Is this a basis for the dual space?

Question

There is an example on Wikipedia that I don't understand and I'd appreciate some help.

They define $\mathbb R^\infty$ to be the space of all sequences that are zero except for finitely many indexes. The space of all sequences is $\mathbb R^{\mathbb N}$. This is the dual of $\mathbb R^\infty$. What I don't understand is this:

''The dimension of $\mathbb R^\infty$ is countably infinite, whereas $\mathbb R^{\mathbb N}$ does not have a countable basis.''

The set of sequences $e_i$ that are zero expcept at $i$ where $e_i$ equals one obviously is a basis for $\mathbb R^\infty$. I mean, I get that basis means every vector can be written as a finite linear combination and obviously the constant $1$ sequence can only be written as the infinite sum $\sum_i e_i$. But the emphasis is cleary on ''does not have a countable basis''. So it probably have an uncountable basis.

What is an example of a set forming a basis for $\mathbb R^{\mathbb N}$?

The set $e_i$ is of course too small and the whole space is of course not linearly independent. My next guess was to consider something like all sequences consisting of $0$ and $1$ only but that's already too large because it's not linearly independent.

Is the set consisting of $e_i$ together with all sequences of both infinitely many $1$s and infinitely many $0$s a basis?

Answer 1

The set that you suggest is not linearly independent. So it cannot be a basis. But it is indeed a generating set.

To see that it is not linearly independent, consider the sequence $u$ which has $1$ exactly at the odd indexed coordinates and $0$ otherwise. Then $u-e_1$ is a sequence which has infinitely many ones and infinitely many zeros. So it is part of your suggested set.

This means that this is not a basis.

To the general question, as Ittay points out, it is consistent that the axiom of choice fails and $\Bbb{R^N}$ does not have a Hamel basis. Therefore it is impossible to "write down" a formula for representing the basis elements (and prove that it is always a basis for the space) like we can with $\Bbb R^3$ or so.

See this thread for some discussion on the topic: 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 (in fact this is equivalent to the axiom of choice). So in particular $\Bbb{R^N}$ has a basis, and of course it is uncountable (if $V^*$ is the algebraic dual, then for infinite dimensional spaces it is always the case that $\dim_F(V)<\dim_F(V^*)$).

But we can also calculate that $\dim_\Bbb R(\Bbb{R^N})=2^{\aleph_0}$. This can be done by using set theoretic theorems, or by noting that $\Bbb{R^N}$ embeds (as a vector space) the spaces $\ell_p$ -- which we can prove (slightly easier) that their dimension is $2^{\aleph_0}$.

Answer 2

It is impossible to exhibit a concrete basis for the space $\mathbb R ^\mathbb N$. Using the axiom of choice (or a slightly weaker version of it) one can prove that it does have a basis. But the construction is not constructive.