A Combinatorial Proof of the Dense Hindman Theorem

TL;DR

This paper provides a direct combinatorial proof of the Dense Hindman Theorem, which traditionally relied on ultrafilters, thus offering a more elementary approach to a complex combinatorial result.

Contribution

It introduces the first purely combinatorial proof of the Dense Hindman Theorem, removing the need for ultrafilter-based arguments.

Findings

Proof is valid for any notion of density satisfying certain properties.

The combinatorial approach simplifies understanding of the theorem.

The method can be adapted to related combinatorial density results.

Abstract

The Dense Hindman's Theorem states that, in any finite coloring of the integers, one may find a single color and a "dense" set $B_1$, for each $b_1\in B_1$ a "dense" set $B_2^{b_1}$ (depending on $b_1$), for each $b_2\in B_2^{b_1}$ a "dense" set $B_3^{b_1,b_2}$ (depending on $b_1,b_2$), and so on, such that for any such sequence of $b_i$, all finite sums belong to the chosen color. (Here density is often taken to be "piecewise syndetic", but the proof is unchanged for any notion of density satisfying certain properties.) This theorem is an example of a combinatorial statement for which the only known proof requires the use of ultrafilters or a similar infinitary formalism. Here we give a direct combinatorial proof of the theorem.