Does the Hausdorff property hold on closed subsets of $\mathbb{R}^n?$

Question

I am trying to prove that given disjoint closed $A,B\subseteq \mathbb{R}^n$, there exist disjoint open $U,V$ containing $A,B$ respectively. In other words that we can take the Hausdorff property to closed subsets of $\mathbb{R}^n$.

I do not know whether this is true or false, but have set about proving that it is true.

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My idea was, for each $x\in A$, to have $\epsilon_x=\inf_{b\in B} \|x-b\|=\|x-\beta_x\|$ for some $\beta_x\in B$ (closure). Define $U_x=\{y:\|x-y\|<\epsilon_x/2\}$. Define similar balls for each element of $B$. Basically the ball around $x\in A$ is disjoint from the ball around $\beta_x$, the closest point to $x$ in $B$.

Have $U=\bigcup_{A} U_a$ and $V=\bigcup_B V_b$.

Now I believe $U,V$ are disjoint but I can't seem to argue it formally. It seems quite obvious by drawing $A,B$ as blobs in a plane, but obviously $A,B$ could be more complicated than nice bounded subsets. So I need to show $U_a\bigcap V_b=\emptyset$ for any $a,b\in A,B$; now if $z$ were in both then $\|z-a\|<\epsilon_a/2$ and $\|z-\beta_a\|>\epsilon_a/2$, and also $\|z-b\|<\epsilon_b/2$...

I feel like the argument is a step away but I can't seem to finish it off. Help?

Answer 1

Yes, this is true in any metric space.

We may assume that $A,B$ are nonempty; if one of them is empty we can just take $\emptyset,X$ as the two open sets. The most obvious construction works fine: let $U$ be the set of points nearer to $A$ than to $B,$ and let $V$ be the set of points nearer to $B$ than to $A.$ That is, take $$U=\{x:d(x,A)\lt d(x,B)\}=\{x:d(x,B)-d(x,A)\gt0\}$$ and let $$V=\{x:d(x,B)\lt d(x,A)\}=\{x:d(x,A)\gt d(x,B)\}$$ where $$d(x,S)=\inf\{d(x,y):y\in S\}.$$ The sets $U,V$ are obviously disjoint. They are open sets because, for a nonempty set $S,$ the function $x\mapsto d(x,S)$ is continuous, as is easily shown using the triangle inequality. The inclusions $A\subseteq U$ and $B\subseteq V$ follow from the assumption that $A,B$ and disjoint and closed; e.g., if $x\in A$ then $x\notin B,$ so $d(x,B)\gt0=d(x,A).$

Answer 2

Note that for all $a \in A$ and $b \in B$ we have $$ \inf_{a' \in A}\|a' - b\| \leq \|a-b\|. $$

Take $a \in A$, $b \in B$. Suppose $z \in U_a \cap V_b$. By the triangle inequality, $$ \|a - b \| \leq \|a - z \| + \|z - b\| < \epsilon_a/2 + \epsilon_b/2 $$ $$ = \frac{\inf_{b' \in B}\|a - b'\|}{2} + \frac{\inf_{a' \in A}\|a' - b\|}{2} $$ $$ \leq \frac{\|a - b\|}{2} + \frac{\|a-b\|}{2} = \|a-b\|. $$ So $\|a-b\| < \|a-b\|$. Contradiction.