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This problem is present in "Supplements to the Exercises in Chapter 1-7 of Walter Rudin's Principles of Mathematical analysis" by Prof. George M. Bergman, which states as follows,

Let $X$ be a $4$-element set $\{w, x, y, z\}$, and let $d$ be the metric on $X$ under which the distance from $w$ to each of the other points is $1$, and the distance between any two of those points is $2$.

  1. Show that no function $f$ of $X$ into a space $\mathbb{R}^{k}$ is distance-preserving, i.e., satisfies $|f(p) - f(q)| = d(p, q)$ for all $p, q \in X$.
  2. The above example has the property that every $3$-point subset of $X$ can be embedded (mapped by a distance-preserving map) into space $\mathbb{R}^{k}$ for some $k$, but the whole $4$-point space cannot be so embedded for any $k$. Can you find a $5$-point metric space, every $4$-point subset of which can be so embedded but such that the whole $5$-point space cannot?

I searched on the internet about this topic and found the Cayley-Menger determinant may be helpful to part 2. However, here is a concrete $5$-point case and I was wondering if there are some intuitive counterexamples? And I also have no clue about part 1, can anyone give me some hints on it. Thanks in advance.

Jose Avilez
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hmeng
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2 Answers2

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  1. Suppose that $f$ exists. Without loss of generality, we may set $f(w) = 0$. Thus, $f(x)$, $f(y)$ and $f(z)$ must be the vertices of an equilateral triangle inscribed on the unit sphere, whose sides have length $2$. Alas, the largest equilateral triangles that you can inscribe on a unit sphere have sides of length $\sqrt 3$.

  2. In general, let $X = \{ w, x_1, \dots, x_n \}$. Turn $X$ into a metric space by setting $d(w, x_i) = 1$ and $d(x_i, x_j) = \lambda$, where $\lambda$ is just large enough that you can inscribe a regular simplex with $n-1$ vertices on the unit sphere, but not a regular simplex with $n$ vertices.

isekaijin
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  • For part 1. I understand your idea, but it seems that we need to prove it for cases $k$ is a positive integer. The idea here to me will be complicated if $k$ is large. Maybe I misunderstand your idea somehow. For part 2, can u elaborate on your last inequality or give me some resourse to check it by myself. – hmeng Jul 18 '21 at 00:37
  • For part 1, when I said “unit sphere”, I meant the unit sphere in $\mathbb R^k$, not the unit sphere in $\mathbb R^3$. In any case, the largest regular $n$-gons that you can embed must lie on a plane containing the center of the sphere, which reduces the problem to the case $k = 2$. – isekaijin Jul 18 '21 at 00:39
  • For part 2, to find the bounds for $\lambda$, compute the distance between two consecutive $n$-th roots of unity, considered as complex numbers. (This is justified because we have reduced the problem to the case $\mathbb R^2 \cong \mathbb C$. – isekaijin Jul 18 '21 at 00:40
  • Oh, I got your part 1. For part 2, I know the procedure you provide, but why can this problem be reduced to find the distance between two consecutive $n$th roots of unity? Sorry for my slow – hmeng Jul 18 '21 at 00:45
  • For rotational symmetry reasons, you may assume that your regular $n$-gon is formed by the $n$-th roots of unity on the complex plane. So you just need to take the norm of the complex number $1 - e^{2\pi i/n}$. – isekaijin Jul 18 '21 at 00:47
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    Wait, you are still talking about the four points located in a plane? I thought it may be a pyramid with a triangular base. In this case, the complex plane is not enough to do it, right? – hmeng Jul 18 '21 at 00:54
  • Oops, you are absolutely right. Use a tetrahedron, rather than a square. I updated my answer accordingly. – isekaijin Jul 18 '21 at 00:54
  • Your idea is inspiring, maybe your direction is right, please allow me more time to think about your way – hmeng Jul 18 '21 at 00:58
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It is easy to prove that every $3$-point of $\{w, x, y, z\}$ can be embedded into $\mathbb{R}^{k}$ for $k=3$ by an equilateral triangle with vertex $x,y,z$ or straight lines $\{w,x,z\}$, $\{w,x,y\}$ and $\{w,y,z\}$.

A counterexample of $5$-point case is as follows, let $\{v^{\prime},w^{\prime},x^{\prime},y^{\prime},z^{\prime}\}$ be the image of $\{v, w, x, y, z\}$ under $f$. Assume, without loss of generality, $v^{\prime}$ is the origin, and we follow the same idea of part 1, and construct the $4$-point set $\{w^{\prime},x^{\prime},y^{\prime},z^{\prime}\}$ as an equilateral figure with length of side $1$ in $\mathbb{R}^{k}$, e.g., a regular $4$-simplex in $\mathbb{R}^{3}$. Let the distance between $v^{\prime}$ and other points are all equal to $r$. By the Jung's theorem\index{Jung's theorem}, the upper bound of radius of the minimum ball enclosing these $4$ points is $\sqrt{\frac{k}{2(k+1)}}$ and this bound is attained by the regular $k$-simplex in $\mathbb{R}^{k}$. Suppose the $5$-point set is Euclidean, then we can always find a regular $k$-simplex in $\mathbb{R}^{k}$ with vertices $\{w^{\prime},x^{\prime},y^{\prime},z^{\prime}\}$ and the center at origin $v^{\prime}$. Then the regular $k$-simplex can be enclosed by a closed ball $\overline{B}(v^{\prime},r)$. Since for $3 \leq k < +\infty$, we have \begin{equation*} \sqrt{\frac{3}{8}} \leq r = \sqrt{\frac{k}{2(k+1)}} <\sqrt{\frac{1}{2}} \end{equation*} by letting $\sqrt{\frac{1}{3}} = r < \sqrt{\frac{3}{8}}$, such $v^{\prime}$ does not exist. Note that every $4$-points set is Euclidean since $\{w^{\prime},x^{\prime},y^{\prime},z^{\prime}\}$ can be a tetrahedron and $\{v^{\prime},x^{\prime},y^{\prime},z^{\prime}\}$ can be a equilateral triangle with center at $v^{\prime}$.

hmeng
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