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I want to show that that the following two notions of separation in a metric space are equivalent

$\textbf{1st Definition:}$ Let $(X,d)$ be a metric space then a subset $E$ of $X$ is said to be separated if there exists two non-empty and disjoint open sets $U_1$ and $U_2$ in $X$ such that $E \subset U_1 \cup U_2$ and $E\cap U_1 \neq \emptyset \neq E \cap U_2$.

$\textbf{2nd Definition:}$ Let $(X,d)$ be a metric space then a set subset $E$ of $X$ is said to be separated if there exists non-empty and disjoint open sets $V_1$ and $V_2$ in metric space $(E,d\restriction _{E\times E})$ such that $E = V_1 \cup V_2$.

$\textbf{My attempt:}$ It's pretty easy to show that $(1st) \implies (2nd)$ but I am stuck at the reverse implication. If a subset $E$ of $X$ is separated according to 2nd defnition then $E = V_1 \cup V_2$ for some non-empty and disjoint open sets $V_1$ and $V_2$ in $E$. Therefore there exists open sets $U_1$ and $U_2$ in the ambient space $X$ such that $V_1 = E \cap U_1$ and $V_2 = E \cap U_2$. $U_1$ and $U_2$ satisfies all the conditions of 1st definition except disjointness. If I take $U_3 = U_1 \cap (\overline{U_2})^{c}$ to replace $U_1$ in order to impose disjointess I find it difficult to prove that $V_1 = E \cap U_3$.

Ryan
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1 Answers1

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First note that you cannot take arbitrary $U_1$ and $U_2$, for example $U_2 = X$ won't be good for your argument.

We can construct $U_1$ and $U_2$ so that they are disjoint. For every $x\in V_1$ consider $r_x = d(x,V_2)/3$. Obviously, $B(x,r_x)\cap V_2 = \emptyset$ (balls are in $X$). We take $U_1 = \bigcup\limits_{x\in V_1} B(x,r_x)$, $U_2$ is constructed similarly. We will show that they indeed are disjoint.

Assume that there exists $z\in U_1\cap U_2$. Then, by definition, $z\in B(x,d(x,V_2)/3)$ for some $x\in V_1$ and $z\in B(y,d(y,V_1)/3)$ for some $y\in V_2$. Without loss of generality assume that $d(x,V_2) \leq d(y,V_1)$. Then, $$d(x,y) \leq d(x,z) + d(z,y) < d(x,V_2)/3 + d(y,V_1)/3 \leq \frac 23 d(y,V_1),$$ which contradicts the definition of $d(y,V_1)$. Therefore $U_1\cap U_2 = \emptyset$.

ajr
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