--- tag: YF766VK21E kind: statement role: proposition name: Incompatibility with the axiom of choice sources: - wikipedia: en:1002300@r1345076451 locator: Axiom of determinacy, section Incompatibility with the axiom of choice note: adopted mention: wikipedia://en/1002300#4Q9B sections: [Statement] inventory: [MRTQ0P31S3] --- ## Statement Under assumption of the axiom of choice, we present two separate constructions of counterexamples to the axiom of determinacy. It follows that the axiom of determinacy and the axiom of choice are incompatible. The set *S*1 of all first player strategies in an ω-game *G* has the same [[cardinality|0G0ASKNA6Z]] as the [[continuum|0WC28W9DNM]]. The same is true for the set *S*2 of all second player strategies. Let *SG* be the set of all possible sequences in *G*, and *A* be the subset of sequences of *SG* that make the first player win. With the axiom of choice we can well order the continuum, and we can do so in such a way that any proper initial portion has lower cardinality than the continuum. We use the obtained well ordered set *J* to index both *S*1 and *S*2, and construct *A* such that it will be a counterexample. We start with empty sets *A* and *B*. Let *α* ∈ *J* be the index of the strategies in *S*1 and *S*2. We need to consider all strategies *S*1 = {s1(*α*)}*α*∈*J* of the first player and all strategies *S*2 = {s2(*α*)}*α*∈*J* of the second player to make sure that for every strategy there is a strategy of the other player that wins against it. For every strategy of the player considered we will generate a sequence that gives the other player a win. Let *t* be the time whose axis has length ℵ0 and which is used during each game sequence. We create the counterexample *A* by transfinite recursion on *α*: 1. Consider the strategy s1(*α*) of the first player. 1. Apply this strategy on an ω-game, generating (together with the other player's strategy) s1(*α*)) a sequence ⟨*a*1, *b*2, *a*3, *b*4, ...,a*t*, *b**t*+1, ...⟩, which does not belong to *A*. This is possible, because the number of choices for ⟨*b*2, *b*4, *b*6, ...⟩ has the same cardinality as the continuum, which is larger than the cardinality of the proper initial portion { *β* ∈ *J* | *β* \< *α* } of *J.* 1. Add this sequence to *B* to indicate that s1(*α*) loses (on ⟨*b*2, *b*4, *b*6, ...⟩). 1. Consider the strategy s2(*α*) of the second player. 1. Apply this strategy on an ω-game, generating (together with the other player's strategy) *s*2(*α*)) a sequence ⟨*a*1, *b*2, *a*3, *b*4, ..., a*t*, *b**t*+1, ...⟩, which does not belong to *B.* This is possible, because the number of choices for ⟨*a*1, *a*3, *a*5, ...⟩ has the same cardinality as the continuum, which is larger than the cardinality of the proper initial portion { *β* ∈ *J* | *β* ≤ *α* } of *J.* 1. Add this sequence to *A* to indicate that *s*2(*α*) loses (on ⟨*a*1, *a*3, *a*5, ...⟩). 1. Process all possible strategies of *S*1 and *S*2 with [[transfinite induction|0ZHQMQ2C1N]] on *α.* For all sequences that are not in *A* or *B* after that, decide arbitrarily whether they belong to *A* or to *B,* so that *B* is the complement of *A.* Once this has been done, prepare for an ω-game *G*. For a given strategy *s*1 of the first player, there is an *α* ∈ *J* such that *s*1 = *s*1(*α*), and *A* has been constructed such that *s*1(*α*) fails (on certain choices ⟨*b*2, *b*4, *b*6, ...⟩ of the second player). Hence, *s*1 fails. Similarly, any other strategy of either player also fails. In this construction, the use of the axiom of choice is similar to the choice of socks as stated in the quote by Bertrand Russell. In a ω-game, the two players are generating the sequence ⟨*a*1, *b*2, *a*3, *b*4, ...⟩, an element in ωω, where our convention is that 0 is not a natural number, hence neither player can choose it. Define the function *f*: ωω → {0, 1}ω such that *f*(*r*) is the unique sequence of length ω with values in {0, 1} whose first term equals 0, and whose sequence of runs (see run-length encoding) equals *r.