Friday, February 3, 2017

linear algebra - Is there a constructive way to exhibit a basis for $mathbb{R}^mathbb{N}$?



Assuming the Axiom of Choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$, the vector space of real sequences?


Answer



"Constructively" "exhibiting" a basis for $\mathbb R^{\mathbb N}$ means "constructively" "exhibiting" a lot of linear functionals on $\mathbb R^{\mathbb N}$; one coordinate functional for each element of the basis. So, in particular, it would mean "constructively" "exhibiting" a linear functional on $\mathbb R^{\mathbb N}$ that is linearly independent of the point-evaluations. Can you "constructively" "exhibit" even one such functional? I think not.


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