
Re: F^I isomorphic to finite(F^I)
Posted:
Mar 8, 2013 3:45 PM


On Fri, 08 Mar 2013 18:27:38 +0200, Kaba <kaba@nowhere.com> wrote:
>6.3.2013 22:36, David C. Ullrich wrote: >> On Wed, 06 Mar 2013 21:34:23 +0200, Kaba <kaba@nowhere.com> wrote: >> >>> 6.3.2013 1:47, Shmuel (Seymour J.) Metz wrote: >>>> In <kh5tht$csg$1@news.cc.tut.fi>, on 03/06/2013 >>>> at 01:03 AM, Kaba <kaba@nowhere.com> said: >>>> >>>>> 2) What could be a basis for F^I? >>>> >>>> Google for "Hamel Basis". >>> >>> Sure, a Hamel basis, but is it possible to give some intuitive >>> construction for the Hamel basis of F^I?:) >> >> No. If I is infinite there _is_ no "construction" of a basis, >> a basis exists by the Axiom of Choice (Zorn's Lemma >> gives a maximal independent set). > >I'm not sure whether this is a good argument against there existing a >"construction" for a specific case. Consider the following example: > >Let I be a set, and F be a field. Let > > B = {b_i : I > F}_{i in I}. > >be such that > > b_i(x) = 1, if x = i > 0, otherwise. > >and > > U = {sum_{i in I} alpha_i b_i : alpha_i in finite(F^I)}, > >where finite(.) again denotes only those functions with finite number of >nonzero positions. Then U is a vector space over F, with arbitrary >dimension I, and whose basis B we can construct, without appealing to >the Axiom of Choice. (I think there's a name for this construction, >perhaps a free vector space over B?) Thus, not every >infinitedimensional vector space requires the Axiom of Choice to have a >basis.
I didn't say AC was required for every infinitedimensional vector space! You say you don't think I gave a good argument? I didn't give any argument at all in the case of F^I, good or otherwise.
>The question then is whether F^I is such or not.

