The question about the vector space

As far as I understand, a vector space is a set + two operations (addition and multiplication by a number) + a set of coefficients. The set is closed with respect to operations, operations satisfy axioms, the set of coefficients is a field.

Well, for example, a set of directed line segments on a plane with the traditional operations of addition and multiplication by a number, and the coefficients from $\mathbb{Q}$ are a vector space, isn't it?

If this is a vector space, then the question about the basis arises. After all, an ordinary pair of non-parallel vectors will not be a basis here. It turns out that we got some kind of infinite-dimensional space for nothing?

In general, debunk my delusions, please...

Solution 1:

The vector space $\mathbb{R}^2$ of ordered pairs of real numbers, considered as a vector space over $\mathbb{Q}$, is infinite dimensional. You do not have any misunderstandings or delusions there. It is not difficult to exhibit an infinite linearly independent subset, since in fact we can do that for $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$; so just take the ordered pairs of the form $(\sqrt{p},0)$ with $p$ ranging over all primes to exhibit an infinite $\mathbb{Q}$-linearly independent set of vectors. This suffices to show that this vector space does not have a finite basis.

Whether this vector space has a basis at all, though, depends on your set theory. In standard set theory, the Axiom of Choice guarantees that every vector space has a basis. In fact, the Axiom if Choice is equivalent to the statement that "Every vector space has a basis".

But there are models of set theory without the Axiom of Choice in which $\mathbb{R}$ does not have a basis over $\mathbb{Q}$. In those models, $\mathbb{R}^2$ does not have a basis over $\mathbb{Q}$ either, even though you can exhibit infinitely many linearly independent vectors.

Again, there is nothing to debunk: one can prove in Zermel-Fraenkel Set Theory that $\mathbb{R}^2$ is not finite dimensional over $\mathbb{Q}$.

Solution 2:

Let $\mathbb{K}$ a general field (such as $\mathbb{R}$ or $\mathbb{Q}$). We define vector space $\mathbb{V}$ and we write $(\mathbb{K}, +,\cdot)$ a structure where are defined the following operations:

$+:\, \mathbb{V}\cdot\mathbb{V}\to\mathbb{V}$ that is known as vector addition: $\forall v, w \in \mathbb{V}\,\, v+w$.

$\cdot\,: \mathbb{K}\cdot\mathbb{V}\to \mathbb{V}$, known as scalar multiplication: $\forall v\in \mathbb{V} \forall k\in \mathbb{K}\,\, k\cdot v$.

$\mathbb{Q}$ is a a field, but $\mathbb{N}$ is not. (because it's not close with sum). Thus, $(\mathbb{Q}, +,\cdot)$ is a vector space.