In this question Sizes of bases of vector spaces without the axiom of choice it is said that "It is consistent [with ZF] that there are vector spaces that have two bases with completely different cardinalities". Now I've had a hard time finding a reference for this fact, so my first question would be whether someone knows a proper sources detailing the construction of a vector space with bases of different cardinalities (in some suitable model of $ZF+\neg AC$).
Secondly, is there a general procedure for generating such spaces? (E.g. by starting with a model $M$ of $ZF+\neg AC$ and a vector space $V$ which has no basis in $M$ and then adding several bases of $V$ to $M$ via iterated forcing.)
And lastly, are there only "artifical/exotic" examples of such vector spaces or are there also models where some "standard" vector spaces such as $\mathbb{F}_2^\omega$ over $\mathbb{F}_2$ or $\mathbb{R}$ over $\mathbb{Q}$ have bases of different cardinalities. (Pretty sure that $\mathbb{R}$ over $\mathbb{Q}$ is still an open problem but maybe there are some other "nice" examples.)


1 Answer 1


This is not a very thoroughly studied problem. So to start from the end, there is no standard procedure for this sort of construction. We know of one, it can maybe be adapted slightly to get a mildly more general result, but it's not something like "let's add a new vector space without a basis" or "let's add an amorphous set" that has a very well-understood and common constructions.

The original construction is due to Läuchli and you can find it in German in his paper

Läuchli, H., Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, 1-18 (1962). ZBL0108.01002.

It also appears as an exercise in Jech "The Axiom of Choice" as Problem 10.5 with an elaborate explanation of the proof.

What you suggest is, in principle, a valid approach. Start with no bases, add one, then add another one with a different cardinality. Unfortunately, we don't really understand the mechanism of adding subsets to models of $\sf ZF$ as well, so it's not clear as to how to do that without:

  1. Making the space well-orderable to begin with; or
  2. collapsing the two bases to have the same cardinality after all; or
  3. giving the space a well-orderable basis; or
  4. one of many other unforeseen problems that can crop up when you add sets to your universe.

Frankly, I do not understand the construction, mainly because Jech's explanation of it is a bit wishywashy to my taste (it is marked with an asterisk, which at least says that it's not very obvious).

  • 1
    $\begingroup$ Should've known you'd be the first one answer, this type of question seem to be right up your alley haha. Thanks for the reference, the construction seems to be indeed very technical, but I think I get the gist. So do I understand correctly that this is currently pretty much the only known way of constucting (rather artificial) examples of vector spaces with bases of different cardinalities? $\endgroup$ Aug 18, 2021 at 15:46
  • $\begingroup$ To my knowledge, yes. $\endgroup$
    – Asaf Karagila
    Aug 18, 2021 at 16:52
  • $\begingroup$ Wow quite surprising we seem to know so little about this type of problem... Thanks a lot! $\endgroup$ Aug 18, 2021 at 16:59
  • $\begingroup$ Not very surprising, I'm afraid... $\endgroup$
    – Asaf Karagila
    Aug 18, 2021 at 17:00

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