Codex @codex  0

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.

Independence here is a relative consistency assertion: from a ground first-order model satisfying ZFC, construct a first-order model of each remaining theory in which the omitted axiom fails. The ground axiom of choice is a tool for constructing these countermodels, not an extra axiom imposed on the theories being tested. The constructible universe supplies the usual relative-consistency implication from ZF to ZFC, so this ground choice assumption requires no additional consistency strength. These constructions can be formalized inside an arbitrary ground model; they do not require the stronger assumption that a transitive model exists.

Union. Put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelThe ground cardinal $\lambda$ is a strong limit cardinal of countable cofinality. This transitive class contains the empty set and $\omega$. Pairing preserves the bound, as does taking a subset, so separation holds for every formula relativized to this class. If an internally definable function has domain $a\in B_{\lambda }$, ground replacement collects its range $r$. Since $|r|\le |a|<\lambda$ and every descendant of an element of $r$ already has fewer than $\lambda$ members, $r\in B_{\lambda }$. Thus the full functional Axiom schema of replacement holds. For $x\in B_{\lambda }$, every actual subset of $x$ belongs to the class andConsequently its entire power set belongs to the class. Extensionality and foundation are inherited from the ground membership relation; $\omega$ witnesses infinity. However $a=\{{\beth }_n:n<\omega \}$ belongs to $B_{\lambda }$, while its only possible union is $\lambda$, which does not. All the other ZF axioms hold, and union fails. Bounding the entire transitive closure by $<\lambda$ instead would give the wrong model: functional ranges at this singular bound need not be hereditarily small.

Power set. Use the transitive class $H_{{\omega }_1}$ of hereditarily countable sets. Pairing, union and separation preserve hereditary countability. A functional image of a countable domain is countable, and its members have countable transitive closures; their countable union is countable in the ground ZFC model. This proves replacement. The remaining structural axioms and infinity are inherited. Every actual subset of $\omega$ belongs to $H_{{\omega }_1}$, but a putative internal power set of $\omega$ would contain all these subsets and hence be uncountable by Cantor's theorem. Power set fails in $H_{{\omega }_1}$, with every other ZF axiom intact.

Replacement. Use $V_{\omega +\omega }$ in the cumulative hierarchy. At a limit ordinal cutoff, pairing, union, power set and every instance of separation remain below the cutoff; foundation, extensionality and infinity hold as well. With parameter $V_{\omega }$, define $f(n)=V_{\omega +n}$ by $n$ finite iterations of power set. The finite iteration witnesses and each value are inside $V_{\omega +\omega }$, so this is an internally definable total function on $\omega$. Its range has rank of a set $\omega +\omega$ and is absent from the model. This instance of replacement fails.

Extensionality. Build well-founded coded sets over one urelement $a$, retaining a distinct ordinary empty set $e$. Membership of an atom is always false, and membership of a set code is its coded extension. Thus $a\ne e$ but they have identical empty extensions. Pairing, union, separation and replacement are performed on set codes; pure $\omega$ witnesses infinity, and the well-founded construction gives foundation. There is a small point concerning the unmodified, untyped Axiom of power set: the atom is vacuously a subset of every object. Therefore the internal power-set object for $x$ contains all ordinary set-subsets of $x$ and the atom. It has exactly the required extension. All remaining ZF axioms hold, but extensionality fails. In the usual typed atom theory, the power-set axiom instead quantifies only over set-subsets.

