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Topic: F^I isomorphic to finite(F^I)
Replies: 11   Last Post: Mar 8, 2013 3:45 PM

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David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: F^I isomorphic to finite(F^I)
Posted: Mar 8, 2013 3:45 PM
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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.
> 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

I didn't say AC was required for every infinite-dimensional 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.

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