Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 10 / 3

Take $\mathbb{N}=\{1,2,\dots \}$. We first prove the Ellis–Numakura lemma in the one-sided continuity convention needed here. Let $S$ be a nonempty compact Hausdorff space with an associative multiplication for which each map $x\mapsto xa$ is continuous. Among its nonempty closed subsemigroups, there is an inclusion-minimal one, $M$: a decreasing chain has nonempty intersection by compactness, and that intersection is a closed subsemigroup, so Zorn's lemma applies.

Fix $a\in M$. The set $Ma$ is nonempty and compact, hence closed in the Hausdorff space, and is a subsemigroup becauseMinimality gives $Ma=M$. Thus the set $T=\{x\in M:xa=a\}$ is nonempty. It is closed by the stated continuity, and it is a subsemigroup, since $(xy)a=x(ya)=xa=a$ for $x,y\in T$. Minimality again gives $T=M$, so $a\in T$ and $\boxed{a^2=a}$, giving a semigroup idempotent. If one instead uses the opposite one-sided continuity convention, the same proof uses $aM$ and $\{x:ax=a\}$; no joint continuity is required.

Apply this lemma to the Stone-Čech compactification of the natural numbers with its given addition. In the addition on the Stone-Čech compactification of the natural numbers convention,The continuous variable is $p$ when $q$ is fixed. The granted compactness, Hausdorff property and associativity therefore imply there exists an idempotent ultrafilter $p$ with $p+p=p$.

On the positive integers such a $p$ is nonprincipal: a principal ultrafilter at $n$ adds to itself to give the one at $2n$, which is different. Hence every cofinite set belongs to $p$. If zero is included in one's convention for $\mathbb{N}$, use the same compact semigroup argument on the closed space of nonprincipal ultrafilters: it is nonempty by compactness of the infinite discrete set's compactification, and the displayed addition sends two free ultrafilters to a free ultrafilter. This avoids the trivial principal idempotent at zero.

To deduce Hindman's theorem, let $A$ be the unique colour class belonging to $p$. DefineIdempotence gives $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma also gives $A^{\ast }-n\in p$ whenever $n\in A^{\ast }$. Here is its proof: $A-n\in p$, and applying idempotence to $A-n$ givesIntersecting this set with $A-n$ produces exactly $A^{\ast }-n$.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_r$ have been chosen with every nonempty finite sum in $A^{\ast }$. The finite intersectionbelongs to $p$, and is therefore nonempty. Choose $x_{r+1}$ from it. The new sums are $x_{r+1}$ and $s+x_{r+1}$, and they all lie in $A^{\ast }$. Induction yieldsThis proves every finite colouring of the positive integers admits a monochromatic finite-sums set generated by a strictly increasing infinite sequence, the required Hindman theorem.

No. The ten residue classes modulo $10$ form a finite partition, so an ultrafilter $p$ contains exactly one class $C_r$. The addition formula shows that $p+p$ contains $C_{2r}$: for $n\in C_r$, the translate $C_{2r}-n$ is $C_r$ and belongs to $p$. If $p$ is an idempotent ultrafilter, uniqueness of its selected class gives $2r\equiv r(\mathrm{mod}10)$, hence $r=0$.

Thus every idempotent ultrafilter contains the multiples of $10$, and cannot also contain their complement. This is the zero-residue constraint for idempotent ultrafilters, valid for every finite modulus.

Yes. Use the sequence $x_j=2\cdot 10^{j-1}$, $j\ge 1$. Every nonempty finite sum has only digits $0$ and $2$ in decimal notation, with leading digit $2$, so none is a power of $10$, including $10^0=1$.

We still need an idempotent ultrafilter containing these sums; merely exhibiting an infinite set is insufficient. PutThe nonempty clopen sets $\overline{F_r}$ are nested, so compactness gives a nonempty compact $K$. This is the tail finite-sums semigroup. To verify closure under addition, take $p,q\in K$ and fix $r$. For any $s\in F_r$, choose a finite representation of $s$ and then an index $t$ beyond every index in that representation. Disjointness of supports gives $F_t\subseteq F_r-s$, so $F_r-s\in q$. Consequentlyand $F_r\in p+q$. This holds for every $r$, proving $p+q\in K$. The Ellis–Numakura lemma now supplies an idempotent ultrafilter in $K$. It contains $F_1$, a subset of the non-powers of $10$, so by upward closure it contains the whole set of non-powers of $10$.