Many-Sorted First-Order Model Theory Lecture 10 9 th July, 2020 1 / 48
Ehrenfeucht-Fra¨ ıss´ e games: back-and-forth equivalence 2 / 48
A theorem of Cantor’s Theorem 1 Let A and B be dense linear orders without endpoints. If A and B are both countable, then A ∼ = B . Proof. ◮ We construct an isomorphism f : A → B step by step. Fix some enumerations ( a i : i < ω ) and ( b i : i < ω ) of elements of A and B . ◮ Step 0. Put f ( a 0 ) = b 0 . Then f preserves and reflects < . ◮ Step 1. Consider b 1 . If b 0 < b 1 , pick any a j with a j < a 0 ; if b 1 < b 0 , pick any a k with a 0 < a j . Extend f putting f ( a j ) = ( b 1 ). ◮ Now, assume inductively that f constructed so far preserves and reflects < . ◮ Step i + 1, for even i . Consider b i . If b i = f ( a j ) for some j , we are done. Otherwise, let the elements we have chosen so far from B be b i 0 , . . . , b i k , and wlog we have b i 0 < · · · < b i r < b i < b i r +1 < · · · < b i k . ◮ Then we have f − 1 ( b i r ) < f − 1 ( b i r +1 ). Pick a ℓ with f − 1 ( b i r ) < a ℓ < f − 1 ( b i r +1 ), and extend f putting f ( a ℓ ) = b i . Still f preserves and reflects < . ◮ Step i + 1, for odd i . Proceed analogously with A in place of B . ◮ The map f constructed this way (formally f = � i <ω f i where f i is the partial map constructed at stage i ) is an isomorphism. 3 / 48
A theorem of Cantor’s Theorem 1 Let A and B be dense linear orders without endpoints. If A and B are both countable, then A ∼ = B . Proof. ◮ We construct an isomorphism f : A → B step by step. Fix some enumerations ( a i : i < ω ) and ( b i : i < ω ) of elements of A and B . ◮ Step 0. Put f ( a 0 ) = b 0 . Then f preserves and reflects < . ◮ Step 1. Consider b 1 . If b 0 < b 1 , pick any a j with a j < a 0 ; if b 1 < b 0 , pick any a k with a 0 < a j . Extend f putting f ( a j ) = ( b 1 ). ◮ Now, assume inductively that f constructed so far preserves and reflects < . ◮ Step i + 1, for even i . Consider b i . If b i = f ( a j ) for some j , we are done. Otherwise, let the elements we have chosen so far from B be b i 0 , . . . , b i k , and wlog we have b i 0 < · · · < b i r < b i < b i r +1 < · · · < b i k . ◮ Then we have f − 1 ( b i r ) < f − 1 ( b i r +1 ). Pick a ℓ with f − 1 ( b i r ) < a ℓ < f − 1 ( b i r +1 ), and extend f putting f ( a ℓ ) = b i . Still f preserves and reflects < . ◮ Step i + 1, for odd i . Proceed analogously with A in place of B . ◮ The map f constructed this way (formally f = � i <ω f i where f i is the partial map constructed at stage i ) is an isomorphism. 4 / 48
Intuitions from the proof ◮ The construction in the proof can be viewed as a game between two players, say, Abelard and Helo¨ ıse. ◮ Abelard chooses an element from A or B , and Helo¨ ıse must respond by choosing an element from the other structure while maintaining a partial isomorphism. ◮ Abelard can choose any element whatever, so he resembles ∀ (some writers call him ∀ belard). ◮ Helo¨ ıse must choose carefully, to preserve isomorphism, so she resembles ∃ (some writers call her ∃ lo¨ ıse; initial H is silent anyway). ◮ Helo¨ ıse wins if she can maintain isomorphism throughout the game (in this case forever as the game is infinite). ◮ “Maintaining partial isomorphism” can be cashed out as: sequences chosen so far satisfy precisely the same atomic sentences – in the signature expanded by (names of) the elements of these sequences . 5 / 48
Ehrenfeucht-Fra¨ ıss´ e games Comparing models by games A method conceptually due to Fra¨ ıss´ e (1950), and formulated in game theoretic terms by Ehrenfeucht (1961). Two player game of perfect information ∀ ∃ Spoiler Duplicator Abelard Helo¨ ıse ∀ belard ∃ loise Abelard and Helo¨ ıse as depicted in the 14th centure manuscript Roman de la Rose . 6 / 48
Game EF γ ( A , B ) Definition 2 (Game EF γ ( A , B ) for an ordinal γ ) ◮ Fix similar structures A and B . The game EF γ ( A , B ) is played as follows. ◮ At each round ∀ belard selects some a from A or some b from B . ◮ ∃ loise responds by selecting an element from the other structure. ◮ These choices constitute a play. At the end of the play, sequences a = ( a i : i < γ ) and b = ( b i : i < γ ) have been chosen. ◮ ∃ loise wins the play if � a � A ∼ = � b � B under the map a i �→ b i . ◮ ∃ loise wins the game, if she can win every play, regardless of the moves ∀ belard makes. If this is the case, we say that ∃ loise has a winning strategy. Definition 3 If ∃ has a winning strategy for EF ω ( A , B ), we say that A and B are back-and-forth equivalent, and write A ∼ ω B . 7 / 48
