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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.

Use the product topology on $\mathcal{C}=[k{]}^{\mathbb{Z}}$, with $[k]$ discrete, and the left shift $(\mathcal{L}c)(j)=c(j+1)$. A basic cylinder set specifies finitely many coordinates. This is a compact metric space, and $\mathcal{L}$ is a homeomorphism. Write $Y=\overline{\{{\mathcal{L}}^{n}c:n\ge 0\}}$ for the forward orbit closure. A minimal point means that $Y$ is a minimal dynamical system, equivalently that every point of $Y$ has a dense forward orbit in $Y$; it need not be a fixed point.

There is a convention needed in the source: the bounded-gaps property concerns every finite integer interval $I$, hence every finite word over an alphabet. Literally allowing $I=\mathbb{Z}$ would make the displayed condition impossible for finite $U$, although a constant coloring is a minimal point. Under the standard finite-word interpretation, the property is precisely uniform recurrence.

Suppose first that $Y$ is a minimal dynamical system. Its nonempty compact forward-invariant subset $\mathcal{L}(Y)$ must equal $Y$. As the ambient left shift is invertible, all integer translates of $c$ belong to $Y$. Let $I=[a,b]\cap \mathbb{Z}$ have length $\ell =b-a+1$, and let $V=\{z\in Y:z{|}_I=c{|}_I\}$. This is a nonempty clopen set. Each forward orbit in $Y$ meets $V$, so $\{{\mathcal{L}}^{-n}V:n\ge 0\}$ covers $Y$. By compactness, finitely many suffice; let $D$ be the largest index in this finite cover. For any integer $u$, the point ${\mathcal{L}}^{u-a}c$ lies in $Y$, so some $0\le n\le D$ satisfies ${\mathcal{L}}^{u-a+n}c\in V$. The prescribed word therefore occurs at positions $[u+n,u+n+\ell -1]$, entirely inside $[u,u+D+\ell -1]$. Taking $M=D+\ell$ proves the bounded-gaps property for every interval of length at least $M$.

Conversely suppose $c$ is uniformly recurrent. Every finite word from any integer translate of $c$ occurs arbitrarily far to the right, so every such translate belongs to $Y$. Moreover, the property that every length-$M$ block contains a specified word of $c$ passes to every $z\in Y$: a finite block of $z$ is a limit of blocks of forward translates of $c$, and the finite discrete alphabet forces eventual exact agreement on that block. Given $I=[a,b]\cap \mathbb{Z}$, look for its word inside $z([a,a+M-1])$. An occurrence starts at $a+n$ for some $n\ge 0$, so ${\mathcal{L}}^{n}z$ agrees with $c$ on $I$. Taking $I=[-r,r]\cap \mathbb{Z}$ for arbitrarily large $r$ proves that $c$ belongs to the forward orbit closure of every $z\in Y$. That closure is a closed forward-invariant subset of $Y$, so it also contains every forward translate of $c$ and hence all of $Y$. Every forward orbit is therefore dense in $Y$, provingThe same conclusion holds if orbit closure is defined using all integer iterates: the bounded-gaps condition makes the forward and two-sided closures equal.

The two Ramsey properties and the topologies. Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, and $[A{]}^{\omega }$ for the infinite subsets of $A$. The positive integers are used throughout. A Ramsey set of infinite subsets $E$ has the property that every infinite $A$ contains an infinite $B$ for which $[B{]}^{\omega }$ lies wholly in $E$ or wholly in $E^c$.

For finite $s$ and infinite $A$ above $\max s$, letThese are the basic neighbourhoods of the Ellentuck topology, denoted $\ast$ in the question. They form a basis: if $X$ belongs to two such sets, their stems are comparable initial segments of $X$; take the longer stem and the intersection of the two tails beyond that stem. This intersection contains the infinite remaining tail of $X$ and yields a basic set contained in both.

