A solution to van Douwen's problem on Bohr topologies (Q5946829)

This paper provides a negative answer to the following question posed by \textit{E. K. van Douwen} [Topology Appl. 34, 69-91 (1990; Zbl 0696.22003)]: if \(G\) and \(H\) are discrete Abelian groups of the same cardinality, does it follow that the Bohr topologies associated to these groups are homeomorphic? We recall here that the Bohr topology associated to an Abelian topological group \(G\) is the weak topology induced by the group of characters of \(G\). NEWLINENEWLINENEWLINEThe counterexample presented in this paper consists of a group \(G_2\) of exponent 2 and a group \(G_3\) of exponent 3, both of cardinality \(\kappa\) with \(\kappa >2^{2^{\mathfrak c}}\). If \(G^{\#}\) denotes the abstract group \(G\) equipped with its Bohr topology, it is proved that every continuous map \(G_2^{\#}\rightarrow G_3^{\#}\) is constant on some infinite subset of \(G_2\) and thus that no homeomorphism can be constructed between \(G_2\) and \(G_3\). The flavour of the proof is largely combinatorial. NEWLINENEWLINENEWLINEThe result in this paper is to be compared with an analogous one based on different ideas that was given independently around the same time by \textit{K. Kunen} [Topology Appl. 90, 91-107 (1998; Zbl 0974.54025)]. Here the counterexample consisted of two countable groups with different prime exponents.