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I have recently noticed$^1$ that in models of ZF+DC+"All sets of real numbers have the Baire property" (e.g. Solovay's model or Shelah's model mentioned by Andreas) there is a very interesting propert …

Some things about vector spaces which are consistent with the failure of choice: Vector spaces may have bases of different cardinality. In particular, this means that the notion of "dimension" is no …

It is consistent with the axioms of $\sf ZF$ that this is impossible. Specifically, if you consider $\Bbb R[x]$, then its dual space is just $\Bbb{R^N}$. And it is consistent with $\sf ZF$ that $\Bbb{ …

Suppose $V$ is a vector space, we say that $\mathcal B$ is a basis for $V$ if: Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$; If $\sum\alpha_i b_i = 0$, where $\a …

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 …

It is a known result by A. Blass that the axiom of choice is equivalent to the assertion that every vector space has a basis. (Rubin's Equivalents of the Axiom of Choice: form B) It is also known tha …

Yes, the axiom of choice is needed. Läuchli has constructed a vector space whose only endomorphisms are scalar multiplication. In such vector space, if $v\neq \alpha w$ there is no such $T$. The pape …

Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two …

One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem: If $V$ is an infinite vector space over a field $F$, and …

It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$". If $\cal P$ holds, then the projection onto $k\ …

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 …

To add the proof for my claim in Todd's answer, which essentially repeats Läuchli's original [1] arguments with minor modifications (and the addition that the resulted model satisfies $DC_\kappa$). W …

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 …