On Monday 27 August 2012 00:12:08 Fred lunnon wrote:
Huh? The same argument would conclude that the (vector space of) reals also has dimension c ( 2^aleph-null ) !! WFL
How so? Oh, I see: there's a one-word error in what I wrote; the conclusion of the first paragraph should have been "our vector space has CARDINALITY max(|F|,|A|)". Everything else is as it should be. Sorry about that. (So what this lets us conclude about the vector space of reals is that its dimension is at most c. Unless, e.g., we consider it as a vector space over Q, in which case its dimension really is c.)
On 8/26/12, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Sunday 26 August 2012 22:44:30 Andy Latto wrote:
Well the question is whether the answer is simply "aleph-null" or simply "2^aleph-null" (though I strongly suspect it is one or the other). The aleph-null vectors that each have a 1 in one position and zeroes everywhere else do not form a basis; for a set to be a basis, every vector has to be a *finite* linear combination of the basis vectors, and any vector with infinitely many non-zero entries is not in the span of these vectors.
A vector space with basis A over field F has cardinality sum {B a finite subset of A} |F|^|B| (more or less). When, as here, |F| is infinite, subject to AC we have |X| |Y| = max(|X|,|Y|), whence our vector space has dimension max(|F|,|A|).
In particular, if |V| = 2^c and |F| = c then |A| = 2^c.
I'm not sure how much of this works without AC.
-- g