* (Such an *f* can be shown to be injective. The image is the subset of {0, 1}ω of sequences that start with 0 and that are not eventually constant. Formally, *f* is the Minkowski question mark function, {0, 1}ω is the Cantor space and ωω is the Baire space.) Observe the [[equivalence relation|0G8VEZ3WTK]] on {0, 1}ω such that two sequences are equivalent if and only if they differ in a finite number of terms. This partitions the set into equivalence classes. Let *T* be the set of equivalence classes (such that *T* has the cardinality of the continuum). Define *g*: {0, 1}ω → *T* that takes a sequence to its equivalence class. Define the complement of any sequence *s* in {0, 1}ω to be the sequence s1 that differs in each term. Define the function *h*: *T* → *T* such that for any sequence *s* in {0, 1}ω, *h* applied to the equivalence class of *s* equals the equivalence class of the complement of *s* (which is well-defined because if *s* and *s**are equivalent, then their complements are equivalent). One can show that*****h*is an involution with no fixed points, and thus we have a partition of*T*into size-2 subsets such that each subset is of the form {*t*,*h*(*t*)}. Using the axiom of choice, we can choose one element out of each subset. In other words, we are choosing "half" of the elements of*T,*a subset that we denote by*U*(where U ⊆ T) such that*t*∈*U*iff*h*(*t*) ∉*U.** Next, we define the subset *A* ⊆ ωω in which **1** wins: *A* is the set of all *r* such that *g*(*f*(*r*)) ∈ *U.* We now claim that neither player has a winning strategy, using a strategy-stealing argument. Denote the current game state by a finite sequence of natural numbers (so that if the length of this sequence is even, then **1** is next to play; otherwise **2** is next to play). Suppose that *q* is a (deterministic) winning strategy for **2**. Player **1** can construct a strategy *p* that beats *q* as follows: Suppose that player **2** response (according to *q*) to ⟨1⟩ is *b*1. Then **1** specifies in *p* that *a*1 = 1 + *b*1. (Roughly, **1** is now playing as **2** in a second parallel game; **1** winning set in the second game equals **2** winning set in the original game, and this is a contradiction. Nevertheless, we continue more formally.) Suppose that **2** response (always according to *q*) to ⟨1 + *b*1⟩ is *b*2, and **2** response to ⟨1, *b*1, *b*2⟩ is *b*3. In constructing *p* for **1**, we only aim to beat *q,* and therefore only have to handle the response *b*2 to **1** first move. Therefore, set *b*3 as **1** response to ⟨1 + *b*1, *b*2⟩. In general, for even *n,* denote **2** response to ⟨1 + b1, ..., b*n*−1⟩ by *b**n* and **2** response to ⟨1, *b*1, ..., *b**n*⟩ by *b**n*+1. Then we specify in *p* that **1** response to ⟨1 + *b*1, *b*2, ..., *b**n*⟩ is *b**n*+1. Strategy *q* is presumed to be winning, and game-result *r* in ωω given by ⟨1, *b*1, ...⟩ is one possible sequence allowed by *q,* so *r* must be winning for **2** and *g*(*f*(*r*)) must not be in *U.* The game result *r*' in ωω given by ⟨1 + *b*1, *b*2, ...⟩ is also a sequence allowed by *q* (specifically, *q* playing against *p*), so *g*(*f*(*r*')) must not be in *U.* However, *f*(*r*) and *f*(*r*') differ in all but the first term (by the nature of run-length encoding and an offset of 1), so *f*(*r*) and *f*(*r*') are in complement equivalent classes, so *g*(*f*(*r*)), *g*(*f*(*r*')) cannot both be in *U,* contradicting the assumption that *q* is a winning strategy. Similarly, suppose that *p* is a winning strategy for **1**; the argument is similar but now uses the fact that equivalence classes were defined by allowing an arbitrarily large finite number of terms to differ. Let *a*1 be **1** first move. In general, for even *n,* denote **1** response to ⟨*a*1, 1⟩ (if *n* = 2) or ⟨*a*1, 1, *a*2, ..., *a*n−1⟩ by *a**n* and **1** response to ⟨*a*1, 1 + *a*2, ... *a**n*⟩ by *a**n*+1. Then the game result *r* given by ⟨*a*1, 1, *a*2, *a*3, ...⟩ is allowed by *p* so that *g*(*f*(*r*)) must be in *U*; also the game result *r*' given by ⟨*a*1, 1 + *a*2, *a*3, ...⟩ is also allowed by *p* so that *g*(*f*(*r*')) must be in *U.* However, *f*(*r*) and *f*(*r*') differ in all but the first *a*1 + 1 terms, so they are in complement equivalent classes, therefore *g*(*f*(*r*)) and *g*(*f*(*r*')) cannot both be in *U,* contradicting that *p* is a winning strategy.