[proofplan] We use [Zorn's Lemma](/theorems/1226) on the poset of linearly independent subsets of $V$, ordered by inclusion. The main technical work is verifying that every chain has an upper bound: given a chain $\{A_i\}_{i \in I}$, its union $A := \bigcup_i A_i$ is linearly independent because any finite linear dependence involves vectors from finitely many $A_{i_j}$, which are totally ordered and hence contained in a single $A_{i_m}$ — contradicting independence of $A_{i_m}$. A Zorn-maximal linearly independent set must span $V$: if some $v \in V$ were outside its span, adjoining $v$ would give a strictly larger linearly independent set. Hence a Zorn-maximal element is a basis. [/proofplan] [step:Set up the poset of linearly independent subsets] Let \begin{align*} X := \{A \subseteq V : A \text{ is linearly independent over } k\}, \end{align*} ordered by set inclusion $\subseteq$. Here a subset $A \subseteq V$ is **linearly independent** if, for every finite collection of distinct vectors $v_1, \ldots, v_n \in A$ and every choice of scalars $\lambda_1, \ldots, \lambda_n \in k$, \begin{align*} \lambda_1 v_1 + \cdots + \lambda_n v_n = 0 \implies \lambda_1 = \cdots = \lambda_n = 0. \end{align*} The empty set $\varnothing$ is vacuously linearly independent, so $\varnothing \in X$ and in particular $X \neq \varnothing$. [guided] A basis is a maximal linearly independent subset: any extra vector would have to be expressible in terms of the existing ones, and conversely any vector is expressible in such terms. So the existence of a basis is equivalent to the existence of a **maximal element** in the poset of linearly independent subsets. This is exactly the kind of statement Zorn's Lemma is designed to produce. The poset is \begin{align*} X := \{A \subseteq V : A \text{ is linearly independent over } k\}, \quad \text{ordered by } \subseteq. \end{align*} The empty set $\varnothing$ is always in $X$ (no non-trivial finite linear dependence has zero summands), so $X$ is non-empty — an important preliminary for Zorn. To apply Zorn's Lemma we must verify: **every chain in $X$ has an upper bound in $X$.** This is the content of Steps 2 and 3. [/guided] [/step] [step:Identify the union of a chain as the candidate upper bound] Let $\mathcal{C} = \{A_i : i \in I\} \subseteq X$ be a non-empty chain (for every $i, j \in I$, either $A_i \subseteq A_j$ or $A_j \subseteq A_i$). Define \begin{align*} A := \bigcup_{i \in I} A_i \subseteq V. \end{align*} Then $A_i \subseteq A$ for every $i \in I$, so $A$ is an upper bound for $\mathcal{C}$ in $\mathcal{P}(V)$. It remains to show $A \in X$, i.e., $A$ is linearly independent. [/step] [step:Prove the chain union is linearly independent by a finite-dependence argument] Suppose for contradiction that $A$ is linearly **dependent**. Then there exist distinct vectors $v_1, \ldots, v_n \in A$ and scalars $\lambda_1, \ldots, \lambda_n \in k$, not all zero, with \begin{align*} \lambda_1 v_1 + \cdots + \lambda_n v_n = 0. \end{align*} For each $j \in \{1, \ldots, n\}$, since $v_j \in A = \bigcup_{i \in I} A_i$, choose $i_j \in I$ with $v_j \in A_{i_j}$. Consider the finite collection $\{A_{i_1}, \ldots, A_{i_n}\}$. Because $\mathcal{C}$ is a chain (i.e., totally ordered by inclusion), this finite subset has a maximum element: by induction on $n$, any finite totally-ordered subset of a poset has a maximum. Let $A_{i_m}$ be this maximum, so $A_{i_j} \subseteq A_{i_m}$ for all $j = 1, \ldots, n$. Then $v_j \in A_{i_j} \subseteq A_{i_m}$ for every $j$, so \begin{align*} v_1, \ldots, v_n \in A_{i_m}. \end{align*} But now the same relation $\lambda_1 v_1 + \cdots + \lambda_n v_n = 0$ with not all $\lambda_j$ zero is a non-trivial linear dependence in $A_{i_m}$ — contradicting $A_{i_m} \in X$. Hence $A$ is linearly independent, so $A \in X$, and $A$ is an upper bound of $\mathcal{C}$ in $X$. [guided] Why does the union of a chain of linearly independent sets remain linearly independent? The essential ingredient is that **linear dependence is a finite phenomenon**: a set is linearly dependent iff **some finite subset** is linearly dependent. Suppose $A = \bigcup_i A_i$ were dependent. Then there is a non-trivial relation $\lambda_1 v_1 + \cdots + \lambda_n v_n = 0$ involving finitely many vectors $v_1, \ldots, v_n \in A$. Each $v_j$ lives in some $A_{i_j}$ — not necessarily the same index for different $j$'s. Here the **chain property** becomes indispensable. In a chain, any two indices can be compared: $A_{i_j} \subseteq A_{i_k}$ or vice versa. Since there are only $n$ indices $i_1, \ldots, i_n$ (a finite number!), among these finitely many sets there is a largest one, say $A_{i_m}$. Every $v_j$ lies in $A_{i_j} \subseteq A_{i_m}$, so the non-trivial relation holds inside $A_{i_m}$. This contradicts $A_{i_m} \in X$ (i.e., $A_{i_m}$ linearly independent). **Why does this argument fail for general collections that are not chains?