Problems from the Lviv topological seminar

Centerpole sets for colorings of abelian groups

“…One of them happens if the number of colors is 1. In this case The other trivial case happens if the Boolean subgroup G [2] = {x ∈ G : 2x = 0} ⊂ G is unbounded in G. In this case, for each finite coloring χ : G → k there is a color i ∈ k such that the set S = G [2] ∩ χ −1 (i) is unbounded. Since S = −S, we conclude that S is an unbounded monochromatic symmetric subset with respect to 0, which means that the singleton {0} is k-centerpole in G and thus…”

“…By the same reason, there is a Borel 2-coloring φ : X → 2 witnessing that the singleton {b} = {e} is not 2-centerpole for Borel colorings of X. Using the colorings φ and χ 0 one can define a (Borel) 3-coloring χ 2 : X → 3 such that χ 2 …”

“…The equality c 4 (Z 4 ) = 12 from the statement (4) of Theorem 2 answers the problem of the calculation of c 4 (Z 4 ) posed in [1] and then repeated in [5,Problem 2.4], [6,Problem 12], and [2,Question 4.5].…”

“…It remains to calculate the values of the cardinal numbers c k (G) and c B k (G) for k ≥ 2 and an abelian topological group G with totally bounded Boolean subgroup G [2].…”

“…By [4], for any discrete abelian group G ν(G) = ⎧ ⎪ ⎨ ⎪ ⎩ max{|G [2]|, log |G|} if G is uncountable or G [2] is infinite, r Z (G) + 1 i fG is finitely generated, r Z (G) + 2 otherwise.…”