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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?

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    $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг Dec 13 '18 at 7:13
  • $\begingroup$ 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). $\endgroup$ – Jean Marie Dec 13 '18 at 7:49
  • $\begingroup$ A reference (taxicab geometry = $\|\|_1$ geometry) : jwilson.coe.uga.edu/EMAT6680Fa06/Sexton/GeoFinalProject/Taxicab/… $\endgroup$ – Jean Marie Dec 13 '18 at 15:39
  • $\begingroup$ @ Jean Marie Thanks! For all the standard metrics I have tried the assertion is true. However, I am unable to prove the general assertion. $\endgroup$ – Jaikrishnan Dec 13 '18 at 15:48
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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 $1<p<\infty$.

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.

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

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