Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 14 / 1

Let $c:\mathbb{N}\to \{1,\dots ,r\}$ be the given finite colouring, with $\mathbb{N}=\{1,2,\dots \}$. Give each unordered pair $\{a,b\}$ the colour $c(a+b)$. The infinite Ramsey theorem for r-sets, in its two-element case, gives an infinite homogeneous set for a colouring $X$. Enumerate it as $x_1<x_2<\cdots$. Every distinct-index sum $x_i+x_j$ has the common pair colour, soThis application of the infinite Ramsey theorem for r-sets uses only unordered pairs.

For the finite colouring $c$, colour each unordered pair $\{a,b\}$, written with $a<b$, by $c(a+2b)$. The infinite Ramsey theorem for r-sets again gives an infinite homogeneous set for a colouring. If its increasing enumeration is $x_1<x_2<\cdots$, thenThe ordering in this pair colouring is essential: it assigns the coefficient $2$ to the larger member.

Use the alternating dyadic colouring excludes symmetric weighted pair sums construction:Thus consecutive dyadic intervals $[2^k,2^{k+1})$ have opposite colours. Suppose an increasing infinite sequence $X=(x_j)$ had all its distinct-index weighted sums in one colour class. Set $\delta (y)=2^{\lfloor {\log }_{2}y\rfloor +1}-y$, the distance to the next strictly larger power of $2$.

If $\delta$ is unbounded on $X$, fix a term $x$ and choose a later term $y>2x$ with $\delta (y)>2x$. Put $L=2^{\lfloor {\log }_{2}y\rfloor }$, so $y=2L-\delta (y)$. ThenThese two weighted sums lie in adjacent dyadic intervals and therefore have opposite colours.

If $\delta$ is bounded by $D$ on $X$, choose a term $x>2D$ and a later term $y>2x$. With the same $L$, we haveFor the upper bounds, $y+2x<2y<4L$ and $2y+x<\frac{5}{2}y<5L<8L$. These weighted sums also lie in adjacent dyadic intervals. In both cases they are $x_i+2x_j$ and $x_j+2x_i$ for two distinct indices, contradicting the common colour class. Hence the symmetric distinct-index conclusion is false, even for this two-colour finite colouring.

Every ultrafilter contains exactly one cell of any finite set partition. Indeed, two disjoint cells cannot both belong to a proper filter, while if none belonged to the ultrafilter, all their complements would belong to it and their finite intersection would be empty.

Apply this property to the two colour classes in (iii). One belongs to any proposed ultrafilter, but neither contains a symmetric weighted-pair configuration from an increasing infinite sequence. This contradicts the proposed property of every member. Therefore no such ultrafilter exists.