On the strength of Hindman’s Theorem for bounded sums of unions Lorenzo Carlucci Department of Computer Science University of Rome I September 2017 Wormshop 2017 Moscow, Steklov Institute Lorenzo Carlucci (Rome I) Moscow, September 2017 1 / 22
Outline Hindman’s Finite Sums Theorem 1 Bounded Sums 2 3 Weak Yet Strong Principles From Hindman to Ramsey 4 Other variants 5 Lorenzo Carlucci (Rome I) Moscow, September 2017 2 / 22
Hindman’s Finite Sums Theorem Theorem (Hindman, 1972) Whenever the positive integers are colored in finitely many colors there is an infinite set such that all non-empty finite sums of distinct elements drawn from that set have the same color. Original proof is combinatorial but intricate. Later proofs are simpler but use strong methods (ultrafilters or ergodic theory). Question, ’80s What is the strength of Hindman’s Theorem? Lorenzo Carlucci (Rome I) Moscow, September 2017 3 / 22
Measures of Strength HT k = ∀ c : N → k ∃ X ⊆ N ( | X | = ℵ 0 and FS ( X ) is mono ) � �� � � �� � instance solution HT = ∀ k HT k Reverse Mathematics: provability in the systems RCA 0 , WKL 0 , ACA 0 , ACA ′ 0 , ACA + 0 , . . . or (mutual) implications over the base theory RCA 0 . Computable Mathematics: complexity of solutions for computable instances . RM and CM: computable reducibility to/from other principles. Lorenzo Carlucci (Rome I) Moscow, September 2017 4 / 22
Lower Bound on Hindman’s Theorem HT ≥ ∅ ( 1 ) , RT 3 2 , ACA 0 Theorem (Blass, Hirst, Simpson 1987) Some computable (resp. computable in X) 2-coloring of N admits 1 only solutions to HT 2 that compute ∅ ( 1 ) (resp. X ′ – the jump of X). RCA 0 + HT 2 ⊢ ACA 0 . 2 Proof is by coding of the Halting Set and formalizes in RCA 0 . Uses the notion of gap, the interval between two successive exponents of a number in base 2. Lorenzo Carlucci (Rome I) Moscow, September 2017 5 / 22
Upper Bound on Hindman’s Theorem 0 , ∅ ( ω + 1 ) ≥ HT ACA + Theorem (Blass, Hirst, Simpson 1987) Any finite computable (resp. computable in X) coloring of N 1 admits a solution to HT computable in ∅ ( ω + 1 ) (resp. in X ( ω + 1 ) ). ACA + 0 ⊢ HT . 2 ACA + 0 is ACA 0 plus ∀ X ∃ Y ( Y = X ( ω ) ) . Proof is by analyzing the original proof by Hindman. Ultrafilter and ergodic proofs give worse bounds (so far). Lorenzo Carlucci (Rome I) Moscow, September 2017 6 / 22
Bounded Sums Question (Blass, 2005) Does the complexity of HT grow with the length of the sums? Is it the case that longer sums require more jumps? FS ( X ) = sums of finitely many distinct elements of X . FS ≤ n ( X ) = sums of 1 , 2 , . . . , n distinct elements of X . HT ≤ n = the restriction of HT to k colors and sums of length ≤ n . k HT ≤ n k , HT ≤ n Lorenzo Carlucci (Rome I) Moscow, September 2017 7 / 22
