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Allow terminal positions of an infinite game of perfect information to carry a losing-player label. A clopen set of infinite winning plays still gives a determined infinite game. At a position where all infinite continuations have one winner, the other player can win only by forcing a terminal loss for that winner. The usual winning-position attractor in an infinite game decides this reachability game. The tree of remaining undecided positions is well-founded: every infinite branch would eventually have its clopen payoff decided by a prefix. Assign the already decided winners upwards by well-founded recursion. Terminal labels are essential; a terminal node created by a covering need not be lost by the player whose turn would have come next.

A clopen up-set is both clopen in the topology and upward closed in the order. The clopen up-sets of a Priestley space form a distributive lattice under intersection and union.

For $A\subseteq \mathbb{N}$, the basic set $\hat{A}=\{U:A\in U\}$ in the Stone-Čech compactification of the natural numbers is a clopen set, since its complement is $\widehat{\mathbb{N}\setminus A}$. Distinct ultrafilters are separated by complementary basic sets. A basic cover with no finite subcover would give complements with the finite intersection property, whose generated filter on a set extends to an ultrafilter missing the entire cover. Thus the space is a compact Hausdorff space.

For an infinite coinfinite set $A\subseteq \mathbb{N}$, let $E_A=[A{]}^{\omega }$ in the space of infinite subsets of the natural numbers. This cone is a nonempty clopen set in the Ellentuck topology, because it is the basic neighbourhood $[\varnothing ,A]$, and any $X\notin E_A$ has a finite initial segment containing a point outside $A$ and hence a basic neighbourhood disjoint from $E_A$.

The cone is a closed set with empty interior in the ordinary topology on infinite subsets: membership of a point outside $A$ witnesses its complement as an open set, while every ordinary finite-stem neighbourhood can be extended by an arbitrarily large point outside $A$. Thus $E_A$ is an ordinary nowhere dense set, but it is neither ordinary open nor Ellentuck nowhere dense. It provides both distinctions between these two topologies.

In the space of infinite subsets of the natural numbers, the families $C_n=\{X:n\notin X\}$ are clopen sets in the ordinary topology on infinite subsets and thus in the finer Ellentuck topology. Their union is $[\mathbb{N}{]}^{\omega }\setminus \{\mathbb{N}\}$. Every basic Ellentuck topology neighborhood of $\mathbb{N}$ contains $\mathbb{N}\setminus \{n\}$ for some sufficiently large $n$, so this union is not a closed set.

A normalized positive linear functional on $C(\Omega )$ for Cantor space gives a finitely additive set function on the algebra of clopen sets. If a countable disjoint union of such sets is itself clopen, compactness gives a finite subcover; all other terms are empty. Finite additivity therefore gives the required countable additivity, making $\ell$ a premeasure. The Caratheodory extension theorem produces a unique Borel probability measure.

A cylinder set in a product of spaces restricts a finite collection of coordinates $F$ and leaves all other coordinates free. It has the form ${\pi }_F^{-1}(A)$, where ${\pi }_F$ is the coordinate projection onto the finite subproduct. Taking $A$ to be a product of open sets gives the basic open cylinders for the product topology. In a finite-alphabet full shift, specifying exact letters at finitely many coordinates gives a clopen set.

For a finite set $s$ and an infinite set $A$ lying above $\max s$, putThese sets form the basic neighborhoods of the Ellentuck topology, also called the star topology. It is finer than the ordinary topology on infinite subsets. Every basic neighborhood is a clopen set in the Ellentuck topology, and a closed set in the ordinary topology.

Every convergent sequence in the Stone-Čech compactification of the natural numbers is eventually constant. If distinct ultrafilters $U_n$ converged, isolate each from the other sequence terms using convergence and finitely many separating clopen sets. Finite disjointification gives disjoint $B_n\in U_n$. The union of the even-indexed $B_n$ then belongs exactly to the even-indexed ultrafilters, contradicting convergence of membership in a clopen set.

Every identity neighbourhood of a compact Hausdorff totally disconnected topological group contains an open normal subgroup. Choose a clopen set $U$ containing the identity inside that neighbourhood. Compactness and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$, hence $VU=U$. The setwise stabilizer $H=\{g:gU=U\}$ is an open subgroup contained in $U$. Its index is finite, so its normal core is an open normal subgroup contained in $U$. These subgroups recover the profinite group as the inverse limit of its finite quotients.

