Huh? The same argument would conclude that the (vector space of) reals also has dimension c ( 2^aleph-null ) !! WFL 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
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