Lower Bounds for bounded sums HT ≤ 3 ≥ ∅ ( 1 ) , RT 3 2 , ACA 0 Theorem (Dzhafarov, Jockusch, Solomon, Westrick, 2017) RCA 0 + HT ≤ 3 ⊢ ACA 0 . 1 3 2 , and RCA 0 + RT 1 + HT ≤ 2 RCA 0 � HT ≤ 2 ⊢ SRT 2 2 . 2 2 SRT 2 2 is the Stable Ramsey’s Theorem (WKL 0 � SRT 2 2 ). Proof of (1): modification of Blass-Hirst-Simpson’s argument. Proof of (2): Given a ∆ 0 2 -set A define a coloring all of whose solutions compute an infinite subset of A or an infinite set disjoint from A . Formalization requires RT 1 (eq. B Σ 0 2 ). Lorenzo Carlucci (Rome I) Moscow, September 2017 8 / 22
Upper bounds for bounded sums? Question (Hindman, Leader and Strauss, 2003) Is there a proof that whenever N is finitely coloured there is a sequence x 1 , x 2 , . . . such that all x i and all x i + x j ( i � = j ) have the same colour, that does not also prove the Finite Sums Theorem? Does HT ≤ 2 imply HT over RCA 0 ? Can we upper bound HT ≤ 2 below ACA + 0 ? Are there natural Hindman-type principles with: Non-trivial lower bounds, and 1 Upper bounds strictly below HT? 2 We call such principles Weak Yet Strong. Lorenzo Carlucci (Rome I) Moscow, September 2017 9 / 22
A brute force proof using Ramsey Given c : N → 2, 1. Use RT 1 2 on N wrt c to get an infinite homset H 1 . 2. Use RT 2 2 on H 1 wrt f 2 ( x , y ) := c ( x + y ) to fix the color of sums of length 2 on an infinite H 2 ⊆ H 1 . . . . k. Use RT k 2 on H k − 1 wrt f k ( x 1 , . . . , x k ) := c ( x 1 + · · · + x k ) to fix the color of sums of length k on an infinite H k ⊆ H k − 1 . This induces a coloring d : [ 1 , k ] → 2, where d ( i ) is the c -color of sums of length i from H k . If k is large, then d has some interesting homogeneous set! E.g. if k ≥ 6 then by Schur’s Theorem there exists a , b > 0 such that d ( a ) = d ( b ) = d ( a + b ) . Lorenzo Carlucci (Rome I) Moscow, September 2017 10 / 22
Hindman-Schur Theorem FS A ( X ) : sums of j -many distinct elements of X for any j ∈ A . Hindman-Schur Theorem : Whenever the positive integers are colored in two colors there exist positive integers a , b and an infinite set H such that FS { a , b , a + b } ( H ) is monochromatic. Theorem (C., 2017) Hindman-Schur Theorem is provable in ACA 0 . A host of similar Hindman-type theorems based on different finite combinatorial principles (e.g., Van Der Waerden, Folkman, etc.). All provable in ACA 0 . What about lower bounds? Lorenzo Carlucci (Rome I) Moscow, September 2017 11 / 22
Hindman-Schur with apartness The Blass-Hirst-Simpson’s lower bound proof works, if we impose that the solution set satisfies the following Apartness Condition, for t = 2. Definition ( t -Apartness) Fix a base t ≥ 2 . A set X ⊆ N satisfies the t-apartness condition if x < x ′ ⇒ µ t ( x ) < λ t ( x ′ ) . λ t ( x ) = least exponent in base t representation of n. µ t ( x ) = maximal exponent in base t representation of n. P with t -apartness = P with t -apartness on the solution set. Theorem (C., Kołodziejczyk, Lepore, Zdanowski, 2017) Hindman-Schur with 2-apartness is equivalent to ACA 0 (over RCA 0 ). Lorenzo Carlucci (Rome I) Moscow, September 2017 12 / 22
