A counterexample to reflection augmentation

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The following proposed lemma is false: if a dissociated set $D\subset(0,M/2)$ has size $d$, then $\{M\}\cup D\cup(M-D)$ contains a dissociated $(d+1)$-subset.

Take

$$D=\{2,4,10,11\},\qquad M=23.$$

The 16 subset sums of $D$ are

$$0,2,4,6,10,11,12,13,14,15,16,17,21,23,25,27,$$

so $D$ is dissociated, and $D\subset(0,23/2)$. Nevertheless,

$$P=\{2,4,10,11,12,13,19,21,23\}$$

has dissociation dimension four. A conceptual finite certificate that every five-subset is dependent follows.

The underlying configuration

Write the points symbolically as

$$\begin{array}{c|ccccccccc} \text{label}&A&B&C&D&E&F&G&H&M\\ \hline \text{value}&a&2a&b&c&a+b&a+c&b+c-a&b+c&a+b+c. \end{array}$$

The counterexample uses $(a,b,c)=(2,10,11)$. The argument below uses only these symbolic identities; it proves the same dependence conclusion whenever the nine displayed points are distinct.

There are four complementary pairs

$$\{A,H\},\quad\{B,G\},\quad\{C,F\},\quad\{D,E\}, \tag{1}$$

each summing to $M$. We shall also use

$$\begin{array}{lll} A+E=B+C,& A+F=B+D,& A+G=C+D,\\ B+H=E+F,& C+H=E+G,& D+H=F+G. \end{array} \tag{2}$$

Reflection $x\mapsto M-x$ exchanges the members of every pair in (1) and preserves equalities between two-element sums.

Five-subsets containing $M$

A five-subset containing $M$ is dependent if it contains either complete pair in (1). Otherwise it contains exactly one element of each pair. Let $r_A,r_B,r_C,r_D\in\{0,1\}$ indicate whether the chosen member is the reflected member, and let $x_A,x_B,x_C,x_D$ be the chosen values. Since $M=A+C+D$, substitution gives

$$(1-r_A-r_C-r_D)M =(-1)^{r_A}x_A+(-1)^{r_C}x_C+(-1)^{r_D}x_D. \tag{3}$$

Unless $r_A=r_C=r_D=1$, every coefficient in (3) belongs to $\{-1,0,1\}$, providing a nontrivial ternary relation.

In the remaining case the chosen set contains $H,F,E$. If its fourth non-$M$ member is $B$, use $B+H=E+F$. If that member is $G$, use $M+H=E+F+G$. Thus every five-subset containing $M$ is dependent.

Five-subsets not containing $M$

Such a set contains a complete pair in (1). If it contains two complete pairs, their equal sums give a dependence. Otherwise it consists of one complete pair and one member of each of the other three pairs.

If its two-element sums are all distinct, (2) restricts the remaining three members as in the following table. The last column supplies a dependence in every surviving case.

For completeness, the restrictions in the table can be checked without enumerating all five-subsets. If $A,H$ are present and $B$ is chosen, the other two choices $(C,E),(F,D),(F,E)$ are excluded by the first, second, and fourth identities in (2), leaving $C,D$. Reflection handles the choice $G$. If $B,G$ are present and $A$ is chosen, the choices $(C,D),(C,E),(F,D)$ are excluded by the third, first, and second identities, leaving $F,E$; reflection handles $H$.

If $C,F$ are present and $A$ is chosen, choosing $B$ is impossible: either choice from $\{D,E\}$ triggers one of the first two identities. Hence $G$ is chosen; the third identity then excludes $D$, leaving $E$. Reflection handles $H$. Interchanging $b,c$ gives the final two rows.

This covers every five-subset of $P$. Hence $d(P)=4$, and the reflection-augmentation lemma fails. Since $|P|=9<2^4$, this example does not contradict the original extremal target.