Infinite-dimensional, countable, rational vector spaces.

Question

Does such a thing exist? More explicitly, can I find a vector space $V$ over $\textbf{Q}$, such that $$\dim_\textbf{Q}V=+\infty$$ and $$\operatorname{card}V=\aleph_0$$?

Answer 1

Take $V = \mathbb{Q}^\infty$ which is the set of all rational sequences $(r_1,r_2,r_3,\ldots)$ --- under the operations of coordinate-wise addition and scalar multiplication --- such that the sequence is eventually zero in the sense that $\exists I \ge 0$ such that $r_i = 0$ if $i \ge I$. If you identify $\mathbb{Q}^n$ with the sequences of the form $(r_1,r_2,r_2,...r_{n-1},r_n,0,0,0,...)$, then you have a nested inclusion of vector spaces $$\mathbb{Q}^1 \subset \mathbb{Q}^2 \subset \mathbb{Q}^3 \subset ... $$ whose union is $\mathbb{Q}^\infty$. You can then use a simple enumeration to show that $\mathbb{Q}^\infty$ is countable, or you can also fall back on the set theoretical theorem saying that a countable union of countable sets is countable.

Answer 2

Take $V$ to be the vector space ${\mathbb Q}[X]$ of polynomials over ${\mathbb Q}$. It's the union of the countable collection of polynomials of degree at most $n$, each themselves countable, so is countable.

Answer 3

$$ \bigcup_{n \in \mathbb{N}} \mathbb{Q}[\pi^n]$$ works