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.

  1. 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?
  2. What is the cardinality of the basis?(I guess that the cardinality is same with $\mathbb{R}$'s.)
  • 1
    $\begingroup$ This should probably answer both your questions. $\endgroup$ – Asaf Karagila Aug 22 '15 at 15:53

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