For a profinite group $G$ and a complete discrete valuation ring $I$ with a continuous ${\mathbb{Z}}_p$-algebra structure, an $I$-valued measure is a bounded $I$-linear functional on continuous $I$-valued functions, with the uniform norm. Equivalently it assigns finitely additive values in $I$ to clopen sets. Locally constant functions are uniformly dense, so integration extends uniquely from finite sums. Countable additivity in the usual real-measure sense is not imposed. The coefficient ring may have infinite residue field and need not be compact.

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.

Assume $A$ is a countable model of $T$ and its automorphism group is an oligomorphic permutation group. Complete types are constant on automorphism orbits, so only finitely many complete $n$-types are realized in $A$, for each $n$.

These realized types are dense in the Stone topology on a type space $S_n(T)$. Indeed, any nonempty basic clopen set $[\theta ]$ is given by a formula consistent with $T$. Completeness implies $T\models \exists \mathbf{x}\theta$, and thus some tuple in $A$ realizes a type in $[\theta ]$. A finite set of points is closed in a Hausdorff space; if it is also dense, it is the whole space. Therefore $S_n(T)$ equals the finite set of types realized in $A$. This proves (iii) implies (ii), and the previous part then yields (i). We have establishedIn particular, the argument does not assume that an arbitrary countable model is saturated or that its type-orbits coincide before that coincidence has been proved.

An inverse system of topological spaces over a directed set $I$ consists of spaces $X_i$ and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=1$ and $f_{ik}=f_{ij}{f}_{jk}$. Its inverse limit is a space $X$ with compatible continuous projections ${\pi }_i$ such that every compatible family of continuous maps $u_i:Y\to X_i$ factors uniquely through a continuous map $u:Y\to X$. To construct it, takewith the subspace topology inherited from the product topology. The map $u(y)=(u_i(y))$ has values in $X$ and is continuous by the definition of the product topology; its coordinates force uniqueness. If $X^{\prime }$ is another inverse limit, the two universal properties give continuous maps $X\to X^{\prime }$ and $X^{\prime }\to X$ whose composites have all the same projections as the identity. Uniqueness makes these composites the identity. The inverse limit exists and is unique up to a unique projection-preserving homeomorphism. An arbitrary inverse limit may be empty; nonemptiness requires additional hypotheses.

Use the usual Hausdorff convention for a topological group. An inverse limit of finite discrete groups is a closed subgroup of their product: each compatibility condition is closed because the finite factors are Hausdorff spaces. The Tychonoff theorem gives compactness, and distinct tuples are separated by a clopen coordinate cylinder, so the limit is totally disconnected. Coordinatewise multiplication and inversion are continuous, giving a profinite group.

Conversely, let $G$ be a compact Hausdorff space and a totally disconnected topological group. We use the standard topological fact that a compact Hausdorff space that is totally disconnected has a basis of clopen sets. Given an identity neighbourhood, choose a clopen identity neighbourhood $U$ inside it. Compactness of $U$ and continuity of multiplication give a symmetric identity neighbourhood $V$ with $VU\subseteq U$: cover $U$ by finitely many neighbourhoods on which multiplication by a sufficiently small identity neighbourhood stays inside $U$, then intersect those identity neighbourhoods. Symmetry gives $vU=U$ for $v\in V$. Consequently $H=\{g:gU=U\}$ is a subgroup containing $V$, hence an open subgroup, and $H\subseteq U$ since $1\in U$. Its open cosets cover the compact group, so there are finitely many. The normal core $N={\bigcap }_{g\in G}gH{g}^{-1}$ is a finite intersection of open subgroups, hence an open normal subgroup contained in $U$. Thus the open normal subgroups form an identity-neighbourhood basis and have trivial intersection.

The canonical group homomorphism $G\to {\underleftarrow{\lim }}_{N}G/N$ is continuous and injective. For a compatible tuple of quotient elements, its inverse-image cosets in $G$ are closed and have the finite intersection property: a finite family is refined by the quotient for the intersection of its normal subgroups. Compactness gives an element in all those cosets, proving surjectivity. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The compact totally disconnected and inverse-limit definitions of a profinite group agree.

