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By separable I mean, given any pair of disjoint closed sets, $A_1$ and $A_2$, a metric space $(X,d)$, there exists disjoint open sets $O_1$ and $O_2$ such that $A_1\subset O_1$ and $A_2\subset O_2$.

I can prove that for any $x\in A_1$, $x$ is separable from $A_2$, because it's easy to prove that $d(x,A_2)=\delta>0$. (If not, you can find a sequence in $A_2$ that converges to $x$, which contradicts with $A_2$ being closed.)

However it is still not obvious that $A_1$ and $A_2$ are separable because each time you find different $\delta$.

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The name for this property is normality. Separability is an entirely different kettle of fish. –  Arthur Fischer Sep 21 '12 at 3:40
Oh. Never mind. I see it. It doesn't matter you find different $\delta$ each time because you can simply union all the open sets you find and you get the big open set that is disjoint with $A_2$. –  Voldemort Sep 21 '12 at 3:41
@ArthurFischer: Oh, That's just a temporary term I made up. And I gave the precise definition of it. –  Voldemort Sep 21 '12 at 3:43
@Arthur However, it standard terminology to refer to the pair $(O_1, O_2)$ as a "separation" of $A_1\cup A_2$, and to say that this pair "separates" $A_1\cup A_2$. The property of normality is determined by one of the "separation axioms", and one often says that $A_1$ and $A_2$ are "separated by" $O_1$ and $O_2$, so Lord Voldemort is on quite firm ground here. –  MJD Sep 21 '12 at 4:20
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1 Answer

up vote 1 down vote accepted

Maybe this is what you had in mind in your comment, but here is one way:

$O_1 = \bigcup_{a\in A_1} B(a,d(a,A_2)/2)$, and

$O_2 = \bigcup_{a\in A_2} B(a,d(a,A_1)/2)$,

where $B(x,d)$ is the open ball of center $x$ and radius $d$.

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This is correct. One little thing I want to add. You should prove that $d(a,A_1)>0$ and $d(a,A_2)>0$, this is the direct consequence of $A_1$ and $A_2$ being close. –  Voldemort Sep 22 '12 at 3:33
@Voldemort True, but I didn't mention that part because you indicated that you knew how to prove it. –  Trevor Wilson Sep 22 '12 at 4:35
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