countable infinite vector space

Question

Good night, can a countable infinite vector space $V$ on the field $\mathbb{C}$ be defined such that it has a finite base?

The spaces that I try to build fail in the property $(\alpha+\beta)v=\alpha v+\beta v$. Thanks.

Answer 1

No, if it's infinite it has at least two different and thereby a non-zero vector $u$. Now there's an injection(*) form the field $\mathbb C$ to $V$ defined as $c\to cu$ so the vector space must have no less cardinality than the field $\mathbb C$ which is uncountable.

Thus for a vector space to be countable it must either be $\left\{\overline0\right\}$ (making it finite) or the over which it's formed must be countable.

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(*) To prove that the map $c\to cu$ is injective you first note that if $cu=du$ then $(c-d)u=\overline0$. If $c\ne d$ then $c-d\ne 0$ so it has an inverse so $$u = 1u = (c-d)^{-1}(c-d)u = (c-d)^{-1}\overline0 = \overline0$$ but this contradicts the assumption that $u\ne\overline0$ so the assumption $c\ne d$ must be false.