Index the finite nonempty sets $\mathcal{P}(N)$ by open normal subgroups, with the transition from $M$ to $N$ when $M\subseteq N$ given by image under $G/M\to G/N$. Images of Sylow subgroups under surjections of finite groups are Sylow subgroups. These transitions are surjective: if $S\le G/N$ is a Sylow subgroup, choose a Sylow subgroup in its full inverse image in $G/M$. It maps onto $S$, and its order has the full $p$-part of $|G/M|$, so it is a Sylow subgroup of $G/M$ too. The transitions respect composition, giving an inverse system of finite sets.

Its inverse limit is nonempty. Indeed, finitely many compatibility conditions can be satisfied by choosing one Sylow subgroup in a quotient refining all the involved quotients and taking its images; compactness of the product of the finite sets then gives a compatible family $(P_N{)}_N$. DefineThis is a closed subgroup of $G$. Its image in $G/N$ is exactly $P_N$: for a prescribed $s\in P_N$, each finite collection of conditions defining $P$ together with ${\pi }_N(g)=s$ is satisfiable in a common refining quotient, because the transitions on the chosen Sylow subgroups are surjective. The finite intersection property again supplies $g\in P$. Hence $P\cong {\underleftarrow{\lim }}_N{P}_N$ is a pro-p group and $PN/N=P_N$ for every $N$.

Finally, if a pro-p subgroup $Q$ contains $P$, its image in every finite quotient is a p-group containing the Sylow subgroup $P_N$, so that image equals $P_N$. Thus every $q\in Q$ belongs to every ${\pi }_N^{-1}(P_N)$, whence $Q\subseteq P$. The constructed $P$ is a maximal pro-p subgroup.

For a profinite abelian group $G$, its Iwasawa algebra is the completed group algebrawhere $U$ runs over open subgroups. The transition map sums the coefficients over the fibers of $G/V\to G/U$ when $V\subset U$. Consequently an element gives compatible values $\mu (gU)\in {\mathbb{Z}}_p$ on cosets, equivalently a finitely additive ${\mathbb{Z}}_p$-valued p-adic measure on the clopen sets of $G$. Conversely these values specify every finite-quotient coefficient. Integration of a continuous function with values in ${\mathbb{Z}}_p$ is obtained by uniform approximation by locally constant functions; the integrals converge because all measure values have $p$-adic absolute value at most $1$. Multiplication in the Iwasawa algebra becomes convolution of p-adic measures. This is its measure realization of an Iwasawa algebra.

The Mahler theorem states that each continuous function $g:{\mathbb{Z}}_p\to {\mathbb{Z}}_p$ has the unique expansion converging in the sense of uniform convergenceIts Mahler coefficients areand $\Vert g{\Vert }_{\infty }={\sup }_n|b_n{|}_p$. The same assertion holds for continuous functions with values in ${\mathbb{Q}}_p$ with coefficients in ${\mathbb{Q}}_p$.

Given $f(T)={\sum }_{n\ge 0}{a}_n{T}^n\in R$, define the bounded linear functionalThe sum converges because $a_n\in {\mathbb{Z}}_p$ and $b_n\to 0$. Its values on indicator functions of clopen sets give a p-adic measure, denoted $\lambda (f)$, withConversely, a p-adic measure determines these coefficients $a_n\in {\mathbb{Z}}_p$, and continuity allows termwise integration of every Mahler expansion. Thus the inverse iswhich is the Amice transform. These two constructions are inverse, proving the canonical bijection $\lambda :R\simeq \Lambda ({\mathbb{Z}}_p)$. In fact it is an isomorphism of rings: the identity $\left(\genfrac{}{}{0ex}{}{x+y}{n}\right)={\sum }_{i+j=n}\left(\genfrac{}{}{0ex}{}{x}{i}\right)\left(\genfrac{}{}{0ex}{}{y}{j}\right)$ identifies convolution of p-adic measures with multiplication of formal power series. For example, $\lambda (1)={\delta }_0$ and $\lambda (1+T)={\delta }_1$, where ${\delta }_a$ is the Dirac measure at $a$.

For the moment identity, let $\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$ denote a Stirling number of the second kind. The finite polynomial identitygives $\int x^{k}d\lambda (f)={\sum }_{n=0}^k{a}_{n}n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. On the other hand, $D=(1+T)d/dT$ satisfies $D^k(1+T{)}^x=x^k(1+T{)}^x$ for every nonnegative integer $x$. Expanding $(1+T{)}^x$ and evaluating at $T=0$ yieldsComparison in the basis of binomial polynomials gives $(D^k{T}^n)(0)=n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. Terms with $n>k$ have zero constant term after at most $k$ differentiations, so no infinite-series interchange is needed. ThereforeThe case $k=0$ says that the total mass is $f(0)$.

