If $V$ is a vector space, then $V$ has a Hamel basis.

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

$$e,e^{\prime }\in V,e⪯e^{\prime }⟺e\subseteq e^{\prime }$$

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

$$c=\bigcup _{e\in C}e$$

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 $H\prec H\cup \{v\}$ so $H$ is not maximal $⟹⟸$

Thus $H$ spans $V$.

Thus every vector space has a Hamel basis

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