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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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Posts: 289
Registered: 5/23/11
Re: F^I isomorphic to finite(F^I)
Posted: Mar 8, 2013 11:27 AM
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6.3.2013 22:36, David C. Ullrich wrote:
> On Wed, 06 Mar 2013 21:34:23 +0200, Kaba <> wrote:

>> 6.3.2013 1:47, Shmuel (Seymour J.) Metz wrote:
>>> In <kh5tht$csg$>, on 03/06/2013
>>> at 01:03 AM, Kaba <> 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
basis. The question then is whether F^I is such or not.


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