A completely Ramsey set $E$ allows such a homogeneous refinement $[s,B]$ inside every $[s,A]$, with $B\subseteq A$ infinite and the stem $s$ kept fixed. Taking $s=\varnothing$ shows that complete Ramseyness implies Ramseyness. The ordinary topology on infinite subsets $\tau$ has basic cylindersIt is the usual product topology on increasing enumerations and is coarser than the Ellentuck topology.

Examples distinguishing the properties. A non-Ramsey example uses the axiom of choice. Let $\kappa =2^{{\aleph }_0}$ and enumerate all infinite sets as $(A_{\alpha }{)}_{\alpha <\kappa }$. Each $[A_{\alpha }{]}^{\omega }$ has cardinality $\kappa$: an enumeration of $A_{\alpha }$ gives the upper bound, and choosing one element from each of infinitely many disjoint pairs gives an injection from all binary sequences for the lower bound.

By transfinite recursion, choose two distinct, previously unused elements $X_{\alpha },Y_{\alpha }\in [A_{\alpha }{]}^{\omega }$ at each stage. Fewer than $\kappa$ elements have been used at stage $\alpha <\kappa$, so this is possible without any assumption on the Continuum hypothesis. Put $R=\{X_{\alpha }:\alpha <\kappa \}$. All the $Y_{\alpha }$ remain outside $R$, since subsequent choices also avoid previously chosen elements. Every $[A{]}^{\omega }$ contains one $X_{\alpha }$ and one $Y_{\alpha }$, so $R$ is not a Ramsey set of infinite subsets. This is the non-Ramsey set from transfinite selection construction.

Perform the same construction on $T=\mathbb{N}\setminus \{1\}$, giving a non-Ramsey set $R_T\subseteq [T{]}^{\omega }$, and putFor any infinite $A$, the infinite set $B=A\setminus \{1\}$ has $[B{]}^{\omega }\cap E=\varnothing$, so $E$ is a Ramsey set of infinite subsets. But no $[\{1\},B]$ with infinite $B\subseteq T$ is homogeneous, because $[B{]}^{\omega }$ meets both $R_T$ and its complement. Therefore $E$ is not a completely Ramsey set.

Every star-open set is completely Ramsey. We give the full fusion proof for open Ellentuck sets, without quoting an infinite Ramsey theorem. Fix a star-open set $O$ and a basic set $[s,A]$. For finite $a\subseteq A$ and an infinite tail $D\subseteq A$ above $\max a$, say $D$ accepts $a$ if $[s\cup a,D]\subseteq O$; it rejects $a$ if no infinite subset of $D$ accepts $a$. Both properties persist under passing to infinite subsets. Every infinite tail can be thinned to decide $a$: take an accepting subset if one exists, and otherwise the original tail rejects it.

First thin $A$ to decide $a=\varnothing$. If it accepts, the required homogeneous neighborhood has already been found. Otherwise keep a rejecting tail. We now fuse to an infinite set $B$ deciding every finite $a\subseteq B$, where the decision for $a$ is made by $B/a=\{b\in B:b>\max a\}$.

To do this, choose $b_1<b_2<\cdots$ successively from an infinite current tail. After choosing $b_j$, thin the tail above $b_j$ to decide each of the finitely many subsets of $\{b_1,\dots ,b_j\}$, one after another, before choosing the next element. For any finite $a\subseteq B$, at the stage when its largest element is selected, the current tail decides $a$ and contains every later selected element. Heredity therefore gives a decision on $B/a$. For $a=\varnothing$, rejection persists from the initial tail.

If $B/a$ rejects $a$, only finitely many $n\in B/a$ can have $B/n$ accepting $a\cup \{n\}$. Otherwise let $D$ be the infinite set of those $n$. Every member of $[s\cup a,D]$ has a first new element $n$, and its remaining tail lies in $B/n$. It therefore belongs to $O$ by acceptance of $a\cup \{n\}$. This would make $D$ accept $a$, contradicting rejection.

