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Oct 30, 2021 at 23:41 comment added John Stillwell I think that the theorem first appeared in Fuchs' Abelian Groups of 1958, as Theorem 47.2. Fuchs attributes the theorem to Łoś.
Jan 18, 2020 at 16:53 comment added YCor This is the shadow of the following more precise statement: the free abelian group $\mathbf{Z}^{(I)}$ is non-reflexive iff $I$ carries a $\{0,1\}$-valued probability defined an all subsets and vanishing on singletons. (This means $I\ge\kappa_0$ for $\kappa_0$ the first measurable cardinal, and that $\kappa_0$ exists.) So this is a result where the set $I$ is part of the input.
Jan 18, 2020 at 11:19 comment added Robert Furber Another related fact with a similar proof is that a product of $\kappa$ bornological locally convex spaces is bornological iff $\kappa$ is less than the first measurable cardinal. A locally convex space $E$ is bornological iff for all locally convex spaces $F$ and bounded linear maps $f : E \rightarrow F$, $f$ is continuous. The proof goes by reducing to products of $\mathbb{R}$ (see Schaefer's Topological Vector Spaces, Chapter II, Exercise 19). The statement in II.8 only mentions inaccessible cardinals, I think because Hanf's work was very recent when it was published.
Jan 17, 2020 at 23:14 comment added Todd Eisworth It's the same situation with "All discrete spaces are realcompact", probably for the same reasons!
Jan 17, 2020 at 6:01 comment added Andreas Blass @TimothyChow I'm sure it's in the book "Almost Free Modules" by Eklof and Mekler, but I can't check immediately because I'm away from home. The result is, if I remember correctly, due to Los, and it says in more detail that the free abelian group on a set $S$ of generators is reflexive iff there is no measurable cardinal $\leq|S|$.
Jan 17, 2020 at 3:53 comment added Timothy Chow This is amazing! Where is this proved?
S Jan 17, 2020 at 2:58 history answered Andreas Blass CC BY-SA 4.0
S Jan 17, 2020 at 2:58 history made wiki Post Made Community Wiki by Andreas Blass