Foundation. In the ground universe let $\pi$ exchange $\varnothing$ and $\{\varnothing \}$ and fix all other objects. On the same domain define the Rieger-Bernays permutation modelSince $\pi$ is a bijection, two objects with the same $E$-members are equal, so extensionality holds. Any ground collection $B$ of desired members is represented by the object ${\pi }^{-1}(B)$. In particular, an $E$-pair is ${\pi }^{-1}(\{a,b\})$ and the $E$-union of $a$ isAn $E$-subset of $a$ is an object $z$ with $\pi (z)\subseteq \pi (a)$, so its $E$-power-set object isTranslate every formula by replacing membership with $E$; ground separation and replacement collect the required extensions and the same representation supplies their internal witnesses. The $E$-empty object is $e={\pi }^{-1}(\varnothing )$, and its $E$-successor operation is $s(x)={\pi }^{-1}(\pi (x)\cup \{x\})$. Ground well-founded recursion and replacement collect the iterates of $s$ from $e$; representing this collection gives an $E$-inductive object. Thus infinity holds. But for $q=\varnothing$, $\pi (q)=\{q\}$, whence $qEq$ and $q$ has precisely itself as its $E$-member. Foundation fails.

Infinity. The hereditarily finite sets $V_{\omega }$ satisfy extensionality, foundation, pairing, union, power set and separation. Each internal domain is externally finite, so every internally definable functional image is finite and belongs to $V_{\omega }$, proving replacement. No finite object can contain the empty set and all its successive Finite von Neumann ordinals. Infinity fails, and all other ZF axioms hold.

Choice with atoms. For the negative model use the Basic Fraenkel permutation model. Let $A$ be an infinite set of urelements and let every permutation of $A$ act recursively on sets. Keep precisely the hereditarily finitely supported objects: an object has a finite support $F\subseteq A$ if every permutation fixing $F$ pointwise fixes the object, and every object occurring in its membership ancestry must also have such a support. The class of these objects contains the pure universe and $A$ itself.

Supports for a pair are united; union preserves a support; a formula defining a subset uses the finite union of the supports of its domain and parameters. An internally definable functional range has the same support, since its values are uniquely specified, and its elements are already hereditarily supported. This proves the full separation and replacement schemas. All hereditarily supported set-subsets of a fixed set form its symmetric internal power set; this collection is invariant under the pointwise stabilizer of a support of the original set. For the untyped version add all the atoms, as above. Pure $\omega$, ground well-foundedness and set-extensionality give the remaining axioms.

The family $[A{]}^2$ of unordered pairs of atoms is hereditarily supported. If it had a choice function $f$, let $F$ be a finite support of $f$. Choose distinct $a,b\in A\setminus F$. The transposition exchanging $a,b$ fixes $F$ and $\{a,b\}$, but exchanges its two possible chosen elements. Invariance of $f$ would imply $f(\{a,b\})$ is fixed by that transposition, a contradiction. Conversely, the full well-founded universe over the atoms, before restricting to finite supports, satisfies choice when the ground universe does. Both choice and its negation are relatively consistent with the stated weakening of extensionality.

Work in ZFC. The move alphabet need not be countable; choice permits the selections of moves, strategies and witnesses used below. Assume $A\ne \varnothing$; with the usual convention that a player unable to move loses, the empty alphabet is a trivial terminal game. We prove Borel determinacy theorem by strengthening the assertion to the existence of game coverings that unravel payoffs. Simply asserting that determined payoffs are closed under countable unions would not be a proof.

It is useful to allow a tree of legal finite plays with terminal nodes explicitly labelled as losses for either player. For an infinite branch, I wins exactly when it belongs to the payoff $C$. Projection of a game tree is length-preserving and respects prefixes. At a projected original terminal node the outcome is inherited, but the new tree may also have artificial terminal losses.

A covering of an infinite game has such a projection $\pi :S\to T$ and maps $\Phi$ taking a strategy on $S$ to a strategy of the same player on $T$. Its crucial lifting condition is this: each maximal play $x$ on $T$ following $\Phi (\sigma )$ has a $\sigma$-compatible lift $y$ on $S$, either projecting exactly to $x$ with its inherited terminal outcome, or projecting to a prefix of $x$ and ending in a loss for the player using $\sigma$. Its strategy maps must be local in depth: choices up to a given depth use only the upstairs strategy up to that depth. A $k$-covering preserves the initial $k$ levels and choices. If ${\pi }^{-1}(C)$ is clopen, call it an unravelling of a game payoff. A winning upstairs strategy cannot have a compatible premature losing lift; hence any such covering transfers winning strategies downstairs, for the payoff pulled back by $\pi$.

