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[parent] Zorn's lemma and bases for vector spaces (Result)

In this entry, we illustrate how Zorn's lemma can be applied in proving the existence of a basis for a vector space. 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. We reprove it here for completion.

Proof. Let $ A$ be a linearly independent subset of $ V$. Let $ \mathcal{S}$ be the collection of all linearly independent supersets of $ A$. First, $ \mathcal{S}$ is non-empty since $ A\in \mathcal{S}$. In addition, if $ A_1\subseteq A_2 \subseteq \cdots$ 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). So by Zorn's Lemma, $ \mathcal{S}$ has a maximal element $ B$. Let $ W=\operatorname{span}(B)$. If $ W\ne V$, pick $ b\in V-W$. If $ 0=rb+r_1b_1+\cdots +r_nb_n$, where $ b_i\in B$, then $ -rb=r_1b_1+\cdots +r_nb_n$, so that $ -rb\in \operatorname{span}(B)=W$. But $ b\notin W$, so $ b\ne 0$, which implies $ r=0$. Consequently $ r_1=\cdots = r_n=0$ since $ B$ is linearly independent. As a result, $ B\cup \lbrace b\rbrace$ is a linearly independent superset of $ B$ in $ \mathcal{S}$, contradicting the maximality of $ B$ in $ \mathcal{S}$. $ \qedsymbol$
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 $ \mathcal{S}$ be the collection of all linearly independent subsets of $ A$. $ \mathcal{S}$ is non-empty as $ \varnothing\in \mathcal{S}$. Let $ A_1\subseteq A_2\subseteq \cdots$ 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, $ \mathcal{S}$ has a maximal element $ B$. In other words, $ B$ is a linearly independent subset $ A$. Let $ W=\operatorname{span}(B)$. Suppose $ W\ne V$. Since $ A$ spans $ V$, there is an element $ b\in A$ 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 proposition, $ B\cup\lbrace b\rbrace$ is linearly independent, which contradicts the maximality of $ B$ in $ \mathcal{S}$. Therefore, $ B$ spans $ V$ and thus a basis for $ V$. $ \qedsymbol$
Proof. Either take $ \varnothing$ 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. $ \qedsymbol$

Remark. The two propositions above can be combined into one: If $ A\subseteq C$ 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 $ A\subseteq B\subseteq C$. The proof again relies on Zorn's Lemma and is left to the reader to try.



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Cross-references: spanning, every vector space has a basis, proposition, argument, spans, spanning set, implies, maximal element, proof, union, chain, addition, supersets, collection, completion, subset, linearly independent, field, vector space, basis, Zorn's lemma
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This is version 6 of Zorn's lemma and bases for vector spaces, born on 2008-06-05, modified 2008-08-21.
Object id is 10659, canonical name is ZornsLemmaAndBasesForVectorSpaces.
Accessed 510 times total.

Classification:
AMS MSC: 15A03 (Linear and multilinear algebra; matrix theory :: Vector spaces, linear dependence, rank)
 13C05 (Commutative rings and algebras :: Theory of modules and ideals :: Structure, classification theorems)
 16D40 (Associative rings and algebras :: Modules, bimodules and ideals :: Free, projective, and flat modules and ideals)
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