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Prove these definitions are equivalent:
Definition $\,(1)\,$: $A\subset X$ is not connected if for open $U, V\subset X\,\,, $$\,\,U\cap\bar{V}=\emptyset\,\,$, $\,\,\bar{U}\cap V=\emptyset\,\,$, $\,\,U\cap A\neq\emptyset\,\,$, $\,\,V\cap A\neq\emptyset\,\,$ and $\,\,A \subset U\cup V$.
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Definition $\,(2)\,$ $A$ is not connected if $U\cap A\neq\emptyset$,$V\cap A\neq\emptyset$, $(U\cap A)\cap (V\cap A)=\emptyset$ but $(U\cap A)\cup (V\cap A)=A$.

Attempt: For $A\subset X$ and open $U,V\subset X$ s.t. $U,V$ disconnect $A$,
(1) $(U\cap A)\cap (V\cap A)=\emptyset\implies$ $U\cap\bar{V}=\emptyset$,$\bar{U}\cap V=\emptyset$
(2) $(U\cap A)\cup (V\cap A)=A\implies$ $U\cup V=A$.

  • 2
    For the first definition you want to add that $U \cap A \neq \emptyset$ and likewise for $V$, and also that $A \subset U \cup V$, not equal, because this would imply $A$ is open in $X$.2012-02-26
  • 1
    In Definition (1), I assume you mean that "... there are open $U,V \subset X$ .... In Definition (2), I assume that $U$ and $V$ are supposed to be open subsets of $X$ (and, again, just that such $U$ and $V$ exist).2012-06-27

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