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?

We know that every vector space has a Hamel basis and also every normed space need not have a Schauder basis. As the normed space $l^\infty$ is not Separable so can't have the Schauder basis, but on the other side $l^\infty$ is also a vector space so what will be the Hamel basis of $l^\infty$?

I am looking for an explicit example for a Hamel basis for $l^{\,p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one can express any element of the vector space as a finite linear combination of these. After some t...

No. There are models of $\mathsf{ZF+\lnot AC}$ in which every linear transformation from a Banach space to a normed space is automatically continuous. In particular this is true for linear functionals. In such models, it follows, every linear functional has to be continuous. An example for thes...

Okay, here is something that I feel is worth posting. We say that a Banach space is a dream space if every linear operator into a normed space is continuous, in particular every linear functional of a dream space is continuous. This means that if we quotient a dream space by a dense subspace th...