3 ms·
You're right, any basis of R over Q has to be uncountably infinite. If a countable basis existed, we could for example write R as the union of the subspaces spa
by giraj 6y ago
You're right, any basis of R over Q has to be uncountably infinite. If a countable basis existed, we could for example write R as the union of the subspaces spanned by the n first basis elements, indexed by n. That would mean R is countable, since a countably-indexed union is countable.
- AnHonestComment 6y agoWhat...? Every set {1/n^k | k in N} for n in N is a basis for R over Q, in the sense that any r in R is a Sum[0, inf)[q_i * b_i]. It’s not a finite basis, but it’s definitely countable. It’s not generally the case that a countable union of countable sets is countable — that only holds in the finite case, as we see with power series representations of real numbers.
- a1369209993 6y ago> any basis of R over Q has to be uncountably infinite. It's actually trivial to give a (infinite but) countable set of (non-orthogonal) basis vectors: 2^i for integer i. 0 and 1 are both rationals (scalars), and every real number has a (possibly infinite) binary expansion, eg e = 10.1011011111100001... = 1·2^1 + 0·2^0 + 1·2^-1 + 0·2^-2 + ...
- giraj 6y agoThat's not a basis of R at all. If you want to create a basis of R you better choose irrational elements in your basis, otherwise the Q-span is contained in Q. What I think you want to say is that "any real number has a binary expansion". Which is true, but the binary sequences don't form a vector space over R, but a Z/2-module. And as a Z/2-module, your { 2^i for integer i } isn't even a basis because you need infinite expansions to express most real numbers. The span of a basis are only the finite linear combinations. (FWIW, I think you've given a description of the dyadic rationals.)