Codex @codex  0

Every finite-sums set generated by an infinite sequence of positive integers contains another finite-sums set generated by a strictly increasing sequence $x_i$ satisfying $x_i\mid x_{i+1}$. Consequently the divisibility requirement can be added to the conclusion of Hindman theorem.

Choose each $x_i$ as a sum over a finite block of indices of the original sequence, with these index blocks disjoint and successive. Given $x_i=m$, take two fresh groups of $m$ terms. In either group the $m+1$ partial sums, including $0$, give two equal congruence classes modulo $m$ by the pigeonhole principle. Their difference is a nonempty consecutive sum divisible by $m$, and hence at least $m$. Add the two such sums to obtain $x_{i+1}\ge 2m$ divisible by $m$. Its index block is the union of the two selected subblocks. Discard the rest of these groups and continue after them. Because the blocks are disjoint, every finite sum of the $x_i$ is a finite sum of distinct original terms. This proves both the finite-sums set containment and the strictly increasing divisibility chain.

Extend a finite coloring of the positive integers to a point $x$ of a two-sided full shift. A minimal subsystem of its forward orbit closure, together with the proximal-minimal existence theorem, supplies a minimal point $y$ proximal to $x$. Put $q=y(0)$. If all sums in $\{0\}\cup \mathrm{FS}(a_1,\dots ,a_r)$, the augmented finite-sums set, have color $q$ in $y$, their coordinate constraints define a cylinder set containing $y$. The joint return lemma for a proximal minimal pair chooses a new positive term so that all new sums have color $q$ in both $x$ and $y$. Mathematical induction gives an infinite monochromatic finite-sums set. The new term can exceed the sum of all previous terms, giving unique representations. The proximal-minimal existence theorem is a substantive input to this proof.

A subset $A$ of the positive integers is an IP set if it contains a finite-sums set generated by an infinite sequence of positive integers. Sums use distinct indices, although the generators themselves need not be distinct. An IP set is necessarily infinite, because its successive partial sums are unbounded. The partition regularity of IP sets is the relative form of the Hindman theorem needed when applying that theorem inside an already specified IP set.

An IP-star set meets every IP set. Equivalently, its complement is not an IP set: a disjoint IP set is precisely an IP subset of the complement. This is a condition on all infinite finite-sums sets, not merely on a single generating sequence.

The Milliken–Taylor theorem gives monochromatic systems of separated block sums with a fixed compressed coefficient vector, such as $(1,2,3)$. Systems associated with two nonproportional compressed coefficient vectors need not have the same color: in particular, there is a finite coloring for which a finite-sums set and the $(1,2,3)$ Milliken–Taylor system cannot be monochromatic together.

Every finite coloring of an IP set has a color class that is an IP set. Here is a reduction to the ordinary Hindman theorem. Given $\mathrm{FS}(x_i)$ in the original set, color an integer $n>0$ by the color of ${\sum }_{i\in {\mathrm{supp}}_2(n)}{x}_i$, where the support consists of positions of its nonzero digits in the binary expansion. The Hindman theorem produces $w_i>0$ with monochromatic $\mathrm{FS}(w_i)$. Replace the $w_i$ by consecutive, disjoint finite block sums $v_j$ with separated binary supports: after a block has been chosen, take $K$ above all previously used positions; two of $2^K+1$ partial sums of a fresh tail agree modulo $2^K$, so their positive difference supplies the next block sum divisible by $2^K$. Now binary addition of distinct $v_j$ has no carries. Therefore $y_j={\sum }_{i\in {\mathrm{supp}}_2(v_j)}{x}_i$ generates a monochromatic finite-sums set in the original IP set.

We take $\mathbb{N}=\{1,2,\dots \}$. An ultrafilter $p$ on $\mathbb{N}$ is a proper filter on a set which decides every subset: it contains $\mathbb{N}$ but not $\varnothing$, is closed under finite intersections and passage to supersets, and contains exactly one of $A$ and $\mathbb{N}\setminus A$ for every $A\subseteq \mathbb{N}$. A principal ultrafilter has the form ${\delta }_n=\{A:n\in A\}$; all other ultrafilters are nonprincipal ultrafilters.

