[proofplan] We apply [Zorn's Lemma](/theorems/1226) to the collection of linearly independent subsets of a vector space $V$ over a field $\mathbb{K}$, ordered by inclusion. The chain condition is verified by showing that the union of a chain of linearly independent sets is linearly independent. A maximal linearly independent set is then shown to span $V$, hence is a basis. [/proofplan] [step:Define the partial order on linearly independent subsets] Let $V$ be a vector space over a field $\mathbb{K}$. Define \begin{align*} \mathcal{I} := \{A \subset V : A \text{ is linearly independent}\}, \end{align*} partially ordered by inclusion $\subset$. The set $\mathcal{I}$ is nonempty: if $V = \{0\}$, then $\varnothing \in \mathcal{I}$ (the empty set is linearly independent by convention, and $\operatorname{span}(\varnothing) = \{0\} = V$, so $\varnothing$ is a basis). If $V \neq \{0\}$, choose any $v \in V \setminus \{0\}$; then $\{v\}$ is linearly independent, so $\{v\} \in \mathcal{I}$. [/step] [step:Verify the chain condition: unions of chains of linearly independent sets are linearly independent] Let $\mathcal{C} = \{A_i\}_{i \in I}$ be a chain in $\mathcal{I}$ (i.e., for any $i, j \in I$, either $A_i \subset A_j$ or $A_j \subset A_i$). Define $A := \bigcup_{i \in I} A_i$. We claim $A \in \mathcal{I}$. Let $v_1, \ldots, v_m \in A$ and suppose $\alpha_1 v_1 + \cdots + \alpha_m v_m = 0$ for scalars $\alpha_1, \ldots, \alpha_m \in \mathbb{K}$. For each $j \in \{1, \ldots, m\}$, there exists $i_j \in I$ with $v_j \in A_{i_j}$. Since $\mathcal{C}$ is a chain and $\{i_1, \ldots, i_m\}$ is a finite subset of $I$, one of the sets $A_{i_1}, \ldots, A_{i_m}$ contains all others. Denote this largest set by $A_{i_*}$. Then $v_1, \ldots, v_m \in A_{i_*}$, and since $A_{i_*}$ is linearly independent, $\alpha_1 = \cdots = \alpha_m = 0$. Therefore $A$ is linearly independent, and $(A, \subset)$ is an upper bound for $\mathcal{C}$ in $\mathcal{I}$. [guided] The key observation is that linear independence is a *finitary* condition: a set is linearly independent if and only if every *finite* subset is linearly independent. When we take a union of a chain, any finite collection of vectors from the union is already contained in a single member of the chain (because finitely many sets from a chain always have a largest). Since that single member is linearly independent, so is the finite collection. Since every finite subset of $A$ is linearly independent, $A$ itself is linearly independent. This finitary character is essential. Not all properties pass to unions of chains. For example, "being a finite set" does not: the union of a chain of finite sets can be infinite. Linear independence, being testable on finite subsets, is preserved. [/guided] [/step] [step:Apply Zorn's Lemma and show the maximal element is a basis] By [Zorn's Lemma](/theorems/1226), $\mathcal{I}$ has a maximal element $\mathcal{B}$. By construction, $\mathcal{B}$ is linearly independent. We claim $\operatorname{span}(\mathcal{B}) = V$. Suppose for contradiction that there exists $w \in V \setminus \operatorname{span}(\mathcal{B})$. We show $\mathcal{B} \cup \{w\}$ is linearly independent, contradicting maximality of $\mathcal{B}$. Suppose $\alpha_0 w + \alpha_1 b_1 + \cdots + \alpha_m b_m = 0$ for distinct $b_1, \ldots, b_m \in \mathcal{B}$ and scalars $\alpha_0, \alpha_1, \ldots, \alpha_m \in \mathbb{K}$. If $\alpha_0 \neq 0$, then \begin{align*} w = -\frac{\alpha_1}{\alpha_0} b_1 - \cdots - \frac{\alpha_m}{\alpha_0} b_m \in \operatorname{span}(\mathcal{B}), \end{align*} contradicting $w \notin \operatorname{span}(\mathcal{B})$. Therefore $\alpha_0 = 0$, and the equation reduces to $\alpha_1 b_1 + \cdots + \alpha_m b_m = 0$. Since $\mathcal{B}$ is linearly independent, $\alpha_1 = \cdots = \alpha_m = 0$. Hence all coefficients are zero, confirming $\mathcal{B} \cup \{w\}$ is linearly independent. But then $\mathcal{B} \cup \{w\} \in \mathcal{I}$ and $\mathcal{B} \subsetneq \mathcal{B} \cup \{w\}$, contradicting the maximality of $\mathcal{B}$. Therefore $\operatorname{span}(\mathcal{B}) = V$, and $\mathcal{B}$ is a basis for $V$. [guided] The maximal linearly independent set $\mathcal{B}$ must span all of $V$. The proof is by contradiction: if some $w \in V$ lies outside $\operatorname{span}(\mathcal{B})$, then $w$ cannot be expressed as a finite linear combination of elements of $\mathcal{B}$. This means adding $w$ to $\mathcal{B}$ preserves linear independence. To see this formally, suppose $\alpha_0 w + \alpha_1 b_1 + \cdots + \alpha_m b_m = 0$. If $\alpha_0 \neq 0$, we can solve for $w$, contradicting $w \notin \operatorname{span}(\mathcal{B})$. So $\alpha_0 = 0$, and the remaining equation $\alpha_1 b_1 + \cdots + \alpha_m b_m = 0$ forces all $\alpha_i = 0$ by linear independence of $\mathcal{B}$. Therefore $\mathcal{B} \cup \{w\}$ would be a strictly larger linearly independent set, contradicting maximality. This means $\operatorname{span}(\mathcal{B}) = V$, so $\mathcal{B}$ is both linearly independent and spanning — i.e., a basis. Note that the Axiom of Choice enters through Zorn's Lemma. Without choice, it is consistent that there exist vector spaces with no basis (in models of ZF where the axiom of choice fails). A classical example is that in Solovay's model, the vector space $\mathbb{R}$ over $\mathbb{Q}$ has no Hamel basis. [/guided] [/step]