```Date: Mar 8, 2013 11:27 AM
Author: Kaba
Subject: Re: F^I isomorphic to finite(F^I)

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 non-zero 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 infinite-dimensional 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
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