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[parent] every vector space has a basis (Theorem)

This result, trivial in the finite case, is in fact rather surprising when one thinks of infinite dimensionial vector spaces, and the definition of a basis: just try to imagine a basis of the vector space of all continuous mappings $ f\colon\mathbb{R}\to\mathbb{R}$. The theorem is equivalent to the axiom of choice family of axioms and theorems. Here we will only prove that Zorn's lemma implies that every vector space has a basis.

Theorem 1   Let $ X$ be any vector space over any field $ F$ and assume Zorn's lemma. Then if $ L$ is a linearly independent subset of $ X$, there exists a basis of $ X$ containing $ L$. In particular, $ X$ does have a basis at all.
Proof. Let $ \mathcal {A}$ be the set of linearly independent subsets of $ X$ containing $ L$ (in particular, $ \mathcal {A}$ is not empty), then $ \mathcal {A}$ is partially ordered by inclusion. For each chain $ C\subseteq\mathcal {A}$, define $ \widehat {C}=\cup C$. Clearly, $ \widehat {C}$ is an upper bound of $ C$. Next we show that $ \widehat {C}\in\mathcal {A}$. Let $ V:=\{v_1,\ldots,v_n\}\subseteq\widehat {C}$ be a finite collection of vectors. Then there exist sets $ C_1,\ldots, C_n\in C$ such that $ v_i\in C_i$ for all $ 1\leq i\leq n$. Since $ C$ is a chain, there is a number $ k$ with $ 1\leq k\leq n$ such that $ C_k=\bigcup\limits _{i=1}^nC_i$ and thus $ V\subseteq C_k$, that is $ V$ is linearly independent. Therefore, $ \widehat {C}$ is an element of $ \mathcal {A}$.

According to Zorn's lemma $ \mathcal {A}$ has a maximal element, $ M$, which is linearly independent. We show now that $ M$ is a basis. Let $ \< M\> $ be the span of $ M$. Assume there exists an $ x\in X\setminus\< M\> $. Let $ \{x_1,\ldots,x_n\}\subseteq M$ be a finite collection of vectors and $ a_1,\ldots,a_{n+1}\in F$ elements such that

$\displaystyle a_1x_1+\cdots+a_nx_n-a_{n+1}x=0.$    

If $ a_{n+1}$ was necessarily zero, so would be the other $ a_i$, $ 1\leq i\leq n$, making $ \{x\}\cup M$ linearly independent in contradiction to the maximality of $ M$. If $ a_{n+1}\neq 0$, we would have
$\displaystyle x=\frac{a_1}{a_{n+1}}x_1+\cdots+\frac{a_n}{a_{n+1}}x_n,$    

contradicting $ x\notin\< M\> $. Thus such an $ x$ does not exist and $ X=\< M\> $, so $ M$ is a generating set and hence a basis.

Taking $ L=\emptyset$, we see that $ X$ does have a basis at all. $ \qedsymbol$



"every vector space has a basis" is owned by GrafZahl. [ full author list (2) | owner history (1) ]
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See Also: Zorn's lemma, axiom of choice, Zermelo's well-ordering theorem, Hausdorff's maximum principle, Kuratowski's lemma

Other names:  every vector space has a Hamel basis

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properties of bases (Result) by CWoo
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Cross-references: generating set, contradiction, span, maximal element, number, vectors, collection, upper bound, chain, inclusion, subset, linearly independent, field, implies, Zorn's lemma, axioms, axiom of choice, equivalent, continuous mappings, basis, vector spaces, infinite, finite
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This is version 6 of every vector space has a basis, born on 2002-09-30, modified 2005-11-17.
Object id is 3494, canonical name is EveryVectorSpaceHasABasis.
Accessed 11461 times total.

Classification:
AMS MSC: 15A03 (Linear and multilinear algebra; matrix theory :: Vector spaces, linear dependence, rank)
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