= Solution
<Hindman theorem> states that every <finite coloring> of $\mathbb N$ admits an infinite sequence $x_1,x_2,\ldots$ whose <finite-sums set> is monochromatic.
Identify the <Stone-Čech compactification of the natural numbers> $\beta\mathbb N$ with the compact Hausdorff space of <ultrafilter>[ultrafilters] on $\mathbb N$, equipped with <addition on the Stone-Čech compactification of the natural numbers>. This makes $\beta\mathbb N$ a compact Hausdorff left-topological <semigroup>. For completeness, the <Ellis–Numakura lemma> gives an idempotent in every such semigroup: by the <Hausdorff space> property and <compact space>[compactness], the intersection of a descending chain of nonempty compact subsemigroups is nonempty, so Zorn's lemma gives a minimal one $K$. For $a\in K$, the compact subsemigroup $K+a$ equals $K$. Hence the nonempty compact subsemigroup
$$
\{x\in K:x+a=a\}
$$
is also $K$, and in particular $a+a=a$. Choose the resulting <idempotent ultrafilter on the natural numbers> $\mathcal U$.
One color class $A$ belongs to $\mathcal U$. For $x\in\mathbb N$, write
$$
A-x=\{y:x+y\in A\},
$$
and define
$$
A^*=\{x\in A:A-x\in\mathcal U\}.
$$
The identity $\mathcal U+\mathcal U=\mathcal U$ implies $A^*\in\mathcal U$. It also implies that $A^*-x\in\mathcal U$ for every $x\in A^*$: both $A-x$ and the set of $y$ for which $A-(x+y)$ belongs to $\mathcal U$ lie in $\mathcal U$, and their intersection is $A^*-x$.
Choose $x_1\in A^*$. Having chosen $x_1,\ldots,x_n$ with every nonempty finite sum in $A^*$, choose
$$
x_{n+1}\in A^*\cap
\bigcap_{s\in\operatorname{FS}(x_1,\ldots,x_n)}(A^*-s).
$$
This is possible because it is a finite intersection of members of the <ultrafilter> $\mathcal U$. Every old finite sum remains in $A^*$, and every new one has the form $s+x_{n+1}$ and also lies in $A^*$. By <mathematical induction>, all nonempty finite sums lie in $A^*\subseteq A$, proving Hindman's theorem.
Now put
$$
S_A=\{x_n:n\in A\}
$$
for each $A\in\mathcal U$. If $A_1,\ldots,A_r\in\mathcal U$, then their intersection belongs to $\mathcal U$ and is nonempty, while
$$
S_{A_1\cap\cdots\cap A_r}\subseteq S_{A_1}\cap\cdots\cap S_{A_r}.
$$
Thus the sets $S_A$ have the <finite intersection property>. Their <closure>[closures] are closed subsets of the compact interval $[0,1]$, so their total intersection contains some $x$. Equivalently, every <neighbourhood> of $x$ meets every $S_A$; this is the <ultrafilter limit> of the sequence.
The point is unique. If distinct points $x,y$ both had this property, choose disjoint neighborhoods $U,V$. The index set $I_U=\{n:x_n\in U\}$ must belong to $\mathcal U$, because otherwise its complement would belong to $\mathcal U$ and the associated $S_A$ would miss $U$. Similarly $I_V\in\mathcal U$. Their intersection is empty, contradicting the definition of an ultrafilter.
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