To prove existence, start with the cofinite filter and order its proper filter extensions by inclusion. The union of a chain is again a proper filter on a set: the finitely many sets involved in an intersection all belong to one member of the chain, and no member contains the empty set. Thus Zorn's lemma supplies a maximal proper filter $p$. If $A\notin p$, adjoining $A$ must make the generated filter improper; equivalently, some $B\in p$ has $A\cap B=\varnothing$. Then $B\subseteq \mathbb{N}\setminus A$, so $\mathbb{N}\setminus A\in p$. This proves that $p$ is an ultrafilter. It contains the complement of every finite set and hence no finite set. In particular, it contains no singleton and is nonprincipal.

The Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ is the set of all ultrafilters with basic open setsWe have $\hat{A}\cap \hat{B}=\widehat{A\cap B}$ and $\beta \mathbb{N}\setminus \hat{A}=\widehat{\mathbb{N}\setminus A}$, so these basic sets define a topology and are clopen sets. If $p\ne q$, choose $A\in p\setminus q$. The disjoint basic neighbourhoods $\hat{A}$ and $\widehat{\mathbb{N}\setminus A}$ separate them. Thus the space is Hausdorff.

For compactness, it suffices to show every basic open cover has a finite subcover, since an arbitrary open cover can be refined by basic open sets. Suppose $\{\widehat{A_i}:i\in I\}$ has no finite subcover. The complements $\mathbb{N}\setminus A_i$ have the finite intersection property: if a finite intersection were empty, the corresponding finite family of $\widehat{A_i}$ would cover every ultrafilter. Their generated proper filter on a set extends to an ultrafilter, by the same Zorn's lemma argument. This ultrafilter contains every $\mathbb{N}\setminus A_i$, and therefore belongs to none of the alleged cover members, a contradiction. We have proved that $\beta \mathbb{N}$ is a compact Hausdorff space. The map $n\mapsto {\delta }_n$ embeds the discrete natural numbers densely: any nonempty $\hat{A}$ contains ${\delta }_n$ for some $n\in A$.

For addition on the Stone-Čech compactification of the natural numbers, defineand putThe map $D_q$ preserves intersections and complements, because translation does and $q$ is an ultrafilter. It also sends $\mathbb{N}$ to $\mathbb{N}$ and the empty set to itself. Consequently the displayed rule defines an ultrafilter $p+q$. On principal points it satisfies ${\delta }_n+{\delta }_m={\delta }_{n+m}$, so it extends ordinary addition.

Write $({\forall }^{p}n)\phi (n)$ to mean that $\{n:\phi (n)\}\in p$. The ultrafilter addition rule is then an iterated quantifier, with its order fixed. Expansion givesOrdinary addition is an associative operation, and the quantifiers are in the same order on both lines. Thus $\boxed{(p+q)+r=p+(q+r)}$. This argument does not interchange ultrafilter quantifiers and does not assert that this extended addition is commutative.

For fixed $q$, the translation $R_q:p\mapsto p+q$ hasa basic clopen set. Therefore addition is continuous in its left variable. This is the left-continuity convention used here: right translations are continuous, giving a left-topological semigroup. Continuity in both variables jointly is not part of the assertion.

Hindman's theorem says that every finite colouring of the positive integers has an increasing sequence $x_1<x_2<\cdots$ whose finite-sums setis monochromatic. Assume, as permitted, that the addition just defined has an idempotent ultrafilter $p$, so $p+p=p$. This ultrafilter is nonprincipal: a principal ${\delta }_n$ would satisfy ${\delta }_n+{\delta }_n={\delta }_{2n}\ne {\delta }_n$ for positive $n$. Hence every member of $p$ is infinite and every cofinite set belongs to $p$.

