# proving definitions of connectedness are equivalent

Prove these definitions are equivalent:
Definition $\,(1)\,$: $A\subset X$ is not connected if for open $U, V\subset X\,\,, $$\,\,U\cap\bar{V}=\emptyset\,\,, \,\,\bar{U}\cap V=\emptyset\,\,, \,\,U\cap A\neq\emptyset\,\,, \,\,V\cap A\neq\emptyset\,\, and \,\,A \subset U\cup V. -and- Definition \,(2)\, A is not connected if U\cap A\neq\emptyset,V\cap A\neq\emptyset, (U\cap A)\cap (V\cap A)=\emptyset but (U\cap A)\cup (V\cap A)=A. Attempt: For A\subset X and open U,V\subset X s.t. U,V disconnect A, (1) (U\cap A)\cap (V\cap A)=\emptyset\implies U\cap\bar{V}=\emptyset,\bar{U}\cap V=\emptyset (2) (U\cap A)\cup (V\cap A)=A\implies U\cup V=A. - For the first definition you want to add that U \cap A \neq \emptyset and likewise for V, and also that A \subset U \cup V, not equal, because this would imply A is open in X. – Henno Brandsma Feb 26 '12 at 21:52 In Definition (1), I assume you mean that "... there are open U,V \subset X .... In Definition (2), I assume that U and V are supposed to be open subsets of X (and, again, just that such U and V exist). – arjafi Jun 27 '12 at 14:56 ## 1 Answer Trying to ununanswer another unanswered question. An idea: try to prove both definitions are equivalent to the following third one: Definition \,(0)\,: \,A\subset X\, is not connected iff there exists a continuous suprajective function \,f:A\longrightarrow \{0,1\}\, , with the latter space having the discrete topology. \underline{(0)\,\Longrightarrow\,(1)\wedge (2)}\,: Define \,U:=f^{-1}(\{0\})\,\,,\,\,V:=f^{-1}(\{1\})\, . Since \,f\, is continuous and \,\{0\}\,,\,\{1\}\, are open and closed, we have that \,A\,,\,B\, are open and closed , so \,U=\overline U\,,\,V=\overline V\, and, of course, \,\emptyset= U\cap V=\overline U\cap V=U\cap\overline V\, , so clearly \,A\subset U\cup V\, and \,A\cap U\neq \emptyset\neq A\cap V\, , lest \,f\, is not onto. \underline{(1)\wedge (2)\Longrightarrow (0)}\,: Define f:A\longrightarrow \{0,1\}\, by$$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 '14 at 6:45