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Statement

Let $k$ be a positive prime power and put $m=4k$. There is a finite set $X_m\subset\mathbb R^{d_m}$, where

$$d_m=\binom m2-1,$$

with the following properties:

1. $|X_m|=\frac12\binom m{m/2}$;
2. after a common rescaling, $\operatorname{diam}(X_m)=1$; and
3. every subset of $X_m$ having diameter strictly less than $1$ contains at most $2\binom{m-1}{m/4-1}$ points.

Consequently, if $f(d)$ denotes the Borsuk partition function, then

$$f(d_m)\ge \frac{\frac12\binom m{m/2}} {2\binom{m-1}{m/4-1}} =\frac{\binom m{m/2}}{\binom m{m/4}}.$$

Construction and proof

Let $V=[m]$ and let $W=\binom V2$ be the edge set of the complete graph on $V$. For every unordered bipartition $P=\{A,B\}$ with $|A|=|B|=m/2=2k$, define

$$S_P=S(A,B):=\{\{a,b\}\in W:a\in A,\ b\in B\}.$$

Thus $S_P$ is the edge set of the complete bipartite graph across the cut. The unordered cut has exactly two ordered descriptions, $(A,B)$ and $(B,A)$, and its crossing-edge set determines those two sides: for each vertex, its non-neighbors in $S_P$ are exactly the other vertices on its side. Hence the family $\mathcal K$ of all such $S_P$ has

$$|\mathcal K|=\frac12\binom m{m/2}.$$

Every member has size $|S_P|=|A||B|=m^2/4$. Represent $S_P$ by its incidence vector $x_P\in\{0,1\}^{W}$. All these vectors lie in the affine hyperplane

$$\sum_{e\in W}x_e=m^2/4,$$

whose dimension is $|W|-1=\binom m2-1=d_m$. An affine isometry identifies this hyperplane with $\mathbb R^{d_m}$.

Consider two cuts $P=\{A,B\}$ and $Q=\{C,D\}$. After choosing the labels, write $r=|A\cap C|$. The four cells determined by the two bipartitions have sizes

$$r,\quad 2k-r,\quad 2k-r,\quad r.$$

An edge crosses both cuts precisely when its endpoints lie in the two opposite cells of one of the two matching pairs. Hence

$$|S_P\cap S_Q| =r^2+(2k-r)^2 =2(r-k)^2+2k^2.$$

The minimum possible intersection is therefore $2k^2=m^2/8$, attained exactly when $r=k$; such distinct cuts exist by choosing two $2k$-sets with intersection $k$. Since all incidence vectors have the same weight,

$$\|x_P-x_Q\|_2^2 =|S_P\mathbin\triangle S_Q| =\frac{m^2}{2}-2|S_P\cap S_Q|.$$

Thus the minimum intersection corresponds to the maximum squared distance $m^2/4$, and this value is attained. The diameter of the incidence configuration is $m/2$; multiplying every vector by $2/m$ makes the diameter one without changing which pairs attain it.

It remains to bound a subset $\mathcal L\subseteq\mathcal K$ of diameter strictly smaller than the full configuration. Choose one side $A_P$ of each cut $P\in\mathcal L$. If distinct chosen sides $A_P,A_Q$ had $|A_P\cap A_Q|=k=m/4$, then the displayed calculation would put $x_P,x_Q$ at the diameter, a contradiction. The chosen sides therefore form a family of $m/2$-subsets with the intersection size $m/4$ forbidden. Applying the exact Frankl–Wilson interface gives

$$|\mathcal L|\le 2\binom{m-1}{m/4-1}.$$

If a partition of $X_m$ has $q$ parts of smaller diameter, counting points in those parts gives

$$\frac12\binom m{m/2} \le q\,2\binom{m-1}{m/4-1}.$$

Finally,

$$\binom{m-1}{m/4-1}=\frac14\binom m{m/4},$$

which proves the asserted ratio.

Source wording

This is a complete expansion of Section 2 on physical PDF p. 2 (journal p. 61). The printed sentence says that two sets with minimum intersection “realize the minimal distance.” The incidence-vector identity above shows that they realize the maximal distance; the next printed sentence also uses the minimum-intersection condition. The proof here follows the displayed construction and records that one-word source defect explicitly.