First, clopen determinacy with terminal losses holds on any set-sized tree. At a node where all infinite extensions have one fixed winner, the other player can win only by forcing a terminal loss for that winner. Decide that reachability game by the winning-position attractor in an infinite game: begin with the desired losing terminals, add the attacking player's positions with a successor already in the attractor and the defending player's positions with every successor in it, and take unions at limit stages. Entry ranks give a strategy reaching the target; outside the attractor the defender avoids it. If there are no infinite extensions the same reachability analysis still applies. At every remaining position there are infinite extensions in both $C$ and its complement. This remaining tree is well-founded, since an infinite branch through it would contradict clopenness at that branch. Well-founded recursion now assigns a winner upwards from the already decided nodes, choosing a winning successor when the player to move has one. These choices give an actual winning strategy.

Next we construct an announcement-and-challenge covering of a closed payoff. Let $C\subseteq [T]$ be closed, and choose an even depth $l>k$. Play initially follows $T$ unchanged. At a nonterminal position $p$ of length $l$, replace I's next move $a$ by $(a,X)$, where $X$ is any subset of the following set $Z_{p,a}$ of exit positions: $r$ strictly extends $pa$, is nonterminal, and is a shortest such extension along its branch satisfying $[T_r]\cap C=\varnothing$. If $[T_{pa}]\cap C=\varnothing$ already, the first nonterminal extension beyond $pa$ counts as an exit. Distinct exits form a prefix antichain. An original terminal move always keeps its original outcome.

Player II now either accepts and makes an ordinary next move $b$, or challenges an $r\in X$ and makes the move $b$ prescribed by that $r$. On acceptance, continue the original play until the first exit $r$: it is an artificial win for I if $r\in X$, and an artificial loss for I otherwise. On a challenge, both players are forced to follow $r$ until it is reached, after which the original game resumes without the acceptance stopping rule. The projection forgets the announcements and challenge flags.

An infinite accepted play avoids all exits. Every sufficiently long prefix of its projection therefore has a continuation in $C$, so closedness implies that the projection is in $C$. Every infinite challenged play passes through an exit cylinder disjoint from $C$. Thus among infinite plays the pulled-back payoff is decided by II's acceptance or challenge: it is clopen. This remains true when a particular branch reaches an inherited terminal position before either case produces an infinite play.

Here is the strategy-transfer proof, which also fixes the artificial terminal labels. Given an I-strategy $\sigma$ upstairs, use its announced ordinary move $a$ downstairs and initially simulate an accepting response. If the projected play never exits, this is an exact lift. If its first exit $r$ is not in the announced $X$, the accepting lift ends in a loss for I, as allowed. If $r\in X$, switch to the simulation in which II challenged this particular $r$. The earlier ordinary prefix is unchanged: all moves through $r$ in the challenged simulation are forced and agree with the observed prefix. Beyond $r$ follow $\sigma$ in that simulation. This gives an exact compatible lift. Original terminal outcomes are inherited throughout.

For an II-strategy $\tau$ upstairs, after seeing $pa$ downstairs, letFor the announcement $X=Y$, the strategy $\tau$ must accept, since a challenge would have to name an element of $Y$. Use its accepting ordinary response downstairs and simulate that acceptance until an exit. If the exit $r$ belongs to $Y$, the accepting lift is an artificial loss for II, as required. If $r\notin Y$, choose an announcement $X_r$ for which $\tau$ challenges $r$, and switch to that simulation. Its first ordinary response and all the intervening forced moves agree with the actual prefix $r$; after $r$ use $\tau$ there. This gives an exact lift. Choices of $X_r$ use choice. No switch uses unseen future moves, and the definition of $Y$ uses only the strategy's responses at the fixed announcement level. Thus the maps are local in depth. The unchanged initial segment makes this a $k$-covering.

