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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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    Thanks @RahulNarain, I fixed up that example.2012-10-22

2 Answers 2

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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:
          Fig.6a

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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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    @RahulNarain: You are right. I was trying to get a hyperbola from the end points, but it doesn't work.2012-10-20