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I was studying linear algebra when a question came to my mind.

Let's say that a set is countable if it is equipotent to the set of natural numbers. This said, let's consider a vector space over a countable field and having countable dimension. It is certainly infinite, but is it countable?

For a concrete example, one can take as field the one of rational numbers.

Amanda Wealth
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  • Yes; think about fixing a countable basis and looking at the possible linear combinations of elements of that basis. Crucially, each basis element can only be used finitely many times, so some vector spaces you might think are countably-dimensional are in fact continuum-dimensional (e.g. "$\mathbb{Q}^\infty$" as a $\mathbb{Q}$-vector space, if thought of in the most obvious way). That said, I strongly suspect that this question is a duplicate, hence my comment rather than answer (and I'll delete this comment if/when a dupe is found). – Noah Schweber Mar 26 '25 at 20:15
  • @NoahSchweber I'm sorry I'm not able to get your point. It's clearly possible to embed the finite subsets of the set of natural numbers into the vector space (using appropriate linear combinations), but I miss a proof of the general case, whatever the answer is. – Amanda Wealth Mar 26 '25 at 20:20
  • If you accept that the countable union of countable sets is countable, then you can argue as follows: If ${v_n}{n \in \mathbb{N}}$ is a basis of the $K$-vector space $V$, then $V = \bigcup{n\in \mathbb{N}} \langle v_1,\dots,v_n\rangle_K$. – Leobeth Mar 26 '25 at 21:14
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    See https://math.stackexchange.com/questions/3282420/prove-that-the-cardinality-of-a-vector-space-equals-to-the-cardinality-of-its-ba – Jens Schwaiger Mar 26 '25 at 21:17
  • @NoahSchweber, there may be an interesting issue here if one does not assume the Axiom of Choice, and assumes that the vector space does possess a basis. Neither the answer here nor the answer at the linked question address this. – Mikhail Katz Mar 27 '25 at 10:39

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Let $V$ be a $F$-vector space with countable basis $(x_n)_{n\in \mathbb N}$ and $F$ a countable field. Define $A_n := \operatorname{lin}_F\{x_1, \dots, x_n\}$ the linear subspace generated by the first $n$ vectors of the basis. Clearly it is countable, as a finite product of countable sets (~a finite dimensional $F$-vector space). Hence the union $A:=\bigcup_{n\in \mathbb N} A_n$ is countable as a countable union of countable sets. Since any vector $v\in V$ is a finite linear combination of some $x_i$, it is contained in some $A_n$. Thus $A=V$, a countable set.

Here you can find reasoning why countable unions of countably many sets is still countable. It simply is the pattern:

$A_1, A_2, A_3, A_4, \dots$

$1$

$2, 1$

$3, 2, 1$

$4, 3, 2, 1$

and so on.

Tina
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  • I’m not sure this fully covers the scenarios contemplated by the OP’s question. What about the vector space of all rational-valued infinite sequences? Surely it has continuum cardinality? (since the set of all functions $f : \mathbb{N} \mapsto {0, 1}$ has continuum cardinality) – NikS Mar 27 '25 at 12:12
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    @NikS That's not countable dimension. – Noah Schweber Mar 27 '25 at 14:40