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Chang's theorem says that ${\mathrm{Spec}}_{\rho }(1_A)$ is contained in the span of a dissociated set of size $O({\rho }^{-2}\log ({\alpha }^{-1}))$.

Use normalized Fourier analysis on a finite abelian group. If $A\subseteq {\mathbb{F}}_2^n$ has density $\alpha$, its large spectrum at level $\rho >0$ isThe Chang theorem states that this spectrum is contained in a subspace of dimensionEquivalently, it is contained in the span of a dissociated set of that size.

Choose $ϵ=1/6$ and a sufficiently large absolute constant $k$. Part (iii)'s size bound givesLetand choose a maximal dissociated set $\Lambda \subseteq \Gamma$. The entropy form of the Chang theorem givesand maximality gives $\Gamma \subseteq \mathrm{Span}(\Lambda )$.

Put $B=B(\Lambda ,1/(6|\Lambda |))$. If $y\in B$ and $\gamma \in \Gamma$, expressing $\gamma$ as a product of characters in $\Lambda$ and their inverses givesFor $f=1_{-A}\ast {\mu }_A$, Parseval identity givesAlso $\hat{\mu }(\gamma )=|\widehat{{\mu }_Y}(\gamma ){|}^{2k}$. Fourier inversion theorem therefore yieldsonce $k$ is large enough.

If $\Lambda$ is empty, interpret $B$ as $G$; the same Fourier estimate, using only the second term, is even stronger.

Part (iii) gives $\Vert f\ast \mu -f{\Vert }_{\infty }\le 1/6$. Applying this at $x$ and $x+y$ and using the last estimate givesFinally $f(0)=1$. Hence $f(y)\ge 1/3>0$ for every $y\in B$, while the support of $f$ is $A-A$. Consequently

For $q=1,\dots ,5$, choose the first $q$ elements of $\{1,2,4,8,16\}$. Every sum of a subset has a unique binary representation, so distinct subsets cannot have the same sum. To see this directly, take the largest power of two belonging to exactly one of two proposed subsets. It exceeds the sum of all smaller powers, and therefore cannot be cancelled by their difference. Thus these chosen sets are dissociated sets and do not have the two-sum property.

If cardinality zero is included, the empty set has only one subset, so it cannot supply two distinct ones either. There is a counterexample at every cardinality from $0$ through $5$.

Let $S$ be a finite set of positive integers and let $T={\sum }_{s\in S}s$. Every sum of a subset is an integer in $\{0,\dots ,T\}$. If $2^{|S|}>T+1$, the pigeonhole principle forces two distinct subsets to have equal sums. Thus $S$ cannot be a dissociated set.

Cancelling their common intersection even produces two disjoint subsets with the same sum. Positivity means neither remaining subset can be empty. Conversely, equality of two distinct subset sums gives a nonzero relation with coefficients in $\{-1,0,1\}$, exactly the obstruction to being a dissociated set.

For an $r$-element subset of $\{1,\dots ,n\}$, the largest possible $T$ is $r(2n-r+1)/2$. Thereforeis a sufficient condition forcing a collision for every such subset. This counting bound is sufficient rather than necessary. Powers of two show how a collision-free collection can instead use the binary representation to make all sums different.