Is the rational span countable?

Question

Let $\{x_n\}_{n\in\mathbb{N}}$ be a subset of an arbitrary vector space. Is the $span_\mathbb{Q} \{x_n\}_{n\in\mathbb{N}}$ countable? I know that unions over countable index sets are countable and tried to write: $$span_\mathbb{Q} \{x_n\}_{n\in\mathbb{N}} = \left\{ \sum_{i=1}^m \lambda_ix_i | m\in \mathbb{N}, x_i\in\{x_n\}_{n\in\mathbb{N}}, \lambda_i\in \mathbb{Q} \right\}\\ = \bigcup_{n\in\mathbb{N}} \bigcup_{\{x_i\}_{i=1}^m \subset\{x_n\}_{n\in\mathbb{N}}} \bigcup_{\{\lambda_j\}_{j=1}^m \subset \mathbb{Q}} \left\{ \sum_{i=1}^m \lambda_ix_i\right\}.$$ I don't know if the last two unions are formed over a countable index sets. Does may be anyone has an idea or can give me reference for this? Thanks in advance.

Edit: For completeness: We need a theorem which states that all finite sequences with entries form a countable set are countable. This is Theorem V.2.8, P.171 in Kazimierz Kuratowski and Andrzej Mostowski, Set theory: with an introduction to descriptive set theory, 2., compl. rev. ed., Studies in Logicand the Foundations of Mathematics V. 86, North-Holland PublishingCompany, Amsterdam, New York, Oxford, 1976 (eng)

Answer 1

Each of your unions is indeed over a countable index set

• the naturals, naturally
• the finite subsets (in your case, of fixed cardinality $m$) of $\{x_i\}_{i\in \Bbb N}$, or equivalently of $\Bbb N$
• $\Bbb Q^m$

Answer 2

Yes. First, note that $\mathbb{Q}[x]$ is a $\mathbb{Q}$-vector space with a countable basis $\{x^n : n \in \mathbb{N}\}$ (here $0 \in \mathbb{N}$). We also know that $\mathbb{Q}[x]$ is countable (since there are countably many elements of degree $\leq k$ for all $k \in \mathbb{N}$). Now, for any countable subset $\{v_n\}_{n \in \mathbb{N}}$ of a $\mathbb{Q}$-vector space, there is a unique surjective linear transformation $$T : \mathbb{Q}[x] \to \operatorname{span}_{\mathbb{Q}}\{v_n\}_{n \in \mathbb{N}}$$ such that $T(x^n) = v_n$ for all $n \in \mathbb{N}$. This shows that $\lvert \operatorname{span}_{\mathbb{Q}}\{v_n\}_{n \in \mathbb{N}} \rvert \leq \lvert \mathbb{Q}[x] \rvert = \aleph_0$, as desired.