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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

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    $\begingroup$ 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 $\neg(43\Rightarrow93)$. $\endgroup$
    – joriki
    Dec 30, 2023 at 23:22
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    $\begingroup$ 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.) $\endgroup$
    – joriki
    Dec 30, 2023 at 23:22

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We just don't have that many examples of models where the reals have a Hamel basis but no well-ordering. Mind you, we know of a number of those, but that's a small number indeed.

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.

So a good strategy is to start with a model where a Hamel basis exists and add via forcing (or symmetric extensions) a countable family of countable sets whose union is uncountable. Note that the above means that we are not adding reals, so the countable sets need to already exist in the model, adding the sequence of these sets is not a problem, since the union is uncountable, this additional sequence will not code any new reals.

The natural knee-jerk reaction is to try and use the Cohen model, which famously has a Hamel basis (although that is quite a recent advance), but in the Cohen model the union of a well-orderable family of well-orderable sets is well-orderable, so that will not work. It might be possible to use the Cohen model as our ground and add the aforementioned sequence.

Of course, another alternative is to simply prove directly that some model where there is a countable family of countable sets of reals whose union is uncountable does have a Hamel basis.

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