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10 votes
Accepted

Properties of vector spaces without AC

(I completely revamped my answer, the previous version (link) had a consistency result with a particular example, this feels much better as it establishes a full equivalence result instead.) The answ …
Asaf Karagila's user avatar
  • 39.9k
6 votes
Accepted

Linear space with (Hamel) basis and the axiom of choice

No. It is not possible. Suppose that $V$ is a specified vector space, then it is consistent that the axiom of choice fails very far above $V$ in the hierarchy of sets (the von Neumann hierarchy). In …
Asaf Karagila's user avatar
  • 39.9k
10 votes
Accepted

Relation between well-orderings of $\mathbb{R}$, and bases over $\mathbb{Q}$

The problem is generally open. However, recently Liuzhen Wu, Liang Yu, Ralf Schindler and Mariam Beriashvili posted a preprint in which they prove the consistency of the existence of a Hamel basis and …
Asaf Karagila's user avatar
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13 votes
Accepted

Axiom(s) of choice and bases of vector spaces

There is no fully elementary proof that you are looking for. The reason is that the axiom of regularity is needed in these proofs. Multiple Choice does not imply Choice without it, and the only proofs …
Asaf Karagila's user avatar
  • 39.9k
15 votes
Accepted

Examples of vector spaces with bases of different cardinalities

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 …
Asaf Karagila's user avatar
  • 39.9k
5 votes

Sets of cardinalities of bases without choice

Sizes of bases of vector spaces without the axiom of choice shows that assuming $\sf BPI$ we have that every two bases have the same cardinality. This means that $BS(V)$ is either empty, or a singleto …
Asaf Karagila's user avatar
  • 39.9k