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Statement. Let $0<a_1<a_2<\cdots$ satisfy $a_{n+1}\ge2a_n$. Then $A=\{a_n:n\ge1\}$ is Sidon: if $i\le j$, $k\le\ell$, and $a_i+a_j=a_k+a_\ell$, then $i=k$ and $j=\ell$. Repeated summands are allowed.

Source and scope. This is the elementary lacunarity step used by the public constructions and its discussion, read in the dated source record. The proof makes that step explicit. This Sidon lemma is distinct from the rational-vector-space theorem in the separately filed Baumgartner paper.

Proof. If $j>\ell$, then

$$a_i+a_j>a_j\ge a_{\ell+1}\ge2a_\ell\ge a_k+a_\ell,$$

contradicting equality. Interchanging the two pairs rules out $\ell>j$. Thus $j=\ell$, and cancellation gives $a_i=a_k$. Strict increase gives $i=k$. Positivity supplies the strict inequality even when a successive gap is exactly a factor of two. $\square$

Bears on. Problem 198: the lemma supplies the Sidon property in each of the three constructions answering the question negatively; on its own it does not answer it.