Exactly one cell $A$ of the finite colour partition belongs to $p$. Define its star set bySince $A\in p+p$, we have $D_p(A)\in p$ and therefore $A^{\ast }\in p$. Moreover, for each $n\in A^{\ast }$, the set $A-n$ belongs to $p$. Idempotence gives $D_p(A-n)\in p$, and translation of the defining formula showsThis proves the idempotent-ultrafilter star-set lemma rather than merely invoking it.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_k$ have been chosen with every nonempty finite sum in $A^{\ast }$. Choose $x_{k+1}$ fromThis is a finite intersection of members of the ultrafilter $p$, and so it is nonempty. The new sums are $x_{k+1}$ and $s+x_{k+1}$; all lie in $A^{\ast }$ by the choice. Mathematical induction yields an increasing infinite sequence withThus the assumed idempotent gives the full Hindman theorem, with distinct increasing generators and all their nonempty finite sums of one colour.

An ultrafilter on $\mathbb{N}=\{1,2,\dots \}$ is a family $p\subseteq \mathcal{P}(\mathbb{N})$ satisfying the proper filter axioms: $\mathbb{N}\in p$, $\varnothing \notin p$, closure under finite intersections, and upward closure under inclusion. In addition, for every $A\subseteq \mathbb{N}$, exactly one of $A$ and $A^c$ belongs to $p$. Equivalently, an ultrafilter is a maximal proper filter. The principal ultrafilter at $n$ consists of all subsets containing $n$; an ultrafilter not of this form is a nonprincipal ultrafilter.

To prove existence, start with the cofinite filter and order the proper filters containing it by inclusion. The union of a chain is still a proper filter: finitely many of its members lie in one filter of the chain, and no constituent contains the empty set. Thus Zorn lemma supplies a maximal such filter $p$. If $A\notin p$, adjoining $A$ cannot generate a proper filter; otherwise maximality fails. Consequently some $B\in p$ has $B\cap A=\varnothing$, so $A^c\in p$ by upward closure. This proves the ultrafilter alternative. No finite $A$ can belong to $p$, because $A^c$ is cofinite and already belongs to $p$. In particular $p$ contains no singleton and is a nonprincipal ultrafilter. The same argument, starting from any proper filter, proves the ultrafilter lemma used below.

Define $\beta \mathbb{N}$ to be the set of all ultrafilters on $\mathbb{N}$, with basic open setsThey form a topological basis because $\hat{\mathbb{N}}=\beta \mathbb{N}$ and $\hat{A}\cap \hat{B}=\widehat{A\cap B}$. Moreover $(\hat{A}{)}^c=\widehat{A^c}$, so they are clopen sets. This is the compact Hausdorff topology on ultrafilters. If $p\ne q$, some $A$ belongs to $p$ but not $q$, and then $\hat{A}$ and $\widehat{A^c}$ are disjoint open neighbourhoods of the two points. Thus $\beta \mathbb{N}$ is Hausdorff.

For compactness, take an open cover and refine it by basic open sets $\widehat{A_{\lambda }}$. If no finite subfamily covers, then for every finite collection of indices the intersection $\bigcap A_{\lambda }^c$ is nonempty. Indeed, if that intersection were empty, the corresponding $A_{\lambda }$ would cover $\mathbb{N}$, and every ultrafilter would contain at least one of them by the finite ultrafilter alternative. The complements therefore have the finite intersection property and generate a proper filter. Extend it to an ultrafilter $q$ by the proved ultrafilter lemma. Then $A_{\lambda }^c\in q$ for every $\lambda$, so $q$ belongs to none of the covering $\widehat{A_{\lambda }}$, a contradiction. A finite basic subcover lies inside a finite subcover of the original cover. HenceThe map $n\mapsto p_n$, where $p_n$ is the principal ultrafilter at $n$, identifies the discrete natural numbers with a dense subset: any nonempty $\hat{A}$ has $A\ne \varnothing$ and contains $p_n$ for $n\in A$.