Game EF γ ( A , B ) Definition 2 (Game EF γ ( A , B ) for an ordinal γ ) ◮ Fix similar structures A and B . The game EF γ ( A , B ) is played as follows. ◮ At each round ∀ belard selects some a from A or some b from B . ◮ ∃ loise responds by selecting an element from the other structure. ◮ These choices constitute a play. At the end of the play, sequences a = ( a i : i < γ ) and b = ( b i : i < γ ) have been chosen. ◮ ∃ loise wins the play if � a � A ∼ = � b � B under the map a i �→ b i . ◮ ∃ loise wins the game, if she can win every play, regardless of the moves ∀ belard makes. If this is the case, we say that ∃ loise has a winning strategy. Definition 3 If ∃ has a winning strategy for EF ω ( A , B ), we say that A and B are back-and-forth equivalent, and write A ∼ ω B . 8 / 48
Back-and-forth equivalence Lemma 4 Let ( a , b ) be a play of EF γ ( A , B ) . The following are equivalent: 1. ∃ wins the play ( a , b ) . 2. A | = ϕ ( a ) ⇔ B | = ϕ ( b ) , for every atomic formula ϕ ( x ) . Proof. By Diagram Lemma (Lecture 5, Lemma 12). Theorem 5 (Isomorphism by back-and-forth game) For countable structures A and B , we have A ∼ = B if and only if A ∼ ω B . Proof. ◮ ( ⇒ ) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ ( ⇐ ) If A ∼ ω B , list the elements of A and B , and let ∀ choose alternately the first “fresh” element of A and B . By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play ( a , b ). Thus, � a � A ∼ = � b � B . ◮ Since A and B are countable, all elements are listed in ( a , b ). So, A ∼ = B . 9 / 48
Back-and-forth equivalence Lemma 4 Let ( a , b ) be a play of EF γ ( A , B ) . The following are equivalent: 1. ∃ wins the play ( a , b ) . 2. A | = ϕ ( a ) ⇔ B | = ϕ ( b ) , for every atomic formula ϕ ( x ) . Proof. By Diagram Lemma (Lecture 5, Lemma 12). Theorem 5 (Isomorphism by back-and-forth game) For countable structures A and B , we have A ∼ = B if and only if A ∼ ω B . Proof. ◮ ( ⇒ ) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ ( ⇐ ) If A ∼ ω B , list the elements of A and B , and let ∀ choose alternately the first “fresh” element of A and B . By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play ( a , b ). Thus, � a � A ∼ = � b � B . ◮ Since A and B are countable, all elements are listed in ( a , b ). So, A ∼ = B . 10 / 48
Back-and-forth equivalence Lemma 4 Let ( a , b ) be a play of EF γ ( A , B ) . The following are equivalent: 1. ∃ wins the play ( a , b ) . 2. A | = ϕ ( a ) ⇔ B | = ϕ ( b ) , for every atomic formula ϕ ( x ) . Proof. By Diagram Lemma (Lecture 5, Lemma 12). Theorem 5 (Isomorphism by back-and-forth game) For countable structures A and B , we have A ∼ = B if and only if A ∼ ω B . Proof. ◮ ( ⇒ ) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ ( ⇐ ) If A ∼ ω B , list the elements of A and B , and let ∀ choose alternately the first “fresh” element of A and B . By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play ( a , b ). Thus, � a � A ∼ = � b � B . ◮ Since A and B are countable, all elements are listed in ( a , b ). So, A ∼ = B . 11 / 48
Back-and-forth equivalence Lemma 4 Let ( a , b ) be a play of EF γ ( A , B ) . The following are equivalent: 1. ∃ wins the play ( a , b ) . 2. A | = ϕ ( a ) ⇔ B | = ϕ ( b ) , for every atomic formula ϕ ( x ) . Proof. By Diagram Lemma (Lecture 5, Lemma 12). Theorem 5 (Isomorphism by back-and-forth game) For countable structures A and B , we have A ∼ = B if and only if A ∼ ω B . Proof. ◮ ( ⇒ ) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ ( ⇐ ) If A ∼ ω B , list the elements of A and B , and let ∀ choose alternately the first “fresh” element of A and B . By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play ( a , b ). Thus, � a � A ∼ = � b � B . ◮ Since A and B are countable, all elements are listed in ( a , b ). So, A ∼ = B . 12 / 48
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