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.

--

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