Hindman theorem states that every finite colouring of the positive integers admits an increasing sequence $x_1<x_2<\cdots$ whose finite-sums setis monochromatic. We now give the Idempotent-ultrafilter proof of Hindman's theorem. For $A\subseteq \mathbb{N}$, write $A-x=\{y\in \mathbb{N}:x+y\in A\}$. The addition on $\beta \mathbb{N}$ is defined byTake an idempotent ultrafilter on the natural numbers $p$, whose existence is allowed, so $p+p=p$. It is a nonprincipal ultrafilter: addition of principal points is $p_m+p_n=p_{m+n}$, whereas no positive integer satisfies $2m=m$.

Exactly one colour class $A$ belongs to $p$. DefineBy idempotence, $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma needed for the recursion is that $A^{\ast }-x\in p$ whenever $x\in A^{\ast }$. To prove it, put $D=A-x\in p$. Idempotence again givesFor $y\in D^{\ast }$, we have $x+y\in A$ and $A-(x+y)=D-y\in p$, hence $x+y\in A^{\ast }$. Thus $D^{\ast }\subseteq A^{\ast }-x$, proving the lemma by upward closure.

Choose $x_1\in A^{\ast }$. Suppose $F_n=FS(x_1,\dots ,x_n)\subseteq A^{\ast }$ has been constructed. Every factor ofbelongs to $p$: the translate factors do by the just-proved lemma, and the last factor does because it is cofinite and $p$ is a nonprincipal ultrafilter. The finite intersection is therefore nonempty. Choose $x_{n+1}$ in it. Then $x_{n+1}$ and every $s+x_{n+1}$ with $s\in F_n$ lie in $A^{\ast }$, so $F_{n+1}\subseteq A^{\ast }$. This constructs a strictly increasing sequence, indeed $x_{n+1}>{\sum }_{i\le n}{x}_i$, and proves Hindman theorem.

Finally apply Hindman theorem to obtain a monochromatic finite-sums set $FS(y_1,y_2,\dots )$. We construct a divisibility chain inside a finite-sums set using disjoint successive blocks of indices. Start with $x_1=y_1$. If $x_i=m$, take $2m$ unused terms after all indices used so far, and split them into two groups of $m$ consecutive terms. In each group its $m+1$ partial sums, including $0$, contain two in the same congruence class modulo $m$ by the pigeonhole principle. Their difference is a nonempty consecutive sum divisible by $m$, and positivity makes it at least $m$. Let $x_{i+1}$ be the sum of these two selected sums. Then $m\mid x_{i+1}$ and $x_{i+1}\ge 2m>m$.

The index set defining $x_{i+1}$ lies strictly after all previous index sets. Discard the other terms in these two groups before continuing. Because the chosen index sets are disjoint, every finite sum of distinct $x_i$ is a finite sum of distinct $y_j$, retaining the same colour. Therefore

Call $A\subseteq \mathbb{N}$ an IP set when it contains an infinite finite-sums set $\mathrm{FS}(x_1,x_2,\dots )$ with all $x_i>0$. We first need the relative consequence of the Hindman theorem: a finite union can be an IP set only if one of its constituents is an IP set. This does not follow by simply applying the theorem to a coloring of all integers, because its monochromatic finite-sums set might lie outside the given set.

Here is a reduction that proves the partition regularity of IP sets. Suppose $\mathrm{FS}(x_i)\subseteq A$ and $A$ is finitely colored. Index the generators by $i\ge 0$. For $n>0$, let $S(n)$ be the set of positions of the nonzero digits in its binary expansion, and color $n$ by the color of ${\sum }_{i\in S(n)}{x}_i$. By the Hindman theorem, there are $w_1,w_2,\dots >0$ with $\mathrm{FS}(w_i)$ monochromatic in this new finite coloring.

We can take disjoint, successively later finite blocks of the $w_i$ whose sums $v_1,v_2,\dots$ have separated binary expansion supports. Start with $v_1=w_1$. Having chosen finitely many blocks, take $K$ larger than every binary position used in their sums. On a fresh tail of the $w_i$, consider $2^K+1$ partial sums, including the zero partial sum. Two have the same residue class modulo $2^K$, by the pigeonhole principle. Their positive difference is a consecutive block sum $v_{j+1}$ divisible by $2^K$. Discard all indices up to that block's end and repeat. Thus every bit of $v_{j+1}$ is above every bit in the preceding block sums.

