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147(1): 現代数学の系譜 雑談 ◆yH25M02vWFhP 02/04(火)16:34 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.
省7
191: 現代数学の系譜 雑談 ◆yH25M02vWFhP 02/05(水)10:50 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
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