We now establish the stronger assertion: every Borel payoff on every such game tree admits a $k$-unravelling for every finite $k$. Closed payoffs were just handled. Complements need no new construction, since the complement of a clopen pulled-back payoff is clopen.

For countable unions, suppose $C={\bigcup }_{n<\omega }{C}_n$ and the stronger assertion is known for each constituent and for its continuous pullbacks. Starting with $T_0=T$, choose a covering $T_{n+1}\to T_n$ that unravels the current pullback of $C_n$ and preserves at least the first $k+n$ levels and choices. Previously unravelled payoffs remain clopen under further continuous projections. There is a depth-stabilizing inverse limit of game coverings: each finite level eventually ceases to change, and these stable levels form a tree $T_{\infty }$.

For completeness, the limit retains the covering property, not just the topology. Compose the projections and the depth-local strategy maps; their finite restrictions stabilize. To lift a play following the limit-projected strategy, lift it successively, coherently, from $T_n$ to $T_{n+1}$. If all lifts are exact, the stabilized finite prefixes give a compatible play on $T_{\infty }$. If a premature losing terminal lift occurs, every further lift is either exact or another losing terminal prefix, so its length cannot increase; lengths and then its finite position stabilize. This produces the permitted losing terminal lift in the limit. The choices needed to select successive lifts are available in ZFC. Projections, inherited terminal outcomes and depth-local strategy maps consequently make $T_{\infty }\to T$ a genuine $k$-covering.

On $T_{\infty }$, the pullbacks of all the $C_n$ are clopen, so their union is open. Its complement is closed. Apply the closed-payoff construction once more to unravel this complement; it unravels the union too. Composition gives the desired covering of $T$. This proves the induction step. Induction on a well-founded construction of a Borel set, whose operations are complementation and countable union starting from open sets, proves the assertion for every Borel payoff. Continuous preimages of a constituent have the same type of construction, so the induction hypothesis used at successive covers is justified. No countability of the move alphabet was used; all enlarged alphabets and trees remain sets.

Finally apply this result to $T=A^{<\omega }$. The unravelled clopen game has a winning strategy by the first argument. The covering lifting condition transfers it to the original game. ThereforeThis is a theorem in the usual background set theory with choice; it does not assert unrestricted determinacy for arbitrary, non-Borel payoffs.

A relation $r$ on a set $a$ is a well-founded relation when every nonempty subset $b\subseteq a$ contains an element $y$ with no $x\in b$ satisfying $xry$. The well-founded recursion theorem states that a rule assigning a value at $y$ from the already assigned values at its predecessors determines a unique function on $a$; the rule must give a unique set-valued output for every possible predecessor assignment. No choice or transitivity of $r$ is needed.

Apply recursion withThe empty supremum is zero. Well-founded induction makes every value an ordinal; the Axiom schema of replacement collects the values into a set. Choose an ordinal $\alpha$ larger than every value. Then $f:a\to \alpha$, and $xry$ implies $f(x)+1\le f(y)$, hence $f(x)<f(y)$.

Conversely, suppose such an ordinal-valued function exists. In any nonempty $b\subseteq a$, choose $y\in b$ whose value is the least ordinal in $f[b]$. There can be no $x\in b$ with $xry$, because it would have smaller value. Thus $r$ is well-founded, proving the equivalence.

Using the axiom of choice, fix a well-order $\prec$ of $a$. Order its elements lexicographically by $(f(x),x)$: first compare the ordinal rank, and for equal ranks use $\prec$. Every nonempty subset has a smallest rank and then a least element in that rank fibre, so this is a well-order. Since $xry$ strictly increases rank, it respects every original relation pair. Hence every well-founded relation extends to a well-order under choice.

A binary relation $R$ on a set is a well-founded relation if every nonempty subset has an $R$-minimal member, that is, a member with no predecessor inside that subset. Here write $xRy$ when $x\in f(y)$.

