Well, you first need to know that the vector space has a basis in the first place. I mean, it does, if you assume the Axiom of Choice -- this is probably the first Zorn's Lemma argument that most math majors ever see. (E.g. here: http://soffer801.<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> wordpress.com<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> /2011/09/24/every-vector-space-has-a-basis/<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> ) I can certainly see that there's no countable basis. Given any countable set of vectors, here's a way to construct a vector that can't be written as finite sum of them. Pick any value for the first coordinate. For each vector in your putative basis, there's at most one scalar multiple of it that matches the vector we're constructing -- which means there are at most countably many values for the second coordinate that wouldn't require a second basis vector be added to the linear combination. Keep going this way: there are always uncountably many values for the nth coordinate that force the linear combo to need at least n vectors. The vector you get at the end can't be formed by any finite linear combo at all. If you're willing to assume the Continuum Hypothesis, then we're done: there are only c many points, some subset is a basis, and no countable subset works. So c it is. But without CH, I don't know what else to say. On Sun, Aug 26, 2012 at 5:44 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Sun, Aug 26, 2012 at 4:58 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Er --- why should the answer be not simply "oo" ( |N or aleph-null ) ?
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.
Andy
Admittedly, just how this might be justified in terms of a formal
definition
of "dimension" for a vector space is not something I have thought about.
WFL
On 8/26/12, Dan Asimov <dasimov@earthlink.net> wrote:
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
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