The Apartness Condition Imposing apartness is a self-strenghtening of Hindman’s Theorem: RCA 0 ⊢ HT ≡ HT with apartness . For restricted versions we have the following: Proposition (C., Kołodziejczyk, Lepore, Zdanowski, 2017) RCA 0 + HT ≤ n 2 k ⊢ HT ≤ n with 3-apartness. k Proof: Give c : N → 2, let d : N → 4: � if n = 3 t + . . . , c ( n ) d ( n ) := if n = 2 · 3 t + . . . . 2 + c ( n ) If FS ≤ 2 ( H ) is monochromatic for d then: all elements have same first coefficient. Then: 1 no two elements of H can have the same first exponent. 2 Lorenzo Carlucci (Rome I) Moscow, September 2017 13 / 22
Bounded Unions Let FUT ≤ n and FUT = n the versions of Hindman’s Theorem in terms of k k unions instead of sums. Proposition For each n , kt ≥ 2 , HT ≤ n with t-apartness is equivalent to FUT ≤ n over k k RCA 0 . Moreover, these principles are mutually strongly computably reducible. The same equivalences hold for HT = n with t-apartness and k FUT = n k . Lorenzo Carlucci (Rome I) Moscow, September 2017 14 / 22
Restricted Hindman and Polarized Ramsey Recall that Dzhafarov et alii proved RCA 0 + HT ≤ 2 + RT 1 ⊢ SRT 2 2 We improve by showing that RCA 0 + HT ≤ 2 ⊢ IPT 2 2 Definition (Dzhafarov and Hirst, 2011) 2 : For all f : [ N ] 2 → 2 there exists a pair of infinite sets ( H 1 , H 2 ) IPT 2 such that all increasing pairs { x 1 , x 2 } with x i ∈ H i get the same f-color. RT 2 2 ≥ IPT 2 2 > SRT 2 2 Lorenzo Carlucci (Rome I) Moscow, September 2017 15 / 22
Restricted Hindman and Polarized Ramsey 2 is strongly computably reducible to HT ≤ 2 In fact we get that IPT 2 4 : any f : [ N ] 2 → 2 of IPT 2 2 computes an instance c : N → 2 of HT ≤ 2 s.t. 4 any solution to HT ≤ 2 for c computes a solution to IPT 2 2 for f . 4 Theorem (C., 2017) RCA 0 + HT ≤ 2 2 ≤ sc HT ≤ 2 ⊢ IPT 2 2 . Moreover, IPT 2 4 . 4 HT = n = restriction of HT k to sums of exactly n elements. k In fact we show: Theorem (C., 2017) RCA 0 + HT = 2 with t-apartness ⊢ IPT 2 2 . Moreover, IPT 2 2 ≤ sc HT = 2 with 2 2 t-apartness. N.B. RT 2 2 proves HT = 2 with t-apartness. 2 Lorenzo Carlucci (Rome I) Moscow, September 2017 16 / 22
IPT 2 2 ≤ sc HT = 2 with 2-apartness 2 Given f : [ N ] 2 → 2, let g : N → 2: � if n = 2 t , 0 g ( n ) := f ( λ ( n ) , µ ( n )) otherwise. Let H = { h 1 < h 2 < h 3 < . . . } be an infinite and 2-apart set such that g is constant on FS = 2 ( H ) . Then λ ( h 1 ) ≤ µ ( h 1 ) < λ ( h 2 ) ≤ µ ( h 2 ) < λ ( h 3 ) ≤ µ ( h 3 ) < . . . So if H 1 := { λ ( h 1 ) , λ ( h 3 ) , λ ( h 5 ) , . . . , } H 2 := { µ ( h 2 ) , µ ( h 4 ) , µ ( h 6 ) , . . . , } Then ( H 1 , H 2 ) is a solution to IPT 2 2 for f . Lorenzo Carlucci (Rome I) Moscow, September 2017 17 / 22
Sums of length 2 and ACA 0 HT ≤ 2 ≥ ∅ ( 1 ) , RT 3 2 , ACA 0 Recall that Dzhafarov et alii proved RCA 0 + HT ≤ 3 ⊢ ACA 0 . Theorem (C., Kołodziejczyk, Lepore, Zdanowski, 2017) RCA 0 + HT ≤ 2 ⊢ ACA 0 . Proposition (C., Kołodziejczyk, Lepore, Zdanowski, 2017) For t ≥ 2 , RCA 0 + HT ≤ 2 with t-apartness ⊢ ACA 0 . 2 Lorenzo Carlucci (Rome I) Moscow, September 2017 18 / 22
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