$\underline{Thm}$ (By the axiom of choice)
Let $S$ be a linear independent subset of a vector space $V$.
Then there exists a maximal linear independent subset $\beta$ of $V$ with $S \subset \beta$.
$\underline{Proof}$
Let $\mathcal{F}$ be the set of all linear independent subsets of $V$ containing $S$.
Let $\mathcal{C} \subset \mathcal{F}$ be any chain.
Define $\displaystyle L_{\mathcal{C}}$ = $\underset{A \in \mathcal{C}}{\cup A}$. Then $A \subset L_{\mathcal{C}} \; \forall A \in \mathcal{C}$.
We check $L_{\mathcal{C}} \in \mathcal{F}$. Since $S \subset A\;\forall A \in \mathcal{C}$, we have $S \subset L_{\mathcal{C}}$.
It remains to check that $L_{\mathcal{C}}$ is linear independent.
Suppose $u_{1},\cdots,u_{n} \in L_{\mathcal{C}},\; a_{1},\cdots,a_{n} \in \mathcal{F}$. and $a_{1}u_{1}+\cdots+a_{n}u_{n}=0$.
Then $\forall i \in \{1,\cdots,n \}$, $u_{i} \in A_{i}$ for some $A_{i} \in \mathcal{C}$.
Since $\mathcal{C}$ is a chain in $\mathcal{F}$, $\exists k \in \{ 1, \cdots, n \}$ such that $u_{i} \in A_{i} \subset A_{k} \; \forall i \in \{ 1, \cdots, n \}$.
Since $A_{k}$ is linear independent, we have $a_{1}=\cdots=a_{n}=0$.
Thus, $L_{\mathcal{C}}$ is linear independent, so that $L_{\mathcal{C}}=\mathcal{F}$.
By Zorn's lemma, $\mathcal{F}$ has a maximal set, say $\beta$. This $\beta$ is a desired set.
$\underline{Coro}$
Every vector space has a basis.
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