Every vector space has a basis

Let $𝑉$ be a vector space, $𝐼$ be a linearly independent set in $𝑉$ and $𝑆$ be a spanning set in $𝑉$ containing $𝐼$ . Then there exists a basis $B$ for $𝑉$ for which $𝐼 \subseteq B \subseteq 𝑆$ .¹ #m/thm/linalg Hence any linearly independent set belongs to some basis, and every spanning set contains a basis.

Proof

Consider the set $A$ of all linearly independent subsets of $𝑆$ containing $𝐼$ , which is inhabited since $𝐼 \subseteq 𝑆$ . Clearly $A$ forms a complete lattice. If $C = {𝐼 𝑘} 𝑘 \in 𝐾$ is a chain, then the union $𝑈 = ⋃ 𝑘 \in 𝐾 𝐼 𝑘$ is linearly independent and satisfies $𝐼 \subseteq 𝑈 \subseteq 𝑆$ . Hence the hypothesis of Zorn’s lemma is satisfied so assuming choice $A$ has a maximal element $B$ which is linearly independent.

This proof relies on Zorn’s lemma and hence the axiom of choice.

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2008. Advanced Linear Algebra ↩

Every vector space has a basis

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Every vector space has a basis

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