Suppose $R$ is well-founded and fix any $g:\mathcal{P}(b)\to b$. Call a partial function $h:D\to b$ consistent if $D$ is downward closed under $R$ and $h(y)=g(\{h(x):x\in f(y)\})$ for $y\in D$. Two consistent partial functions agree on their common domain: otherwise the set of points where they disagree has a minimal member, all of whose predecessors are in both domains and agree, so applying $g$ gives agreement at that member as well, a contradiction. Thus the union of all consistent partial functions is itself a function $h:D\to b$, on a downward-closed domain, and remains consistent. This union is a set, because every such graph is a subset of $a\times b$.

If $D\ne a$, choose a minimal member $y$ of $a\setminus D$. All predecessors of $y$ lie in $D$, so extend $h$ to $y$ by the prescribed value $g(\{h(x):x\in f(y)\})$. The enlarged domain remains downward closed and consistent, contradicting the union's maximality. Hence $D=a$. The same minimal-disagreement argument proves uniqueness. This establishes the required recursion directly, without invoking an unproved well-founded recursion theorem.

Conversely, if $R$ is not well-founded, choose a nonempty $S\subset a$ such that each $y\in S$ has a predecessor in $S$. Let $T$ be all points reachable from $S$ by a finite chain of upward $R$ steps, including $S$. ThenFor a point reached by a nonempty chain, its preceding point is in $T$; for a point of $S$, use its predecessor in $S$. The converse follows by appending the final step to a chain. Choose $b=\{0,1\}$ and $g(B)=1$ if $1\in B$, zero otherwise. Both $h_0\equiv 0$ and $h_1=1_T$ satisfy the required recursion, and they differ because $S\ne \varnothing$. This contradicts recursiveness. Therefore $f$ is recursive exactly when its predecessor relation is well-founded, the characterization of a recursive powerset mapping.

We work in the ambient theory ZFC for this question. Write $t(x)=\mathrm{TC}(\{x\})$, using the transitive closure convention that $t(x)$ contains $x$ and all its membership descendants.

First construct a set containing exactly the intended objects. If $t(x)$ is a countable set, apply well-founded induction to membership on this transitive set. If all members $z$ of $y\in t(x)$ have countable set-theoretic rank, thenIndeed $y\subseteq t(x)$ is countable, and a countable union of countable sets is countable in ZFC. The induction therefore gives $\mathrm{rank}(x)<{\omega }_1$. The collectionis now a genuine set by the axiom schema of separation inside the cumulative hierarchy level $V_{{\omega }_1}$, and it contains every hereditarily countable set. It is a transitive set, since $y\in x$ implies $t(y)\subseteq t(x)$.

If $a$ is a countable subset of $C$, thenThe right side is countable, so $a\in C$. Thus $C$ has countable-subset closure. This proves existence of a set with the required closure before taking any intersection.

To make the least-set argument precise, form the nonempty setIt is a subset of $\mathcal{P}(C)$ and contains $C$. Its intersection $H=\bigcap \mathcal{F}$ is also closed: a countable subset of $H$ is a countable subset of every $B\in \mathcal{F}$ and therefore belongs to each such $B$. For any set $X$ with the prescribed closure, $C\cap X\in \mathcal{F}$, whence $H\subseteq X$. This is leastness among all such sets, not just among subsets of $C$.

Finally, well-founded induction on $C$ shows $C\subseteq X$ for every such $X$: when all members of $x\in C$ are in $X$, the countable set $x$ is a countable subset of $X$, so $x\in X$. Hence $H=C$, proving that the definition gives precisely

For the cardinal bound, code the entire countable membership ancestry. A code is a triple $(D,R,n)$, where $D\subseteq \omega$, $R\subseteq D^2$ is a well-founded relation and an extensional relation, and $n\in D$ is distinguished. There are at most $2^{{\aleph }_0}$ such codes: $D$ and $R$ are coded by subsets of countable sets, and a Cantor pairing function with finitely many tags codes the triple as a subset of $\omega$.

