## Over a countable field, any two bases of an infinite dimensional vector space have the same cardinality

Let $F$ be a field whose cardinality is countable (that is, finite or countably infinite). Let $V$ be an infinite dimensional vector space over $F$ having basis $B$. prove that $V$ and $B$ have the same cardinality.

Note that $V = \oplus_B F$. Certainly we have $\mathsf{card}\ B \leq \mathsf{card}\ V$, since $B \subseteq V$.

Let $X$ be an infinite set. Given a natural number $n$, denote by $\mathcal{P}_n(X)$ the set of all subsets of $X$ having cardinality $n$. Note that $\mathsf{card}\ X \leq \mathsf{card}\ \mathcal{P}_n(X) \leq \mathsf{card} X^n = \mathsf{card}\ X$, since $X$ is infinite. Denote by $\mathcal{P}_F(X) = \bigcup_\mathbb{N} \mathcal{P}_n(X)$ the set of all finite subsets of $X$; we also have $\mathsf{card}\ X \leq \mathsf{card}\ \mathcal{P}_F(X) = \mathsf{card}\ \bigcup_\mathbb{N} \mathcal{P}_n(X) = \sum_\mathbb{N} \mathsf{card}\ X$ $= \mathsf{card}\ X$. (This sum denotes a cardinal sum.) Thus the set of all finite subsets of $X$ has the same cardinality as $X$.

Now every element of $V$ is uniquely an $F$-linear combination of finitely many elements of $B$. Thus we have $\mathsf{card}\ V \leq \mathsf{card}\ \bigcup_{T \in \mathcal{P}_F(B)} \prod_T F$ $\leq \sum_{\mathsf{card}\ B} \mathsf{card}\ F = \mathsf{card}\ B$.

Thus we have $\mathsf{card}\ V = \mathsf{card}\ B$.