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单词 ZornsLemmaAndBasesForVectorSpaces
释义

Zorn’s lemma and bases for vector spaces


In this entry, we illustrate how Zorn’s lemma can be applied in proving the existence of a basis for a vector spaceMathworldPlanetmath. Let V be a vector space over a field k.

Proposition 1.

Every linearly independent subset of V can be extended to a basis for V.

This has already been proved in this entry (http://planetmath.org/EveryVectorSpaceHasABasis). We reprove it here for completion.

Proof.

Let A be a linearly independent subset of V. Let 𝒮 be the collectionMathworldPlanetmath of all linearly independentMathworldPlanetmath supersetsMathworldPlanetmath of A. First, 𝒮 is non-empty since A𝒮. In additionPlanetmathPlanetmath, if A1A2 is a chain of linearly independent supersets of A, then their union is again a linearly independent superset of A (for a proof of this, see here (http://planetmath.org/PropertiesOfLinearIndependence)). So by Zorn’s Lemma, 𝒮 has a maximal elementMathworldPlanetmath B. Let W=span(B). If WV, pick bV-W. If 0=rb+r1b1++rnbn, where biB, then -rb=r1b1++rnbn, so that -rbspan(B)=W. But bW, so b0, which implies r=0. Consequently r1==rn=0 since B is linearly independent. As a result, B{b} is a linearly independent superset of B in 𝒮, contradicting the maximality of B in 𝒮.∎

Proposition 2.

Every spanning set of V has a subset that is a basis for V.

Proof.

Let A be a spanning set of V. Let 𝒮 be the collection of all linearly independent subsets of A. 𝒮 is non-empty as 𝒮. Let A1A2 be a chain of linearly independent subsets of A. Then the union of these sets is again a linearly independent subset of A. Therefore, by Zorn’s lemma, 𝒮 has a maximal element B. In other words, B is a linearly independent subset A. Let W=span(B). Suppose WV. Since A spans V, there is an element bA not in W (for otherwise the span of A must lie in W, which would imply W=V). Then, using the same argument as in the previous propositionPlanetmathPlanetmathPlanetmath, B{b} is linearly independent, which contradicts the maximality of B in 𝒮. Therefore, B spans V and thus a basis for V.∎

Corollary 1.

Every vector space has a basis.

Proof.

Either take to be the linearly independent subset of V and apply proposition 1, or take V to be the spanning subset of V and apply proposition 2. ∎

Remark. The two propositions above can be combined into one: If AC are two subsets of a vector space V such that A is linearly independent and C spans V, then there exists a basis B for V, with ABC. The proof again relies on Zorn’s Lemma and is left to the reader to try.

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