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

We prove the Borel determinacy theorem by reducing each Borel game to a clopen game while preserving winning strategies. The construction permits arbitrary set-sized alphabets; no enumeration of the move set is used. Assume $A\ne \varnothing$. For an empty alphabet the game ends at its initial position, with its assigned terminal outcome, and is trivially determined.

Allow intermediate game trees $T$ to have terminal positions with assigned winners. Write $[T]$ for the space of infinite branches, with its prefix-cylinder topology; the payoff $B\subseteq [T]$ governs infinite plays, while the assigned winners govern finite plays. Question 4's attractor proof still gives open and clopen determinacy: include I-winning terminals among the initial targets and make the opposing terminals absorbing losses.

A covering of an infinite game consists of a new tree $\tilde{T}$, a prefix-preserving map $\pi :\tilde{T}\to T$ preserving lengths, and a strategy map $\Phi$ for each player. Old terminal outcomes must be inherited. The strategy map is local in depth: choices before depth $n$ depend only on the given strategy before depth $n$. Its essential lifting requirement is that each original play $x$ following $\Phi (\sigma )$ has a lift $\tilde{x}$ following $\sigma$ whose projection is $x$, or whose projection is an initial segment of $x$ and which ends with a loss for the player using $\sigma$. Infinite projections are taken by union of prefixes. A $k$-covering is identical to the original tree, terminal assignments and strategy choices through depth $k$.

These conditions transfer winning strategies for every payoff. Indeed, if $\sigma$ wins the covered game with payoff ${\pi }^{-1}(B)$, a lift of a play following $\Phi (\sigma )$ cannot be an early loss for its player. The lift therefore projects to the whole original play. Infinite membership in $B$ agrees under projection, and an original terminal loss would also be a loss in the game covering. Hence the original play is won. Call a game covering an unravelling of a game payoff when ${\pi }^{-1}(B)$ is clopen in $[\tilde{T}]$.

We first construct a $k$-covering unravelling any closed $C\subseteq [T]$. Choose an even $m\ge k$ and leave the first $m$ levels unchanged. At a nonterminal position $p$ of length $m$, I's next move will carry auxiliary information. For each legal ordinary move $a$, let $Z$ be the set of first nonterminal positions strictly beyond $pa$ whose infinite continuations miss $C$: $q\in Z$ means $q$ extends $pa$, $[T_q]\cap C=\varnothing$, and no intervening strict extension of $pa$ has this property. These positions are pairwise incomparable. I plays $(a,X)$, where $X\subseteq Z$.

If $pa$ was already terminal, inherit its outcome. Otherwise II has two types of response. On accepting, II plays $(\mathrm{accept},b)$ with $b$ an ordinary legal move, and subsequent moves are ordinary. When a position $q\in Z$ is reached, end the covered game with a win for I if $q\in X$ and for II if $q\notin X$. Other old terminals retain their winners. On challenging, II plays $(\mathrm{challenge},q,b)$ with $q\in X$ and $pab$ an initial segment of $q$; subsequent play is forced along $q$ until reaching it and then proceeds ordinarily in $T_q$. Original terminal outcomes remain in force. Forget the auxiliary data to define $\pi$.

For an infinite accepted play, no exit in $Z$ was reached. Its projection belongs to $C$: if it did not, closedness would give a prefix cylinder disjoint from $C$, and the first such nonterminal extension strictly beyond $pa$ would have ended the play. An infinite challenged play projects through an exit $q$ and therefore misses $C$. Thus, among infinite plays, the inverse image of $C$ is exactly the acceptance branch of this finite protocol. It is clopen, decided at depth $m+2$.

Here are both strategy-transfer maps; these are necessary because continuous projection by itself does not preserve determinacy. Given a strategy $\sigma$ for I in the covered tree, follow its ordinary moves while simulating acceptance of its announcement $X$. If an exit $q\notin X$ is reached, continue arbitrarily: that play has an early covered loss for I as a lift. If $q\in X$ is reached, switch to the continuation prescribed by $\sigma$ under the challenge of $q$. This is consistent with the already observed prefix, since the challenged game forced every intermediate move along $q$. If no exit is reached, acceptance itself supplies the full lift. Old terminal plays lift in the same way. This defines $\Phi (\sigma )$ and verifies the lifting requirement for I.

