2
$\begingroup$

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 $V_1 = E \cap U_3$.

$\endgroup$
0
0
$\begingroup$

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 $U_1\cap U_2 = \emptyset$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.