Thin $B$ a second time to $C=\{c_1<c_2<\cdots \}$ so that every finite $a\subseteq C$ is rejected. Start with the rejected empty set. At a finite stage, all subsets of the selected elements are rejected. For each of these finitely many subsets, avoid its finitely many next elements that would give an accepted extension, and choose the next $c_j$ beyond all excluded elements. Since $B$ decides every finite extension, each new extension is rejected. Induction gives the asserted property of $C$.

We claim $[s,C]\cap O=\varnothing$. If $X\in [s,C]\cap O$, openness provides a basic set $[t,D]\subseteq O$ containing $X$. Extend $t$, if necessary, to an initial segment $u$ of $X$ containing $s$. Write $u=s\cup a$ with finite $a\subseteq C$. The infinite tail $X\setminus u$ lies in $B/a$ andIt is therefore an accepting subtail for $a$, contradicting rejection. We have found either an accepting neighborhood or a neighborhood disjoint from $O$. Thus

Basic neighborhoods are clopen. Consider $[s,A]$ and a point $X$ outside it. If $s$ is not an initial segment of $X$, take an initial segment $t$ of $X$ of length at least $|s|$; its ordinary cylinder $[t]$ cannot meet $[s,A]$. Otherwise some element of $X\setminus s$ is outside $A$; take $t$ through that element, and again $[t]$ is disjoint from $[s,A]$.

Thus the complement of $[s,A]$ is already open in the ordinary topology on infinite subsets, and hence in the finer Ellentuck topology. A basic neighbourhood is open by definition, so

The epigraph isAssume $E$ is $\tau$-sequentially lower semicontinuous. If $(u_n,t_n)$ lies in the epigraph and converges to $(u,t)$ in the product topology, thenThe limit remains in the epigraph, so it is a sequentially closed set.

Conversely, suppose the epigraph is a sequentially closed set and sequential lower semicontinuity fails along $u_n\stackrel{\tau }{\to }u$. Choose a finite real number $a$ with $\mathrm{liminf}E(u_n)<a<E(u)$. There is a subsequence along which $E(u_{n_j})\le a$. Thus $(u_{n_j},a)$ lies in the epigraph and converges to $(u,a)$, which does not lie there. This contradiction provesThis is a sequential statement; identifying it with ordinary closedness requires an appropriate assumption on the topology.

If $A\subseteq X$, the subspace topology on $A$ isIf $\sim$ is an equivalence relation on $X$ and $q:X\to X/\sim$ is the natural projection, the quotient topology declares $V\subseteq X/\sim$ open exactly when $q^{-1}(V)$ is open in $X$. For topological spaces $X$ and $Y$, the product topology on $X\times Y$ has as a basis all sets $U\times V$ with $U$ open in $X$ and $V$ open in $Y$.

Write $D(f)=\mathbb{C}\setminus f^{-1}(0)$. The zero and constant-one polynomials give $D(0)=\varnothing$ and $D(1)=\mathbb{C}$. Also,so $\tau$ is closed under finite intersections. Every nonzero one-variable complex polynomial has only finitely many roots, and every finite subset $\{a_1,\dots ,a_m\}$ is the zero set of ${\prod }_j(z-a_j)$. Consequently $\tau$ is exactly the cofinite topology: its open sets are the empty set and the complements of finite sets. An arbitrary union of such sets is again empty or has finite complement, so $\tau$ is a topology.

The product topology on $X\times Y$ is the topology with basisThe proposed complement need not be open. TakeThen $g^{-1}(0)$ is the diagonal. If its complement were open, a point such as $(0,1)$ would have a basic neighbourhood $U\times V$ contained in that complement. But $U$ and $V$ are nonempty cofinite subsets of $\mathbb{C}$, so $U\cap V\ne \varnothing$. For $t\in U\cap V$, the point $(t,t)$ belongs both to $U\times V$ and to the diagonal, a contradiction. Hence

The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

A base covers the space and refines intersections. On $X^k$, products of opens form a base. The weighted metric satisfies the metric axioms and has exactly this topology; convergence is coordinatewise. For $X^{\mathbb{N}}$ the metric $d(f,g)={\sum }_{n\ge 1}{2}^{-n}\min \{1,\rho (f(n),g(n))\}$ induces the stated product topology: finitely many coordinates control a basic neighbourhood and the tail is uniformly small. Thus sequences converge exactly coordinatewise.