For a covered strategy $\tau$ for II, at the projected position $pa$ defineAgainst announcement $Y$, the strategy must accept: a challenge would have to name an element of $Y$, contradicting its definition. Simulate this acceptance, including its ordinary response $b$. If an exit $q\in Y$ is reached, it is an early covered loss for II, so continue arbitrarily. If an exit $q\notin Y$ is reached, choose an announcement $X_q$ for which $\tau$ challenges $q$, and switch to that simulation. The challenged game's forced prefix agrees with the observed prefix, including the earlier ordinary response $b$, so it supplies a full lift following $\tau$. If no exit is reached, the original acceptance simulation gives the full lift. This defines $\Phi (\tau )$ and verifies lifting for II. Choices of $X_q$ use the axiom of choice on a set of possibilities.

Both maps are local in depth. The sets $Y$ and the chosen $X_q$ use only II's response function at depth $m+1$; later choices use only strategy values at the already reached depth. The switches do not query choices at later levels. On irrelevant histories assign fixed legal moves. Thus this is a genuine $k$-covering. Complementing $C$ changes neither the game covering nor its lifting property, and complements of clopen sets are clopen. This proves the base construction for both closed and open payoff sets.

We next give the limiting construction needed for the induction. Coverings compose by composing their position and strategy maps; lifting successively verifies the same lifting condition. Supposeis a sequence of game coverings whose $i$th map leaves depth $k+i$ unchanged. For each finite depth $d$, the trees and terminal assignments through that depth eventually stop changing. Define $T_{\infty }$ using these stabilized levels. Its map to $T_i$ on a position of length $d$ is the composite from any sufficiently late stage; stabilization makes the value independent of the stage. Define strategy maps in the same way, using their locality in depth. These maps respect the tree, terminal outcomes and composition, and leave depth $k$ unchanged.

To check lifting, start with a play of $T_i$ consistent with the projected strategy. Lift the entire play successively through the later game coverings, using the corresponding projected strategy at each stage. If all these lifts are infinite, each projects to the preceding one and each finite prefix eventually stabilizes, giving the desired infinite lift in $T_{\infty }$. If a premature own-player loss occurs, later lifts have nonincreasing finite lengths, and those lengths and their prefixes eventually stabilize. This gives a terminal own-player loss in $T_{\infty }$, which is exactly the allowed exceptional lift. The same reasoning handles original finite terminal plays. Thus $T_{\infty }$ covers every $T_i$. This argument uses level stabilization, not compactness of an infinitely branching tree.

Now induct on the countable Borel hierarchy rank, with the strengthened assertion: for every game tree on any set alphabet and every $k$, a payoff of that rank has a $k$-covering unravelling it. The base case of open or closed sets was proved above. Complements preserve the assertion using the same game covering. For the union step, write $B={\bigcup }_{n<\omega }{B}_n$ with constituent sets of smaller rank. Starting at $T_0=T$, successively choose a $(k+n)$-covering $T_{n+1}\to T_n$ that unravels the inverse image of $B_n$ under the composite map to $T$. This is legitimate because continuous inverse images preserve Borel hierarchy rank, and the induction assertion ranges over all trees and alphabets, including the enlarged ones produced by earlier game coverings.

In the stabilized inverse limit, each pulled-back $B_n$ is clopen, because the later projection to the stage where it was unravelled is continuous. Their union is therefore open. Apply the already proved open-set unravelling construction once more, with preservation depth $k$, and compose with the limit game covering. The result unravels $B$ itself. This proves the strengthened assertion at every successor or limit countable Borel hierarchy rank. Every Borel set arises at some such rank: countable unions take the supremum of countably many countable ranks and then one further stage, and complements do not leave the generated hierarchy.

Apply the assertion to $T=A^{<\omega }$ and $k=0$. The covered payoff is clopen and hence determined by Question 4, including its assigned finite terminal outcomes. Transfer its winning strategy through the game covering. We obtainEvery auxiliary alphabet and strategy collection used above is a set. The argument uses power sets and replacement to build those collections and their iterated game coverings, and the axiom of choice for strategy selections; it does not require $A$ to be countable.