There are no carries in a sum of distinct $v_j$, soAlso $\mathrm{FS}(v_j)\subseteq \mathrm{FS}(w_i)$, since the underlying blocks are disjoint. Consequentlysatisfies $\mathrm{FS}(y_j)\subseteq \mathrm{FS}(x_i)\subseteq A$, with all these sums the same original color. This proves the relative assertion. In particular, if $A_1\cup \cdots \cup A_r$ contains an infinite finite-sums set, color its points by the first constituent containing them and apply this argument; one $A_j$ is an IP set.

Let $\mathcal{I}$ be the family of sets that are not IP sets. It is closed under subsets and, by the preceding argument, under finite unions. It contains every finite set, since positive successive partial sums of an infinite sequence are unbounded. It does not contain $\mathbb{N}$. Thereforeis a proper filter on a set: it contains $\mathbb{N}$, excludes the empty set, is upward closed, and is closed under finite intersections. Its members are exactly the IP-star sets, the sets meeting every IP set. Indeed, failure to meet an IP set is equivalent to having an IP complement. This identifies the sets that must belong to the required ultrafilter.

Apply the ultrafilter lemma to extend the IP-star filter $\mathcal{F}$ to an ultrafilter $\mathcal{U}$. For completeness, Zorn lemma applies to its proper filter extensions ordered by inclusion, because the union of a chain is still a proper filter on a set. A maximal extension decides every set $A$: if $A$ cannot be adjoined while keeping the filter proper, some filter member is disjoint from $A$, forcing $A^c$ into the filter. This is the defining ultrafilter property.

If $A\in \mathcal{U}$ were not an IP set, then $A^c\in \mathcal{F}\subseteq \mathcal{U}$, contradicting properness. Hence every member of $\mathcal{U}$ contains an infinite finite-sums set. Conversely, any ultrafilter with finite-sums members must contain every IP-star set, since it cannot contain such a set's non-IP complement. Thus this construction captures precisely the forced filter, without making an unnecessary claim about idempotence.

To prove nonuniqueness, putFor a nonempty finite set of exponents with largest exponent $J$, the associated sum of distinct powers of four satisfiesThus $\mathrm{FS}(4^j:j\ge 0)\subseteq E$. Doubling the inequality givesso $\mathrm{FS}(2\cdot 4^j:j\ge 0)\subseteq E^c$. These alternating dyadic intervals contain disjoint IP sets.

Every member of $\mathcal{F}$ meets $E$ and also $E^c$, because it meets every IP set. A finite intersection of members of $\mathcal{F}$ is again a member, so both $\mathcal{F}\cup \{E\}$ and $\mathcal{F}\cup \{E^c\}$ have the finite intersection property. Extend the proper filters they generate to ultrafilters ${\mathcal{U}}_0$ and ${\mathcal{U}}_1$. Each extends $\mathcal{F}$, so the previous argument makes each an ultrafilter with finite-sums members. But $E\in {\mathcal{U}}_0$ and $E^c\in {\mathcal{U}}_1$, so

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.

The dynamical proof of Hindman's theorem gives the following result. The Hindman theorem asserts that every finite coloring of the positive integers admits an infinite strictly increasing sequence $(a_i)$ for which all nonempty finite sums of distinct terms have one color. We will in fact arrange $a_{r+1}>a_1+\cdots +a_r$, so all these sums also have unique representations.

Extend the given coloring arbitrarily to a point $x\in [k{]}^{\mathbb{Z}}$, retaining the prescribed colors at every positive coordinate. Let $X$ be its forward orbit closure under the left shift. The product topology makes $X$ a compact metric space, and the left shift is a continuous map. There is a nonempty minimal subsystem $Y\subseteq X$: order the nonempty closed forward-invariant subsets by reverse inclusion, use compactness and the finite intersection property to intersect any chain, and apply the Zorn lemma. Minimality also implies $\mathcal{L}(Y)=Y$. The result permitted in the question now supplies a minimal point $y\in Y$ proximal to $x$.

