= Alternating dyadic intervals contain disjoint IP sets
{title2=$\operatorname{FS}(4^j),\ \operatorname{FS}(2\cdot4^j)$}
Let $E$ be the union of integer intervals $[2^{2j},2^{2j+1})$ for $j\geq0$. Every nonempty sum of distinct $4^j$ has largest summand $4^J$ and lies in $[4^J,2\cdot4^J)$, since the sum of all smaller summands is $(4^J-1)/3$. Thus $\operatorname{FS}(4^j)\subseteq E$. Doubling gives $\operatorname{FS}(2\cdot4^j)\subseteq E^c$. Both sides are <IP sets>. Each can be adjoined to the <IP-star filter> while preserving the <finite intersection property>, yielding two distinct <IP ultrafilters>.
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