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147(1): 現代数学の系譜 雑談 ◆yH25M02vWFhP [] 2025/02/04(火) 16:34:09.89 ID:+HgMDnV2(6/11) AAS
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
191: 現代数学の系譜 雑談 ◆yH25M02vWFhP [] 2025/02/05(水) 10:50:53.01 ID:hl9U/ln8(1/5) AAS
>>182 補足
・Hilbert spaceの Hilbert dimension は、下記
"As a consequence of Zorn's lemma, every Hilbert space admits an orthonormal basis; furthermore, any two orthonormal bases of the same space have the same cardinality, called the Hilbert dimension of the space.[94]"
(which may be a finite integer, or a countable or uncountable cardinal number).
・”The Hilbert dimension is not greater than the Hamel dimension (the usual dimension of a vector space).”
”As a consequence of Parseval's identity,[95] 略 ”
・なお、>>146-147 "Proof that every vector space has a basis"では、有限和は 陽には使われていない
なので ”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.”
とやっているので、⊆ による順序は Hilbert space でも そのまま使える
あとは、直交基底と 位相的な収束の話を 色付けすれば、よさそうだ
(参考)
https://en.wikipedia.org/wiki/Hilbert_space
Hilbert space
Hilbert dimension
As a consequence of Zorn's lemma, every Hilbert space admits an orthonormal basis; furthermore, any two orthonormal bases of the same space have the same cardinality, called the Hilbert dimension of the space.[94] For instance, since l^2(B) has an orthonormal basis indexed by B, its Hilbert dimension is the cardinality of B (which may be a finite integer, or a countable or uncountable cardinal number).
The Hilbert dimension is not greater than the Hamel dimension (the usual dimension of a vector space).
As a consequence of Parseval's identity,[95] if {ek}k ∈ B is an orthonormal basis of H, then the map Φ : H → l^2(B) defined by Φ(x) = ⟨x, ek⟩k∈B is an isometric isomorphism of Hilbert spaces: it is a bijective linear mapping such that
⟨Φ(x),Φ(y)⟩l^2(B)=⟨x,y⟩H
for all x, y ∈ H. The cardinal number of B is the Hilbert dimension of H. Thus every Hilbert space is isometrically isomorphic to a sequence space l^2(B) for some set B.
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