# Perpendicular bisector in a different metric

Consider two points $$a, b \in \mathbb{R}^2$$. Then from elementary geometry, the set of points that are equidistant from both $$a$$ and $$b$$ is precisely the perpendicular bisector of the line segment $$ab$$. Now suppose we put some other metric on $$\mathbb{R}^2$$ that generates the standard topology. Then what can we say about the set of points that are equidistant from $$a$$ and $$b$$? Intuitively, it seems to me that this set should be "one-dimensional" and cannot contain an open set. Is this true?

• Can you think of examples of such metrics which are equivalent to the usual metric? It would be helpful if you came up with an example of such a metric , and then tried to plot the perpendicular bisector. Scalar multiples of the usual metric would still give the perpendicular bisector. – астон вілла олоф мэллбэрг Dec 13 '18 at 7:13
• I agree with @астон вілла олоф мэллбэрг : you should take an example of such a metric and then try first by yourself to find out what such a set looks like (and - why not - show the result within your question). – Jean Marie Dec 13 '18 at 7:49
• A reference (taxicab geometry = $\|\|_1$ geometry) : jwilson.coe.uga.edu/EMAT6680Fa06/Sexton/GeoFinalProject/Taxicab/… – Jean Marie Dec 13 '18 at 15:39
• @ Jean Marie Thanks! For all the standard metrics I have tried the assertion is true. However, I am unable to prove the general assertion. – Jaikrishnan Dec 13 '18 at 15:48

## 1 Answer

Very interesting question!

Metrics

For an arbitrary metric on $$\mathbb{R}^2$$ that generates the Euclidean topology, the set of all points equidistant from $$2$$ distinct points does not necessarily have to be "one-dimensional".

Consider the metric $$d(p,q)=\min\{\|p-q\|_2,1\}$$ where $$\|(x,y)\|_2=\sqrt{x^2+y^2}$$ is the Euclidean norm. Do convince yourself that this is a metric. (non-negative, non-degenerate, symmetric, triangle inequality) However, for any two points, there is only a bounded (in the sense of the Euclidean metric) set of points that are not equidistant from them.

Norms

Even if we require our metric to be induced from a norm, the set of equidistant points does not have to be "one-dimensional".

Consider the norm $$\|(x,y)\|_\infty=\max\{|x|,|y|\}$$. Do convince yourself that this is a norm. (non-negative, non-degenerate, scaling, triangle inequality) However, if we take the points $$(1,0)$$ and $$(-1,0)$$ then all points $$(x,y)$$ with $$1-y\leq x\leq1+y$$ are equidistant from $$p$$ and $$q$$.

Strictly convex norms

However, if we require our metric to be induced from a strictly convex norm, then we can prove that the set of equidistant points is homeomorphic to $$\mathbb{R}$$. Strictly convex means there is no line segment in the unit sphere. Algebraically, for all $$p\neq q$$ with $$\|p\|=\|q\|=1$$ we have $$\|p+q\|<2$$. Examples of strictly convex norms are all $$p$$-norms for $$1.

Proof (Not very rigorous, but hopefully convincing enough.)

Let $$\|\cdot\|$$ be a strictly convex norm on $$\mathbb{R}^2$$ and let $$p\neq q\in\mathbb{R}^2$$. Define $$s:=\|p-q\|$$.

Claim 1 For all $$r\geq\frac{s}2$$ on each side (non-strict) of the line through $$p$$ and $$q$$ there is exactly one point equidistant from $$p$$ and $$q$$ with distance $$r$$.

Proof of Claim 1 Assume for the contrary that there are two distinct points $$a$$ and $$b$$ to the right of the line through $$p$$ and $$q$$, both equidistant from $$p$$ and $$q$$ with the same distance $$r$$. We then find that all of the points $$p-a$$, $$p-b$$, $$q-a$$ and $$q-b$$ and their inverses lie in the sphere of radius $$r$$. However, all of the line segments $$(p-a)(p-b)$$, $$(q-a)(q-b)$$, $$(b-p)(a-p)$$ and $$(b-q)(b-q)$$ are different, are parallel, and have the same length. We are forced to conclude there is a line segment inside the sphere of radius $$r$$. This proves Claim 1.

Notice that from the triangle inequality it follows that for $$r<\frac{s}2$$ there are no equidistant points from $$p$$ and $$q$$ with distance $$r$$. Also notice that there is a unique point on the line through $$p$$ and $$q$$ equidistant from $$p$$ and $$q$$, namely $$\frac{p+q}2$$ with distance $$\frac{s}2$$. It follows that we can define a bijection $$f:\mathbb{R}\to E$$, where $$E$$ is the set of points equidistant from $$p$$ and $$q$$. Namely $$f(0)=\frac{p+q}2$$, $$f(t>0)$$ is the unique point on the right of the line through $$p$$ and $$q$$ equidistant from $$p$$ and $$q$$ with distance $$\frac{s}2+t$$, and $$f(t<0)$$ is the unique point to the left of the line through $$p$$ and $$q$$ equidistant from $$p$$ and $$q$$ with distance $$\frac{s}2-t$$.

I don't quite have an idea how you can rigorously prove that $$f$$ is continuous both ways, but this is my intuition. Since the norm is strictly convex, for all $$\epsilon>0$$ the spheres around $$p$$ and $$q$$ through the current equidistant point curve away from each other some $$\delta>0$$ units. Then changing the radius by $$\delta$$ can only move the intersection of the spheres $$\epsilon$$ units.

Final notes

Finally, I want to emphasise how important this strict convexity is. If the norm is not strictly convex, then you can always find two points $$p$$ and $$q$$ for which the equidistance space is not "one-dimensional". Just take the end points of the line segment in the unit sphere.

• +1 very interesting answer. I think you should upvote the question. – Ethan Bolker Dec 13 '18 at 16:15