Nicholas Funari Voltani

Every vector space which has a linearly independent set has a Hamel Basis containing this set

2 min read

up:: Hamel Basis

Let be a Vector Space, and let be a Linearly Independent set. We seek to prove that there is a Hamel basis which will be the “maximal” linearly independent subset of , via Zorn’s Lemma.

Consider the class of linearly independent subsets of , . It is not empty by hypothesis, since .

Zorn’s Lemma’s condition

We need to prove that, for all Totally Ordered subsets of , they have an upper bound.

Thus, let ─ which means it is a collection of linearly independent sets which all are comparable via the partial ordering . A candidate for its upper bound is

A_{\alpha_1} \subseteq \dots \subseteq A_{\alpha_n}

You can't use 'macro parameter character #' in math modein such a way that $\{v_i\}_{i=1}^n \subset A_{\alpha_n}$. Since [[A set is linearly independent iff all its finite subsets are also L.I.]] and $A_{\alpha_n}$ is L.I. per hypothesis, then $\{v_i\}_{i=1}^n$ is also L.I.. Thus, $\mathcal{A}$ is an element of $\mathcal{P}$ and is an upper bound of $\{A_{\alpha_i}\}_{\alpha \in \Lambda}$. # Conclusion By [[Zorn's Lemma]], $\mathcal{P}$ has a maximal element ─ call it $\mathcal{B}$. It is a linearly independent set, and it also spans $V$ ─ since, if it didn't, there'd be some vector $v \in V$ outside of its [[Spanning Set|Span]]. Since [[The union of a linearly independent set with a vector outside of its span is also linearly independent]], we'd have some element $\mathcal{B} \cup \{v\} \in \mathcal{P}$ such that $\mathcal{B} \subseteq \mathcal{B} \cup \{v\}$, which is a contradiction to $\mathcal{B}$ being a maximal element of $\mathcal{P}$. Thus, all vector spaces (finitely generated or not) which have some linearly independent set, will also have a [[Hamel Basis]]. --- ### References - **Um Curso de Álgebra Linear**, Flávio Ulhoa Coelho & Mary Lilian Lourenço. Editora EDUSP.

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