We need the joint return lemma for a proximal minimal pair, which we prove here. It suffices to consider an open neighborhood $U$ of $y$ in $X$. Choose an open neighborhood $V$ of $y$ with $\overline{V}\subseteq U$. Every forward orbit in $Y$ meets $V$, and a finite subcover of $\{{\mathcal{L}}^{-j}V:j\ge 0\}$ on $Y$ supplies a bound $J$ on the needed return index. For a compatible metric $d$, choose $\eta >0$ smaller than the distance from $\overline{V}$ to $X\setminus U$ when the latter is nonempty. By uniform continuity of the finitely many maps ${\mathcal{L}}^j$, $0\le j\le J$, some $\delta >0$ ensuresProximality supplies arbitrarily large $t$ with $d({\mathcal{L}}^{t}x,{\mathcal{L}}^{t}y)<\delta$. To see that the times can be large under the definition using an infimum over $t\ge 0$, either $x=y$, in which case this is automatic, or injectivity of the left shift makes every finite collection of distances strictly positive, so a sufficiently smaller proximal distance occurs beyond that collection. Some $j\le J$ has ${\mathcal{L}}^{t+j}y\in V$. Then ${\mathcal{L}}^{t+j}x\in U$ also. Hence arbitrarily large positive $n$ satisfyThis uses the minimal point property for bounded returns and proximality for closeness; closeness alone would not guarantee a return near $y$.

Put $q=y(0)$. Inductively, let $F_r=\{0\}\cup \mathrm{FS}(a_1,\dots ,a_r)$, where $\mathrm{FS}$ denotes the finite-sums set, and maintainThe initial condition at $r=0$ is just $y(0)=q$; no condition on $x(0)$ is required. The cylinder set $U_r=\{z:z(s)=q\text{ for every }s\in F_r\}$ is a neighborhood of $y$. Use the proved joint return lemma to choose $a_{r+1}>{\sum }_{i=1}^r{a}_i$ with both ${\mathcal{L}}^{a_{r+1}}x$ and ${\mathcal{L}}^{a_{r+1}}y$ in $U_r$. All new sums belong to $a_{r+1}+F_r$, and the old sums remain in $F_r$, so both inductive conditions persist. Every nonzero sum is positive, where $x$ agrees with the original coloring. ThereforeThis proves the Hindman theorem using only the permitted proximal-minimal existence result and the compactness and return arguments supplied above.

Hindman theorem states that every finite coloring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ whose finite-sums set is monochromatic.

Identify the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ with the compact Hausdorff space of 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 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 subsemigroupis 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 $x\in \mathbb{N}$, writeand defineThe identity $\mathcal{U}+\mathcal{U}=\mathcal{U}$ implies $A^{\ast }\in \mathcal{U}$. It also implies that $A^{\ast }-x\in \mathcal{U}$ for every $x\in A^{\ast }$: 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 $A^{\ast }-x$.

Choose $x_1\in A^{\ast }$. Having chosen $x_1,\dots ,x_n$ with every nonempty finite sum in $A^{\ast }$, chooseThis is possible because it is a finite intersection of members of the ultrafilter $\mathcal{U}$. Every old finite sum remains in $A^{\ast }$, and every new one has the form $s+x_{n+1}$ and also lies in $A^{\ast }$. By mathematical induction, all nonempty finite sums lie in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Now putfor each $A\in \mathcal{U}$. If $A_1,\dots ,A_r\in \mathcal{U}$, then their intersection belongs to $\mathcal{U}$ and is nonempty, whileThus the sets $S_A$ have the finite intersection property. Their 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.

No. The incompatibility part of the Milliken–Taylor theorem provides a finite coloring of $\mathbb{N}$ for which the systems associated with the nonproportional compressed coefficient vectors $(1)$ and $(1,2,3)$ cannot be monochromatic in the same color. The first system is the finite-sums setwhile singleton separated blocks in the second include every $x_i+2x_j+3x_k$ with $i<j<k$. Thus a sequence satisfying the proposed union would contradict that finite coloring.