The product topology is generated by the basis of a topology of open rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. For a fixed $y$,Both possibilities are open; checking this basis of a topology proves that $i_y$ is continuous. The analogous map $j_x:Y\to X\times Y$, $j_x(y)=(x,y)$, is continuous by the same argument.

If either factor is empty the product is empty and connected. Otherwise let $h:X\times Y\to \{0,1\}$ be continuous. For each $y$, $h\circ i_y$ is constant since $X$ is connected. Thus $h(x,y)=h(x_0,y)$ for any fixed $x_0\in X$. The right-hand side is $h\circ j_{x_0}$ and is constant in $y$ since $Y$ is connected. Therefore $h$ is constant on the whole product. The given characterization proves $X\times Y$ is connected.

The product topology on $X\times Y$ has as a basis of a topology the rectangles $U\times V$ with $U$ open in $X$ and $V$ open in $Y$; arbitrary unions of such rectangles are its open sets.

For distinct points $(x_1,y_1),(x_2,y_2)$, at least one coordinate differs. If $x_1\ne x_2$, disjoint open neighbourhoods $U_1,U_2$ in the Hausdorff space $X$ give disjoint open neighbourhoods $U_1\times Y,U_2\times Y$ of the two points. If the second coordinates differ instead, use $X\times V_1,X\times V_2$. Thus a product of two Hausdorff spaces is Hausdorff.

The product topology on $X\times Y$ has as a basis all open rectanglesThus a subset $W\subseteq X\times Y$ is open precisely when, around every $(x,y)\in W$, there are open $U\ni x$ and $V\ni y$ with $U\times V\subseteq W$. These rectangles cover the product, and their pairwise intersections are rectangles because $(U\times V)\cap (U^{\prime }\times V^{\prime })=(U\cap U^{\prime })\times (V\cap V^{\prime })$. Hence they define a topology by part (b).

This is equivalently the coarsest topology making the two coordinate projection maps continuous: the preimages of factor-open sets are the open strips $U\times Y$ and $X\times V$, whose finite intersections are the basis rectangles.

Let $\{B_i:i\in I\}$ and $\{C_j:j\in J\}$ be countable bases of the two factors. The familyis countable, since the Cartesian product of two countable sets is countable. It is a basis of a topology for the product topology: given a product-open neighbourhood of $(x,y)$, choose a contained open rectangle $U\times V$, then basis members with $x\in B_i\subseteq U$ and $y\in C_j\subseteq V$. Their product contains $(x,y)$ and is contained in the original neighbourhood. ThereforeIn standard terminology, a finite product of second-countable spaces is second-countable.

Yes: at least one factor topology must be indiscrete. Suppose instead that both have a nonempty proper open set, say $\varnothing \ne U⊊X$ and $\varnothing \ne V⊊Y$. Thenis open in the product topology. Both coordinate projections of $W$ are the full factors, because $U,V$ are nonempty. If $W$ were a rectangle $A\times B$, this would force $A=X$, $B=Y$, hence $W=X\times Y$. But choosing $x\notin U$ and $y\notin V$ gives $(x,y)\notin W$, a contradiction. Thus both factors cannot have nontrivial open sets. This is rectangularity of every product-open set.

Choose $v_0\in V$, using its nonemptiness. For each $u\in U$, the openness of $U\times V$ in the product topology supplies open $O_u\subseteq X$, $W_u\subseteq Y$ such thatSince $v_0\in W_u$, every point of $O_u$ belongs to $U$. Thus $U={\bigcup }_{u\in U}{O}_u$ is open. Interchanging the factors, using a fixed $u_0\in U$, proves $V$ open. Therefore $\boxed{U\in {\mathcal{T}}_X,V\in {\mathcal{T}}_Y}$. Nonemptiness matters: an empty product is open whatever the other factor is.

