# Hamel bases and countable unions

Does the existence of a Hamel basis for $$\mathbb{R}$$ over the rationals (called "FORM 367." in Howard & Rubin) imply that the union of a denumerable family of denumerable subsets of $$\mathbb{R}$$ is denumerable ("FORM 6.")?

Howard & Rubin present many models that satisfy 367. and falsify the more general

FORM 31. $$UT(\aleph_0, \aleph_0, \aleph_0)$$: The countable union theorem: The union of a denumerable set of denumerable sets is denumerable.

but, being permutation models, all of them also satisfy 6.

Herrlich also has some models satisfying 31. and falsifying 367., but all of them satisfy 6., so no help in this case

• An implication table for Howard & Rubin is here. The codes are explained here. The entry for $367\Rightarrow6$ is $0$: "The status of the implication is unknown." As the Herrlich models show, $\neg(6\Rightarrow367)$; this can be traced to $43\Rightarrow8\Rightarrow94\Rightarrow6$, $367\Rightarrow366\Rightarrow93$ and $\neg(43\Rightarrow93)$. Dec 30, 2023 at 23:22
• The reference for $\neg(43\Rightarrow93)$ is on p. $341$ of Howard & Rubin (which, by the way, can be borrowed here). (I posted these as comments because they don't answer your question; I just thought they might be of interest.) Dec 30, 2023 at 23:22

One way to solve this is to note that if $$M$$ is model of $$\sf ZF$$ in which there is a Hamel basis, and $$M\subseteq N$$ is such that $$\Bbb R^M=\Bbb R^N$$, then $$N$$ has a Hamel basis.