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ガロア第一論文と乗数イデアル他関連資料スレ13

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>>147
> つづき
> 
> Proof that every vector space has a basis
> Let V be any vector space over some field F. Let X be the set of all linearly independent subsets of V.
> 
> The set X is nonempty since the empty set is an independent subset of V, and it is partially ordered by inclusion, which is denoted, as usual, by ⊆.
> 
> Let Y be a subset of X that is totally ordered by ⊆, and let LY be the union of all the elements of Y (which are themselves certain subsets of V).
> 
> Since (Y, ⊆) is totally ordered, every finite subset of LY is a subset of an element of Y, which is a linearly independent subset of V, and hence LY is linearly independent. Thus LY is an element of X. Therefore, LY is an upper bound for Y in (X, ⊆): it is an element of X, that contains every element of Y.
> 
> As X is nonempty, and every totally ordered subset of (X, ⊆) has an upper bound in X, Zorn's lemma asserts that X has a maximal element. In other words, there exists some element Lmax of X satisfying the condition that whenever Lmax ⊆ L for some element L of X, then L = Lmax.
> 
> It remains to prove that Lmax is a basis of V. Since Lmax belongs to X, we already know that Lmax is a linearly independent subset of V.
> 
> If there were some vector w of V that is not in the span of Lmax, then w would not be an element of Lmax either. Let Lw = Lmax ∪ {w}. This set is an element of X, that is, it is a linearly independent subset of V (because w is not in the span of Lmax, and Lmax is independent). As Lmax ⊆ Lw, and Lmax ≠ Lw (because Lw contains the vector w that is not contained in Lmax), this contradicts the maximality of Lmax. Thus this shows that Lmax spans V.
> 
> Hence Lmax is linearly independent and spans V. It is thus a basis of V, and this proves that every vector space has a basis.
> 
> This proof relies on Zorn's lemma, which is equivalent to the axiom of choice. Conversely, it has been proved that if every vector space has a basis, then the axiom of choice is true.[9] Thus the two assertions are equivalent.
> (引用終り)
> 以上

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