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

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.

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?

• Closed subsets of Rn are metric spaces, hence normal, satisfying the necessary property. en.wikipedia.org/wiki/Normal_space So the statement is true. May 21 '16 at 22:22
• @William Phew, thanks. Is my attempt a viable route for proving it? May 21 '16 at 22:23
• I'm lazy so haven't evaluated the details exactly, but the gist of it seems like the right direction. Basically the idea is that since the sets are closed and disjoint, their closures do not intersect, and hence they have non-zero distance (epsilon) from each other, so take open sets U and V which cover each of the two sets without going more than epsilon/3 away from their boundary, hence U and V can't intersect. en.wikipedia.org/wiki/… Obviously you will have to work harder to make that precise. May 21 '16 at 22:27
• In the picture in the Wikipedia article I linked to, for example, take open sets around A and B whose farthest points from A and B are no more than 1 anyway; if A and B were closed spheres with radius R, and the distance between them was three, then take the open spheres centered at the centers of A and B and with radius R+1 May 21 '16 at 22:29
• @William. Disjoint closed subsets of a metric space do not necessarily have positive distance from each other. For example,in $R^2$ let $A=\{(x,1/x):x>0\}$ and $B=\{(x,-1/x):x>0\}.$ Then $d( (x,1/x),(x,-1/x))=2/x,$ which has no positive lower bound May 21 '16 at 23:36

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 $d(x,B)\gt0=d(x,A).$
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.