9

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?

Deeponjit Bose
  • 352
mez
  • 10,937

3 Answers3

9

It is consistent that the axiom of choice fails (and so does Zorn's lemma as a consequence) and $\Bbb{R^N}$ does not have a basis. This is a consequence, for example, of all sets of real numbers having the Baire property (something which is false under the axiom of choice, but Solovay and Shelah both gave models where it is in fact true).

Therefore we cannot constructively exhibit such basis, even when assuming the axiom of choice exists.

The dimension of the vector space is $2^{\aleph_0}$, and in order to see this one has to observe two facts:

  1. The cardinality of the space itself is $2^{\aleph_0}$ and therefore a basis cannot have more than $2^{\aleph_0}$ elements to begin with.

  2. Every $\ell_p$ space has an obvious embedding (as a vector space, not as a topological vector space) into $\Bbb{R^N}$. One can show that the dimension of an infinite Banach space cannot be less than $2^{\aleph_0}$, and so we have a subspace which has the maximal possible dimension.

    Or, one can use the argument given here and noting that $\Bbb{R^N}$ is the algebraic dual of $\Bbb R[x]$, the space of polynomials over $\Bbb R$.

The second point can be used to show that it is impossible that $\Bbb{\dim_RR^N}=\aleph_0$.

Community
  • 1
Asaf Karagila
  • 408,303
  • 7
    How about just considering the sequences $(\lambda^k)_{k\in\mathbb{N}}$ for $0 < \lambda < 1$? That exposes $2^{\aleph_0}$ linear independent (Vandermonde) sequences. Seems lower-brow. – Daniel Fischer Nov 18 '13 at 14:24
  • @Daniel: That seems reasonable. I haven't thought about that until now. Kinda useless to swat a fly with your hand when you can just send a MIRV to do the job... :-) – Asaf Karagila Nov 18 '13 at 14:43
3

Another way of showing there is no countable basis is by proving that the space $\mathbb{R}^{\mathbb{N}}$ has a quotient which does not admit a countable basis. This is in fact the hyperreals ${}^\ast\mathbb{R}$. By the saturation property, given any (countable) sequence of nonzero hyperreals, one can find an infinitesimal which is incomparably smaller than all of the hyperreals in the sequence. This shows that there is no countable basis (but of course does not quite give $2^{\aleph_0}$ without CH).

Mikhail Katz
  • 49,567
3

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.

Trevor Wilson
  • 17,236