** If $\{A_i\}$ were merely a directed family or an arbitrary family, the indices $i_1, \ldots, i_n$ might not be totally ordered: no single $A_{i_m}$ need contain all the $v_j$'s simultaneously. The chain hypothesis is exactly what lets us "consolidate" finitely many sets into one. [/guided] [/step] [step:Apply Zorn's Lemma to obtain a maximal linearly independent set] By the previous three steps, every chain $\mathcal{C} \subseteq X$ has an upper bound in $X$. By [Zorn's Lemma](/theorems/1226), the non-empty poset $(X, \subseteq)$ contains a maximal element, i.e., a linearly independent set \begin{align*} B \in X \end{align*} such that for every $A \in X$ with $B \subseteq A$, we have $A = B$. [/step] [step:Show that a maximal linearly independent set spans $V$] We claim $B$ spans $V$. Let $v \in V$. We split into two cases. **Case 1: $v \in B$.** Then $v \in \operatorname{span}(B)$, since $v = 1 \cdot v$ is a linear combination of elements of $B$. **Case 2: $v \notin B$.** Consider the set $B \cup \{v\} \subseteq V$. We have $B \subsetneq B \cup \{v\}$ (strict inclusion since $v \notin B$), so by maximality of $B$ in $X$, the set $B \cup \{v\}$ is **not** in $X$ — that is, $B \cup \{v\}$ is linearly **dependent**. Hence there exist distinct vectors $w_1, \ldots, w_n \in B \cup \{v\}$ and scalars $\mu_1, \ldots, \mu_n \in k$, not all zero, with \begin{align*} \mu_1 w_1 + \cdots + \mu_n w_n = 0. \end{align*} The vector $v$ must appear among $w_1, \ldots, w_n$: otherwise all $w_j \in B$, giving a non-trivial relation in $B$, which contradicts $B$ being linearly independent. Relabel so $w_1 = v$, and write the relation as \begin{align*} \mu_1 v + \mu_2 w_2 + \cdots + \mu_n w_n = 0, \end{align*} where now $w_2, \ldots, w_n \in B$. Moreover $\mu_1 \neq 0$: if $\mu_1 = 0$, then $\mu_2 w_2 + \cdots + \mu_n w_n = 0$ would be a relation in $B$, forcing $\mu_2 = \cdots = \mu_n = 0$ by linear independence of $B$, contradicting "not all zero". So $\mu_1 \neq 0$ is invertible in the field $k$, and we solve: \begin{align*} v = -\mu_1^{-1}(\mu_2 w_2 + \cdots + \mu_n w_n) = \sum_{j=2}^n (-\mu_1^{-1}\mu_j) w_j \in \operatorname{span}(B). \end{align*} In either case $v \in \operatorname{span}(B)$, so $V = \operatorname{span}(B)$. Combined with linear independence of $B$, this shows $B$ is a basis of $V$. [guided] The argument has two parts: (i) use maximality to conclude that adjoining any new vector produces dependence, and (ii) extract an expression for the new vector in terms of the existing basis. **Part (i): dependence of $B \cup \{v\}$.** Since $B$ is maximal in $X$, any **strict** superset of $B$ in $\mathcal{P}(V)$ cannot be in $X$. So for $v \notin B$, the strict superset $B \cup \{v\}$ fails linear independence: some non-trivial relation $\mu_1 w_1 + \cdots + \mu_n w_n = 0$ holds among finitely many $w_j \in B \cup \{v\}$ with not all $\mu_j$ zero. **Part (ii): solving for $v$.** The relation must "use" $v$ — if it used only vectors from $B$, we would have a dependence in $B$ itself, contradicting $B \in X$. So (after relabelling) $w_1 = v$ and the relation reads \begin{align*} \mu_1 v + \mu_2 w_2 + \cdots + \mu_n w_n = 0 \quad \text{with } w_2, \ldots, w_n \in B. \end{align*} Why is the coefficient $\mu_1$ of $v$ nonzero? If $\mu_1 = 0$, the relation reduces to $\mu_2 w_2 + \cdots + \mu_n w_n = 0$ among elements of $B$. Linear independence of $B$ then forces $\mu_2 = \cdots = \mu_n = 0$, contradicting the assumption that some $\mu_j \neq 0$. So $\mu_1 \neq 0$, and being a nonzero element of a **field** $k$, $\mu_1$ is invertible. Dividing by $-\mu_1$: \begin{align*} v = -\mu_1^{-1}\sum_{j=2}^n \mu_j w_j = \sum_{j=2}^n(-\mu_1^{-1}\mu_j) w_j, \end{align*} expressing $v$ as a linear combination of $w_2, \ldots, w_n \in B$. Hence $v \in \operatorname{span}(B)$. Note that the field hypothesis — that $k$ has multiplicative inverses for all nonzero elements — is essential here. For vector spaces over rings that are not fields (i.e., for modules), a maximal linearly independent subset need **not** span; this is one of the fundamental reasons modules behave worse than vector spaces. Combining with the $v \in B$ case, every element of $V$ lies in $\operatorname{span}(B)$, so $V = \operatorname{span}(B)$. Together with linear independence, $B$ is a basis of $V$. [/guided] [/step]