I just learned on Wikipedia that the Baer-Specker group, that is, the group of all integer sequences, is not free abelian. I'm hoping I could be helped to understand why this is true by someone knowledgeable here. It interests me to contrast the uncountable but nonfree-abelian Baer-Specker group with the uncountable free abelian group generated by an uncountable set. Thanks. Although this question had been asked before, and got a good answer(but it didn't give mereal intuition for "why", at least not yet), I think this perhaps shouldn't be closed as a duplicate, because if more intuition were possible(maybe it's not) then this could be usefully continued. The reason I mentioned the uncountable free group generated by an uncountable set was because I thought comparisons and relationships with Baer-Specker could give intuition.
Asked
Active
Viewed 147 times
