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 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 (http://planetmath.org/EveryVectorSpaceHasABasis). 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 (http://planetmath.org/PropertiesOfLinearIndependence)). So by Zorn’s Lemma, $\\mathcal{S}$ has a maximal element $B$ . Let $W=\\operatorname{span}(B)$ . If $W\eq V$ , pick $b\\in V-W$ . If $0=rb+r_{1}b_{1}+\\cdots+r_{n}b_{n}$ , where $b_{i}\\in B$ , then $-rb=r_{1}b_{1}+\\cdots+r_{n}b_{n}$ , so that $-rb\\in\\operatorname{span}(B)=W$ . But $b\otin W$ , so $b\eq 0$ , which implies $r=0$ . Consequently $r_{1}=\\cdots=r_{n}=0$ since $B$ is linearly independent. As a result, $B\\cup\\{b\\}$ is a linearly independent superset of $B$ in $\\mathcal{S}$ , contradicting the maximality of $B$ in $\\mathcal{S}$ .∎

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 $\\mathcal{S}$ 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 $W\eq V$ . 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, $B\\cup\\{b\\}$ is linearly independent, which contradicts the maximality of $B$ in $\\mathcal{S}$ . Therefore, $B$ spans $V$ and thus a basis for $V$ .∎

Corollary 1.

Every vector space has a basis.

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. ∎

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.

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 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 (http://planetmath.org/EveryVectorSpaceHasABasis). 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 (http://planetmath.org/PropertiesOfLinearIndependence)). So by Zorn’s Lemma, $\\mathcal{S}$ has a maximal element $B$ . Let $W=\\operatorname{span}(B)$ . If $W\eq V$ , pick $b\\in V-W$ . If $0=rb+r_{1}b_{1}+\\cdots+r_{n}b_{n}$ , where $b_{i}\\in B$ , then $-rb=r_{1}b_{1}+\\cdots+r_{n}b_{n}$ , so that $-rb\\in\\operatorname{span}(B)=W$ . But $b\otin W$ , so $b\eq 0$ , which implies $r=0$ . Consequently $r_{1}=\\cdots=r_{n}=0$ since $B$ is linearly independent. As a result, $B\\cup\\{b\\}$ is a linearly independent superset of $B$ in $\\mathcal{S}$ , contradicting the maximality of $B$ in $\\mathcal{S}$ .∎ 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 $\\mathcal{S}$ 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 $W\eq V$ . 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, $B\\cup\\{b\\}$ is linearly independent, which contradicts the maximality of $B$ in $\\mathcal{S}$ . Therefore, $B$ spans $V$ and thus a basis for $V$ .∎ Corollary 1. Every vector space has a basis. 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. ∎ 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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