Kaba
Posts:
289
Registered:
5/23/11


Re: F^I isomorphic to finite(F^I)
Posted:
Mar 8, 2013 11:27 AM


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. The question then is whether F^I is such or not.
 http://kaba.hilvi.org

