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While studying mathematical analysis, the textbook gives an definition as following:

Definition 1: A set $E\subset\mathbb{R}^n$ is a connected set that cannot be represented as the union of two disjoint nonempty sets $A,B$ which satisfy $A'\cap B=\phi$ and $A\cap B'=\phi$.

My teacher gives the following definition:

Definition 2: A set $E\subset\mathbb{R}^n$ is a connected set that cannot be represented as a subset of the union of two disjoint open sets $A,B$ which satisfy $A\cap E\neq\phi$ and $B\cap E\neq\phi$.

I try to prove they are equivalent. 1$\Rightarrow$2 is easy, but I find difficulty proving 2$\Rightarrow$1. Is it correct or not? (Sorry for my poor English)

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  • $\begingroup$ What are $A', B'$? $\endgroup$ – D_S Oct 4 '17 at 1:37
  • $\begingroup$ the derived set of $A,B$, all limit points of $A,B$ $\endgroup$ – AbnerYe Oct 4 '17 at 1:39
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They are equivalent. Suppose $A$ and $B$ witness that $E$ is not connected according to definition 1. Let $U=\{x\in\mathbb{R}^n:d(x,A)<d(x,B)\}$ and $V=\{x\in\mathbb{R}^n:d(x,A)>d(x,B)\}$. Here $d(x,A)$ means $\inf \{d(x,y):y\in A\}$. Then $U$ and $V$ are open (exercise), and they are obviously disjoint. Since $A\cap\overline{B}=\emptyset$, $d(x,B)>0$ for any $x\in A$ and so $A\subseteq U$, and similarly $B\subseteq V$. Thus $U$ and $V$ witness that $E$ is not connected according to definition 2.

This argument more generally shows that the definitions are equivalent for subsets of any metric space. For more general topological spaces, though, they are not equivalent (see Definition of disconnected subsets in metric spaces and in more general settings), and definition 1 is the standard definition.

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  • $\begingroup$ Helps a lot! Great appreciate! $\endgroup$ – AbnerYe Oct 4 '17 at 2:11

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