# line equidistant from two sets in the plane

Suppose $A$ and $B$ are disjoint subsets of the plane, both closed, nonempty, and connected. Define $E(A, B)$ as the set of points in the plane equidistant from $A$ and $B$. For example, if $A$ is a point and $B$ is a straight line, $E$ is a parabola.

(1) I think that $E$ is always homeomorphic to a circle or a line. Is that right?

(2) Are there any generalizations? For example, if instead of the plane we take the ambient space to be any $n$-manifold with a metric, is $E$ something like an $(n - 1)$-dimensional CW complex?

And here are some bonus questions...

Are there conditions on $A$ and $B$ which will guarantee that $E$ is a submanifold of codimension 1? For example, in $\mathbb{R}^3$, $E$ is not always a surface even if $A$ and $B$ are connected. (To see this, let $A$ and $B$ be like two forks kissing: for example, $A$ is the union of the three line segments given by the sequence $(-1, 0, 0)$, $(-1, 0, 4)$, $(1, 0, 4)$, $(1, 0, 0)$, and $B$ is given similarly by $(0, 1, 4)$, $(0, 1, 0)$, $(0, -1, 0)$, $(0, -1, 4)$.) But perhaps $E$ is a surface if $A$ and $B$ are separated by a hyperplane.

What do we get in $\mathbb{R}^n$ if $A$ and $B$ are finite?

I think I can prove that if $A$ and $B$ are graphs of continuous functions $\mathbb{R} \rightarrow \mathbb{R}$ , then $E$ is too. Is there a similar result for $C^k$ functions? Is there a nice description of the $E$ function if the $A$ and $B$ functions are, say, polynomials?

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Gosh...not even the ghost of an answer. Do I really need to put a bounty on this one? Surely someone knows something. –  Hew Wolff Oct 16 '12 at 4:22
I think you want to think of $E$ as the zero level set of $e(x) = d(x,A) - d(x,B)$. Then I imagine you might able to prove that the sets $\{x: e(x) < 0\}$ and $\{x: e(x) > 0\}$ are connected, from which nice things should follow. –  Rahul Oct 20 '12 at 4:51
I have to say I don't understand your example in $\mathbb R^3$. When $i=1$, $A_1 = e^{2\pi/4} \not\in S^1$. Perhaps you meant $A_k = e^{2\pi ik/4}$ instead? And how do the four points $A_0A_0'A_2A_2'$ (which I presume are $(1,0,0)$, $(1,0,4)$, $(-1,0,0)$, $(-1,0,4)$ in $\mathbb R^3$) specify a line? –  Rahul Oct 20 '12 at 4:56
Thanks @RahulNarain, I fixed up that example. –  Hew Wolff Oct 22 '12 at 19:44

I don't think any of these references answers exactly your question, but each addresses special cases of your question. If nothing else, it give you the vocabulary to search further: the key term is bisector:

• "Bisectors of Linearly Separable Sets." Lee R. Nackman and Vijay Srinivasan. Discrete Comput Geom 6:263-275 (1991). (Springer link).

• "The Bisector of a Point and a Plane Algebraic Curve." Huahao Shou, Tao Li and Yongwei Miao. Communications in Computer and Information Science, 2011, Volume 164, 449-455, 2011. (Springer link).

• "Bisector Curves of Planar Rational Curves." Gershon Elber , Myung-soo Kim. 1999. (Citeseer link).

Below is Fig. 6(a) from the 3rd paper above:

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Awesome, thank you. –  Hew Wolff Oct 22 '12 at 20:08

Let $A$ be the closed segment from $(0,0)$ to $(0,1)$ plus the closed segment from $(0,0$) to $(0,-1)$ and $B$ the closed segment from $(1,0)$ to $(1,1)$ plus the closed segment from $(1,0)$ to $(2,0)$. I think the equidistant set is homeomorphic to a sans-serif Y.

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I think the equidistant set is the line $x=1/2$ here. If $y<0$, the points on $A$ and $B$ closest to $(x,y)$ are $(0,0)$ and $(1,0)$ respectively, and similarly when $y>1$. –  Rahul Oct 20 '12 at 5:00
@RahulNarain: you are right for the previous version. It was too simple. Try this one... –  Ross Millikan Oct 20 '12 at 5:22
I don't think that works either. Can you draw a picture? Also: if the points to the left of the Y are closer to $A$, and the ones to the right are closer to $B$, what about the points above the fork of the Y? –  Rahul Oct 20 '12 at 6:00
@RahulNarain: You are right. I was trying to get a hyperbola from the end points, but it doesn't work. –  Ross Millikan Oct 20 '12 at 14:34