# Tagged Questions

**6**

votes

**1**answer

212 views

### Sets of cardinalities of bases without choice

For a vector space $V$, let $BS(V)$ be the set of cardinalities (not necessarily $\aleph$s) of bases of $V$. Of course, in ZFC each $BS(V)$ is a singleton, but supposing the axiom of choice fails, it ...

**17**

votes

**1**answer

252 views

### Linear maps between arbitrarily chosen vectors of vector spaces $V$ and $W$

I recently came across this question:
Is the axiom of choice needed to prove the following statement:
Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. ...

**17**

votes

**1**answer

627 views

### Linear Algebra without Choice

We consider the field of "usual" linear algebra.
Q. Which aspects of it can be carried out without the Axiom of Choice?
Q. Do interesting "exotic" phenomena appear in presence of (some instance of) ...

**13**

votes

**1**answer

667 views

### Sizes of bases of vector spaces without the axiom of choice

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 ...

**22**

votes

**1**answer

1k views

### If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?

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 ...

**11**

votes

**2**answers

748 views

### Indecomposable vector spaces and the axiom of choice

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 ...

**30**

votes

**6**answers

2k views

### Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The ...

**35**

votes

**4**answers

3k views

### Does the fact that this vector space is not isomorphic to its double-dual require choice?

Let $V$ denote the vector space of sequences of real numbers that are eventually 0, and let $W$ denote the vector space of sequences of real numbers. Given $w \in W$ and $v \in V$, we can take their ...

**11**

votes

**2**answers

2k views

### AC in group isomorphism between R and R^2

Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of ...