TITLE: Is there a constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$?
QUESTION [28 upvotes]: 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?

REPLY [13 votes]: Assume that every set of real numbers has the Baire property, as well the axiom of dependent choice (it is true if we assume AD, or live in Solovay's model but we can get away with less than large cardinals too):
Since every set of $\mathbb R^\mathbb N$ has the Baire property, and $\mathbb R^\mathbb N$ is a Polish group, every homomorphism of it into itself is continuous [1, Th. 9.10].
Given a Hamel basis has to have cardinality $\frak c$, it defines $2^\frak c$ many endomorphisms of $\mathbb R^\mathbb N$.
Now, given that $\mathbb R^\mathbb N$ is a separable space (by rational sequences which are eventually zero) this means that a continuous function is defined uniquely on the countable dense set, in particular this implies that we can only have $\frak c$ many continuous functions from $\mathbb R^\mathbb N$ to itself.
Contradiction.

Bibliography:

Kechris, A. Classical Descriptive Set Theory. Springer-Verlag, 1994.