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現代数学の系譜11 ガロア理論を読む25 [無断転載禁止]©2ch.net

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2016/10/30(日) 14:06:40.31ID:S5Jl1CaY
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2016/12/03(土) 13:51:14.04ID:6Rgz8i9T
>>623
こんなのが
http://math.sta
ckexch
ange.com/questions/176475/what-is-the-standard-proof-that-dimk-mathbb-n-is-uncountable
linear algebra - What is the standard proof that dim(k^N is uncountable? - Mathematics Stack Exchange: asked Jul 29 '12 at 13:46 Chindea Filip

What is the standard proof that dim(kN)is uncountable?
This is my (silly) proof to a claim on top of p. 54 of Rotman's "Homological algebra".



1 Answer answered Jul 29 '12 at 14:29 Asaf Karagila
One liner argument which uses a much more difficult theorems (swatting gnats with cluster bombs kind of proof):
kN is the algebraic dual of the polynomials in one variable, k[x] which has a countable dimension. If kN had a countable basis then k[x] would be isomorphic to its dual, and since this cannot be we conclude that kN has a basis of uncountable size.

The arguments given in Arturo's answer show that the above is indeed a proof (in particular Lemma 2 with κ=?0 ).
2016/12/03(土) 13:52:09.33ID:6Rgz8i9T
>>627 再投稿

http://math.stackexchange.com/questions/176475/what-is-the-standard-proof-that-dimk-mathbb-n-is-uncountable
linear algebra - What is the standard proof that dim(k^N is uncountable? - Mathematics Stack Exchange: asked Jul 29 '12 at 13:46 Chindea Filip

What is the standard proof that dim(kN)is uncountable?
This is my (silly) proof to a claim on top of p. 54 of Rotman's "Homological algebra".



1 Answer answered Jul 29 '12 at 14:29 Asaf Karagila
One liner argument which uses a much more difficult theorems (swatting gnats with cluster bombs kind of proof):
kN is the algebraic dual of the polynomials in one variable, k[x] which has a countable dimension. If kN had a countable basis then k[x] would be isomorphic to its dual, and since this cannot be we conclude that kN has a basis of uncountable size.

The arguments given in Arturo's answer show that the above is indeed a proof (in particular Lemma 2 with κ=?0 ).
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