What can be said of the structure of a metric space without isosceles triangles?

Question

This is a question that has been bothering me in the back of my head for quite some time.

Suppose we have a metric space $X$ with metric $\mathrm{d}$. By an isosceles triangle we mean a tuple of three points $a, b, c \in X$ such that $\mathrm{d}(a,b)=\mathrm{d}(a,c)$, with all three points distinct.

Now, the question is, what can we say of a space $X$ for which no such triangle exists?

I have only gotten some very weak results. Specifically, considering the non-isosceles property we are assuming the space has, we can define a function $\lambda_x(r)$ for every $x\in X$, such that $\mathrm{d}(\lambda_x(r),x)=r$ wherever it is defined. That is, we can define a function that finds the point at distance $r$ from $x$, whenever such a point exists. This $\lambda$ is continuous at 0, and is not continuous on any open subset of $\mathbb{R}$.

Through some simple manipulations involving this $\lambda$, we can show that $X$ must be totally disconnected.

That is just about the best result I have managed to get. A possible thread to continue on is a result found in Arkhangel'skii & Tkachenko (2008), saying that a locally compact hausdorff space is zero-dimensional iff it is totally disconnected.

Answer 1

You can't hope for a topological characterization of metric spaces with no isosceles triangle, because for any metric space with more than two points you can construct an equivalent metric that gives it lots of isosceles triangles. Namely, for any $\epsilon > 0$ take $d' = \min(d, \epsilon)$.

What might be interesting would be those metrizable spaces that have some metric with no isosceles triangles.

Answer 2

The Cantor space $\mathcal C=\{0,1\}^\mathbb N$ with the metric $$d(x,y)=2^{-\min\{n:x(n)\ne y(n)\}}$$ is totally disconnected.

Yet any three distinct elements form an isosceles triangle. Namely take $b$ and $c$ to be the two that agree on the longest initial segment of $\mathbb N$. Then $d(a,b)=d(a,c)$.

Answer 3

There are papers related to dimension in metric spaces ($dim$ or $ind$) that study this definition.

See e.g.: Ludvik Janos and Harold Martin, Metric characterizations of dimension for separable metric spaces, Proc. Amer. Math. Soc. 70 (1978), 209-212, http://www.ams.org/journals/proc/1978-070-02/S0002-9939-1978-0474229-9/

From the above paper: "We define a metric $d$ on a set $A$ to be star rigid iff whenever $x$, $y$ and $z$ are points of $A$ with $y\not=z$, then $d(x,y)\not=d(x,z)$."

Also, from their abstract: "A subset $B$ of a metric space $(X, d)$ is called a $d$-bisector set iff there are distinct points $x$ and $y$ in $X$ with $B=\{z:d(x,z)=d(y,z)\}$. It is shown that if $X$ is a separable metrizable space, then $dim(X)\le n$ iff $X$ has an admissible metric $d$ for which $dim(B)\le n-1$ whenever $B$ is a $d$-bisector set."

(Comment: Of course, in the above, for star rigid spaces, the set $B$ will be empty, and the dimension of the empty set is $-1$.)

See also the more recent related paper: Yasunao Hattori, Congruence and dimension of nonseparable metric spaces, Proc. Amer. Math. Soc. 108 (1990), 1103-1105 http://www.ams.org/journals/proc/1990-108-04/S0002-9939-1990-1000155-8/

Abstract: "In this paper, we prove that, if a metrizable space $X$ has an admissible metric such that $X$ has no two distinct congruent subsets of cardinality $3$, then $ind(X)\le 1$. We also show that if a non-empty metrizable space $X$ has an admissible star-rigid metric, then $ind(X)=0$. The latter answers a question of L. Janos and H. Martin."

The following paper of mine might also be loosely related (I hope :) Strashimir G. Popvassilev, $(m,n)$-Equidistant Sets in $R^k$,$S^k$, and $P^k$, Discrete & Computational Geometry, 40(2008), 2, 279-288 http://link.springer.com/article/10.1007%2Fs00454-007-9048-4

From the abstract: "We call a metric space $X$ $(m,n)$-equidistant if, when $A⊆X$ has exactly $m$ points, there are exactly $n$ points in $X$ each of which is equidistant from (the points of) $A$. We prove that, for $k≥2$, the Euclidean space $R^k$ contains an $(m,1)$-equidistant set if and only if $k≥m$. Although the sphere $S^2$ is $(3,2)$-equidistant, $S^3$ and $R^4$ contain no $(4,2)$-equidistant sets."

Answer 4

Since you say that metric spaces without isosceles triangles must be totally disconnected, it make sense to consider the Cantor set $C$: there exist uncountably many metrics on $C$ without isosceles triangle, even up to bi-Lipschitz equivalence.

Indeed, choose any $\lambda\in(1/3,1)$ and consider the $\lambda$-middle Cantor subset of the interval $C_\lambda$ (remove the central interval of length $\lambda\cdot 1$ from $[0,1]$, then inductively remove the central interval of length $\lambda\cdot \ell$ from each remaining interval of length $\ell$). Endow this set with the distance induced from the Euclidean distance. Given any pair of points $x,y\in C_\lambda$, their middle point $(x+y)/2$ is not in $C_\lambda$ (consider the smallest interval in the inductive construction of $C_\lambda$ that contains both $x$ and $y$: $(x+y)/2$ must lie in its removed central interval). Now, given that $C_\lambda$ is a subset of the line, the only possible isosceles triangles are of the form $x,y,(x+y)/2$. Moreover, for different $\lambda$ the $C_\lambda$ have different Hausdorff dimension, so are bi-Lipschitz inequivalent.

A question left open and that seems interesting, is to compute the supremum $\bar\alpha$ of all $\alpha$ such that there exists a metric space without isosceles triangles of Hausdorff dimension $\alpha$. In particular, is this supremum finite?

Regarding this question, one can do better than the above construction: for each countable collection of triples of points in the line, Tamás Keleti constructed a compact subset of the line of Hausdorff dimension $1$ not containing any similar copy of any of the element of the collection (Analysis and PDE Vol. 1 (2008), No. 1, 29–33). Applying this to $(0,1,1/2)$ shows that $\bar\alpha\ge 1$. I would be surprised if this turned out to be optimal.