= Solution
Take $\mathbb N=\{1,2,\ldots\}$. 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> because
$$
(xa)(ya)=\big(x(ay)\big)a\in Ma.
$$
Minimality 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 \b[$\boxed{a^2=a}$], 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,
$$
A\in p+q\iff\{n:A-n\in q\}\in p,\qquad A-n=\{t\in\mathbb N:n+t\in A\}.
$$
The continuous variable is $p$ when $q$ is fixed. The granted compactness, Hausdorff property and associativity therefore imply \b[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 $p$. Define
$$
A^*=A\cap\{n:A-n\in p\}.
$$
Idempotence gives $A^*\in p$. The <idempotent-ultrafilter star-set lemma> also gives $A^*-n\in p$ whenever $n\in A^*$. Here is its proof: $A-n\in p$, and applying idempotence to $A-n$ gives
$$
\{t:(A-n)-t\in p\}=\{t:A-(n+t)\in p\}\in p.
$$
Intersecting this set with $A-n$ produces exactly $A^*-n$.
Choose $x_1\in A^*$. Suppose $x_1<\cdots<x_r$ have been chosen with every nonempty finite sum in $A^*$. The finite intersection
$$
A^*\cap\bigcap_{s\in\operatorname{FS}(x_1,\ldots,x_r)}(A^*-s)\cap\{n:n>x_r\}
$$
belongs 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 $A^*$. Induction yields
$$
\boxed{\operatorname{FS}(x_1,x_2,\ldots)=\left\{\sum_{i\in F}x_i:\ \varnothing\ne F\subseteq\mathbb N\text{ finite}\right\}\subseteq A.}
$$
This proves \b[every finite colouring of the positive integers admits a <monochromatic> <finite-sums set> generated by a strictly increasing infinite sequence], the required <Hindman theorem>.
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