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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) $(U\cap A)\cup (V\cap A)=A\implies$ $U\cup V=A$.

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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 $X$. –  Henno Brandsma Feb 26 '12 at 21:52
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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). –  Arthur Fischer Jun 27 '12 at 14:56

1 Answer 1

Trying to ununanswer another unanswered question.

An idea: try to prove both definitions are equivalent to the following third one:

Definition $\,(0)\,$: $\,A\subset X\,$ is not connected iff there exists a continuous suprajective function $\,f:A\longrightarrow \{0,1\}\,$ , with the latter space having the discrete topology.

$\underline{(0)\,\Longrightarrow\,(1)\wedge (2)}\,$: Define $\,U:=f^{-1}(\{0\})\,\,,\,\,V:=f^{-1}(\{1\})\,$ . Since $\,f\,$ is continuous and $\,\{0\}\,,\,\{1\}\,$ are open and closed, we have that $\,A\,,\,B\,$ are open and closed , so $\,U=\overline U\,,\,V=\overline V\,$ and, of course, $\,\emptyset= U\cap V=\overline U\cap V=U\cap\overline V\,$ , so clearly $\,A\subset U\cup V\,$ and $\,A\cap U\neq \emptyset\neq A\cap V\,$ , lest $\,f\,$ is not onto.

$\underline{(1)\wedge (2)\Longrightarrow (0)}\,$: Define $f:A\longrightarrow \{0,1\}\,$ by $$f(x)=\left\{\begin{array}{} 0&,\,\,\,\,\text{if }\,\,x\in A\cap U\\1&,\,\,\,\,\text{if}\,\,\,x\in A\cap V\end{array}\right.$$

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( (1) and (2) => (0) ) does not imply that (1) => 0!! –  user21820 Nov 24 at 6:45

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