a union of two non-empty separated sets $A$ and $B$. Define $U$ to be $X-\overline{B}$ and $V=X-\overline{A}$, where $\overline{A}$ and $\overline{B}$ are the topological closures of $A$ and $B$, respectively. Then, $U$ and $V$ are open sets and if $A$ and $B$ are separated, then $A\subseteq U$ and $B\subseteq V$. Since $E=A\cup B$, we see that $U\cap E\neq \emptyset$, $V\cap E\neq \emptyset$, but $U\cap E\cap V =\emptyset$ and $E=A\cup B\subseteq U\cup V$.

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Minor typo: should be $A \subseteq U$ and $B \subseteq V$ (not $B \subseteq A$). –  mjqxxxx Dec 12 '11 at 15:13
Thanks! Fixed the typo. –  Álvaro Lozano-Robledo Dec 12 '11 at 15:14
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