The following open rectangles form a basis of a topology for the product topology:These open rectangles in a finite product cover $X\times Y$, and intersections remain rectangles since $(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)$. One may equivalently take products of elements from a chosen basis of a topology in each factor.

Yes: at least one factor topology must be indiscrete. Suppose instead that both have a nonempty proper open set, say $\varnothing \ne U⊊X$ and $\varnothing \ne V⊊Y$. Thenis open in the product topology. Both coordinate projections of $W$ are the full factors, because $U,V$ are nonempty. If $W$ were a rectangle $A\times B$, this would force $A=X$, $B=Y$, hence $W=X\times Y$. But choosing $x\notin U$ and $y\notin V$ gives $(x,y)\notin W$, a contradiction. Thus both factors cannot have nontrivial open sets. This is rectangularity of every product-open set.

Choose $v_0\in V$, using its nonemptiness. For each $u\in U$, the openness of $U\times V$ in the product topology supplies open $O_u\subseteq X$, $W_u\subseteq Y$ such thatSince $v_0\in W_u$, every point of $O_u$ belongs to $U$. Thus $U={\bigcup }_{u\in U}{O}_u$ is open. Interchanging the factors, using a fixed $u_0\in U$, proves $V$ open. Therefore $\boxed{U\in {\mathcal{T}}_X,V\in {\mathcal{T}}_Y}$. Nonemptiness matters: an empty product is open whatever the other factor is.

The following open rectangles form a basis of a topology for the product topology:These open rectangles in a finite product cover $X\times Y$, and intersections remain rectangles since $(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)$. One may equivalently take products of elements from a chosen basis of a topology in each factor.

The product topology on $X\times Y$ has the open rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$, as a topological basis. The projection preimages are $p_X^{-1}(U)=U\times Y$ and $p_Y^{-1}(V)=X\times V$, so both projections are continuous.

If $f:Z\to X\times Y$ is continuous, both coordinate compositions are continuous. Conversely, if both are continuous, thenis open for every basic rectangle, proving continuity of $f$. Given paths between two chosen endpoints in each factor, their coordinatewise pair is therefore a continuous path in the product, proving that a product of path-connected spaces is path-connected.

For the compact-neighborhood assertion, fix $x\in K$. For each $y\in L$, choose an open rectangle $U_y\times V_y$ containing $(x,y)$ inside $W$. Compactness of $L$ chooses finitely many $V_{y_j}$ covering $L$. PutThen $x\in U_x$, $L\subseteq V_x$, and $U_x\times V_x\subseteq W$: any point of $V_x$ belongs to one of those rectangles' second factors, while $U_x$ lies in every first factor.

Compactness of $K$ now selects finitely many $U_{x_i}$ covering $K$. SetThese are open, contain $K$ and $L$, and if $(u,v)\in U\times V$, choose an $i$ with $u\in U_{x_i}$ and use $v\in V_{x_i}$. HenceIf $K$ or $L$ is empty, take the corresponding open factor empty and the other factor the whole space; the assertions then hold directly. No Hausdorff assumption on the factors is needed.

The product topology on $X\times Y$ has as a basis of a topology the rectangles $U\times V$ with $U$ open in $X$ and $V$ open in $Y$; arbitrary unions of such rectangles are its open sets.

For distinct points $(x_1,y_1),(x_2,y_2)$, at least one coordinate differs. If $x_1\ne x_2$, disjoint open neighbourhoods $U_1,U_2$ in the Hausdorff space $X$ give disjoint open neighbourhoods $U_1\times Y,U_2\times Y$ of the two points. If the second coordinates differ instead, use $X\times V_1,X\times V_2$. Thus a product of two Hausdorff spaces is Hausdorff.