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Erdős Problem 965

For every 2-coloring of ℝ, is there an uncountable set $A ⊆ ℝ$ such that all sums $a + b$ for $a, b ∈ A, a ≠ b$ have the same colour?

References:

    erdosproblems.com/965

    [Er75b] Erdős, Paul, Problems and results in combinatorial number theory. Journées Arithmétiques de Bordeaux (Conf., Univ. Bordeaux, Bordeaux, 1974) (1975), 295-310.

    [HLS17] Hindman, Neil and Leader, Imre and Strauss, Dona, Pairwise sums in colourings of the reals. Abh. Math. Semin. Univ. Hambg. (2017), 275--287.

    [Ko16] Komjáth, Péter, A certain 2-coloring of the reals. Real Anal. Exchange (2016), 227--231.

    [SWCol] Sokoup Dániel and Weiss, William, Sums and Anti-Ramsey Colourings of ℝ. https://danieltsoukup.github.io/academic/finset_colouring.pdf

@[expose] public sectionnamespace Erdos965

Erdős asks in [Er75b] if for every 2-coloring of ℝ, there is an uncountable set $A ⊆ ℝ$ such that all sums $a + b$ for $a, b ∈ A, a ≠ b$ have the same colour.

In [Ko16] Péter Komjáth constructed a counterexample. The same result was proven independently in [SWCol] by Sokoup and Weiss.

@[category research solved, AMS 3 5, formal_proof using lean4 at "https://github.com/plby/lean-proofs/blob/dfe2d78128b493c572cf525b1b8edf4897fb7664/src/latest/ErdosProblems/Erdos965.lean#L42"] theorem erdos_965 : answer(False) f : Fin 2, A : Set , ¬ A.Countable ∀ᵉ (a A) (b A) (c A) (d A), a b c d f (a + b) = f (c + d) := False (f : Fin 2), A, ¬A.Countable a A, b A, c A, d A, a b c d f (a + b) = f (c + d) All goals completed! 🐙

For every 2-coloring of ℝ, is there some $k ≥ 2$ and an uncountable set $A ⊆ ℝ$ such that all sums $a_1 + \cdots + a_k$ of $k$ distinct elements of $A$ have the same colour?

In fact, in both [Ko16] and [SWCol] a single 2-coloring of ℝ is constructed such that for every $k ≥ 2$ and every uncountable $A ⊆ ℝ$ the sums of $k$ distinct elements of $A$ are not monochromatic.

@[category research solved, AMS 3 5] theorem erdos_965.variants.generalization : answer(False) f : Fin 2, k 2, A : Set , ¬ A.Countable s t : Finset , s A t A s.card = k t.card = k f (s.sum id) = f (t.sum id) := False (f : Fin 2), k 2, A, ¬A.Countable (s t : Finset ), s A t A s.card = k t.card = k f (s.sum id) = f (t.sum id) All goals completed! 🐙end Erdos965