Cardinality of cartesian product of infinite set with countable set Suppose $M$ is an infinite set. I can prove, using Zorn's lemma, that $M$ is a disjoint union of countably infinite sets. Does it follow from there that there is a bijection between $M$ and $M\times A$ if $A$ is a countable set? And how would this fact prove the fact that the cardinality of bases for $\mathbb{R}^n$ and $\mathbb{R}^m$ as vector spaces over $\mathbb{Q}$ is equal?
 A: Hint/Proof Sketch: by the result you cite, take $M = \sqcup_{\alpha \in \Lambda}M_\alpha$ with each $M_\alpha$ countable. I will write $A \simeq B$ when $A$ and $B$ have the same cardinality. Then, 
$$
M \times A = \bigsqcup _{\alpha \in \Lambda}M_\alpha \times A = \bigsqcup _{\alpha \in \Lambda}(M_\alpha \times A) 
$$
Now, $M_\alpha \times A \simeq \mathbb{N} \times \mathbb{N} \simeq \mathbb{N} \simeq M_{\alpha}$ and so by giving bijection $M_\alpha \times A \stackrel{\simeq}{\to} M_\alpha$ for each $\alpha$, you get that 
$$
M \times A = \bigsqcup _{\alpha \in \Lambda}(M_\alpha \times A) \simeq \bigsqcup _{\alpha \in \Lambda}M_\alpha = M
$$
as claimed. As for your second question, if you have a basis $\mathcal{B}$ for $\mathbb{R}$ as a $\mathbb{k}$ vector space, then
$$
\mathcal{B}_n = \{x \in \mathbb{R}^n : (\exists i) \text{s.t $x_i \in \mathcal{B}$ and all other entries are zero} \}
$$
should give a basis for $\mathbb{R}^n$. Think of $\mathcal{B} = \{1\}$ when $\mathbb{k} = \mathbb{R}$. Now, 
$$
\mathcal{B}_n = \bigsqcup_{1 \leq i \leq n}\{x \in \mathcal{B}_n : x_i \in \mathcal{B}\} \simeq \mathcal{B}^n.
$$
