IP set ($\mathrm{FS}(x_1,x_2,\dots )\subseteq A$)

A subset $A$ of the positive integers is an IP set if it contains a finite-sums set generated by an infinite sequence of positive integers. Sums use distinct indices, although the generators themselves need not be distinct. An IP set is necessarily infinite, because its successive partial sums are unbounded. The partition regularity of IP sets is the relative form of the Hindman theorem needed when applying that theorem inside an already specified IP set.

Let $E$ be the union of integer intervals $[2^{2j},2^{2j+1})$ for $j\ge 0$. Every nonempty sum of distinct $4^j$ has largest summand $4^J$ and lies in $[4^J,2\cdot 4^J)$, since the sum of all smaller summands is $(4^J-1)/3$. Thus $\mathrm{FS}(4^j)\subseteq E$. Doubling gives $\mathrm{FS}(2\cdot 4^j)\subseteq E^c$. 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.

An IP-star set meets every IP set. Equivalently, its complement is not an IP set: a disjoint IP set is precisely an IP subset of the complement. This is a condition on all infinite finite-sums sets, not merely on a single generating sequence.

The IP-star sets form a proper filter on a set over the positive integers. Indeed, sets that are not IP sets are closed under taking subsets. They are also closed under finite unions, because the partition regularity of IP sets would otherwise put an IP set in one of the union's constituent sets. Taking complements gives upward closure and closure under finite intersections for the IP-star sets. The full set belongs to this filter and the empty set does not.

Every finite coloring of an IP set has a color class that is an IP set. Here is a reduction to the ordinary Hindman theorem. Given $\mathrm{FS}(x_i)$ in the original set, color an integer $n>0$ by the color of ${\sum }_{i\in {\mathrm{supp}}_2(n)}{x}_i$, where the support consists of positions of its nonzero digits in the binary expansion. The Hindman theorem produces $w_i>0$ with monochromatic $\mathrm{FS}(w_i)$. Replace the $w_i$ by consecutive, disjoint finite block sums $v_j$ with separated binary supports: after a block has been chosen, take $K$ above all previously used positions; two of $2^K+1$ partial sums of a fresh tail agree modulo $2^K$, so their positive difference supplies the next block sum divisible by $2^K$. Now binary addition of distinct $v_j$ has no carries. Therefore $y_j={\sum }_{i\in {\mathrm{supp}}_2(v_j)}{x}_i$ generates a monochromatic finite-sums set in the original IP set.