every vector space has a hamel basis

Proof

Let $E$ be the set of linearly independent subsets of $V$ . Define a partial order on $E$ , $⪯$ via inclusion:

e, e^{'} \in V, e ⪯ e^{'} ⟺ e \subseteq e^{'}

Towards applying Zorn, let $C$ be a chain in $E$ . Define

c = ⋃_{e \in C} e

This is a linearly independent subset:

Let $v_{1}, \dots, v_{n} \in c$ . Thus there exist $e_{1}, \dots, e_{n} \in C$ where for all $j$ we have $v_{j} \in e_{j}$ .
Since $C$ is a chain, there exists some $J$ such that for all $j = 1, \dots, n$ we have $e_{j} ⪯ e_{J}$ (ie $e_{j} \subseteq e_{J}$ ). Thus $v_{1}, \dots, v_{n} \in e_{J}$ .

Now, since $e_{J}$ is a linearly independent subset, this implies that $v_{1}, \dots, v_{n}$ are themselves linearly independent. Thus $c \in E$

Thus for all $e \in C$ , we have $e ⪯ c$ . ie, there is an upper bound of $C$ .

By Zorn's lemma, $E$ has a maximal element $H$ . I claim that $H$ spans $V$ .

Suppose BWOC that $H$ does not span $V$ . Then there exists some $v \in V$ such that $v$ cannot be written as a finite linear combination of elements in $H$ . Then $H \cup {v}$ is linearly independent. But then $H \subseteq H \cup {v}$ and $H ≺ H \cup {v}$ so $H$ is not maximal $⟹ ⟸$

Thus $H$ spans $V$ .

Thus every vector space has a Hamel basis

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every vector space has a hamel basis

References

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