I know that all vector space has a basis. My question is "concrete" example for basis for $\mathbb{R}$-vector space $\mathbb{R}^{\mathbb{N}}:=\left\{f:\mathbb{N}\to\mathbb{R}\right\}$. If I refer cofinite condition, my question is so easy. But, $\mathbb{R}^{\mathbb{N}}$ is not.
- Can we find a concrete basis for $\mathbb{R}^{\mathbb{N}}$, or its existence is just ensured by AC? In short words, the basis is constitutive or not?
- What is the cardinality of the basis?(I guess that the cardinality is same with $\mathbb{R}$'s.)
