TITLE: basis of vector space of real sequences over $\mathbb{R}$
QUESTION [8 upvotes]: It turns out I cannot find a basis for the vector space of all functions from $\mathbb{N}$ to $\mathbb{R}$ (over $\mathbb{R}$).
By Zorn's Lemma, there is a basis. So I guess it cannot be written out constructively?
What is the dimension then? I am thinking $2^{\aleph_0}$. It cannot be countable. If it were, then by a bijection of basis, it will be isomorphic to the set of sequences with finitely non-zero entries. My intuition says it is absurd. How to prove this?

REPLY [2 votes]: Daniel Fischer's comment is probably the simplest route to a full answer, but if we just want to see that the dimension is uncountable then a Baire Category argument is another easy way to do this.  Any countable-dimensional subspace of $\mathbb{R}^\mathbb{N}$ is a union of countably many finite-dimensional subspaces, and every finite-dimensional subspace of $\mathbb{R}^\mathbb{N}$ is closed and nowhere dense in with respect to the product topology.  The space $\mathbb{R}^\mathbb{N}$ is a Baire space, so it is not the union of countably many closed, nowhere dense subsets.