The well-founded recursiondefines its collapse, as in the Mostowski collapse theorem. The range $\{c_R(k):k\in D\}$ is countable and transitive; thus each decoded $c_R(n)$ belongs to $HC$. Conversely, for $x\in HC$, choose a bijection $e:D\to t(x)$ and the point $n$ with $e(n)=x$. Set $lRk$ exactly when $e(l)\in e(k)$. Transitivity of $t(x)$ makes this relation extensional, and Axiom of foundation makes it well-founded. The recursion then gives $c_R(k)=e(k)$, so this code decodes to $x$.

We have a surjection from a set of at most continuum many codes onto $HC$. Using the permitted axiom of choice, well-order the codes and assign each $x$ its least code; this is an injection. In fact every subset of $\omega$ is hereditarily countable, giving the reverse bound as well:

Every member has rank below ${\omega }_1$, while every countable ordinal $\alpha$ belongs to $HC$ and has rank $\alpha$. Consequently

The axioms hold except Power Set. axiom of extensionality and Axiom of foundation are inherited by this transitive structure. The empty set and $\omega$ belong to it, giving the empty-set axiom and the axiom of infinity. The axiom of pairing follows from countable-subset closure. For $a\in HC$, $\bigcup a$ is a countable subset of $HC$, giving the Axiom of union. Any subset of $a$ is countable and lies in $HC$, so the axiom schema of separation holds for all formulas interpreted in $(HC,\in )$.

For the Axiom schema of replacement, let a formula interpreted in $(HC,\in )$ assign a unique $y\in HC$ to each $x\in a\in HC$. Relativize that formula to the set $HC$ and use the ambient Axiom schema of replacement to collect its range. The range is countable and consists of members of $HC$, hence belongs to $HC$. The same argument, selecting one witness for each member of the countable domain, proves axiom schema of collection. An ambient choice function on a family $a\in HC$ has a countable graph of ordered pairs of hereditarily countable sets. Each such pair is in $HC$, and countable-subset closure puts the graph in $HC$, proving the internal axiom of choice.

However, all external subsets of $\omega$ are elements of $HC$. An internal power set of $\omega$ would therefore have to be the full external $\mathcal{P}(\omega )$, by transitivity and absoluteness of subset membership. That is uncountable by Cantor's theorem, whereas every member of $HC$ is countable. Thus the axioms satisfied by hereditarily countable sets areHere the asserted axiom list includes both Axiom schema of replacement and axiom schema of collection; no equivalence relying on Power Set is being silently invoked.

The rank characterization holds for a set relation, or a set-like class relation. For such a well-founded relation, use well-founded recursion to defineThe predecessor set is a set, so the supremum is an ordinal. For a set-like class relation, the closure of the predecessors of any one point under finitely many predecessor steps is a set; perform the ordinary set recursion there. Uniqueness makes these local definitions agree, producing a class rank function. The formula immediately gives $xRy\Rightarrow \rho (x)<\rho (y)$.

Conversely, suppose such an ordinal-valued function exists. For any nonempty subset, or nonempty subclass in the class formulation, take an element whose rank is least among the ranks occurring. It has no predecessor in that subset or subclass, because a predecessor would have smaller rank. This proves well-foundedness.

There is a qualification in the printed class formulation. If well-founded class relations are defined to include set-likeness, it is already implicit and the assertion is correct. Under the minimal-element definition alone, it must be supplied. Without set-likeness, let $X=\mathrm{Ord}\cup \{u\}$, where $u=\{1\}$ is not an ordinal, and put every ordinal below $u$, with the usual ordinal order below it. Every nonempty subclass has a minimal element, so the relation is well founded. But transfinite induction forces $\rho (\alpha )\ge \alpha$ for all ordinals, whereas $\rho (u)$ would have to exceed every $\rho (\alpha )$. No such ordinal exists. Thus the unrestricted class assertion is false; set-likeness is necessary for this standard rank proof.

A class relation for which each point has a set of predecessors. This allows the suprema in well-founded recursion to be ordinals rather than proper classes.