Carefully state Zorn’s lemma and use it to prove that every vector space has a basis.
Solution
1. Statement of Zorn’s Lemma:
1.1. Zorn’s Lemma: Let \((\mathcal{P}, \le)\) be a non-empty partially ordered set. If every non-empty chain (totally ordered subset) \(\mathcal{C} \subseteq \mathcal{P}\) has an upper bound in \(\mathcal{P}\), then \(\mathcal{P}\) contains at least one maximal element.
Zorn’s lemma is equivalent to the axiom of choice over ZF and is taken as an axiom here.
2. Definition of the poset of linearly independent subsets:
2.1. Let \(V\) be a vector space over a field \(F\). Define:\[ \mathcal{P} = \{S \subseteq V \mid S \text{ is linearly independent over } F\}, \]partially ordered by set inclusion \(\subseteq\).
Inclusion is reflexive, antisymmetric, and transitive on subsets of \(V\).
2.2. \(\mathcal{P} \neq \emptyset\) because the empty set \(\emptyset\) is vacuously linearly independent, so \(\emptyset \in \mathcal{P}\).
The empty set admits no nontrivial linear relation.
3. Verification of the chain condition:
3.1. Let \(\mathcal{C} = \{S_i\}_{i \in I}\) be a non-empty totally ordered chain in \(\mathcal{P}\).Define \(U = \bigcup_{i \in I} S_i \subseteq V\).
This is a definition.
3.2. Show \(U \in \mathcal{P}\) (i.e. \(U\) is linearly independent):Suppose \(\sum_{j=1}^k c_j v_j = 0\) for scalars \(c_j \in F\) and distinct vectors \(v_1, \dots, v_k \in U\).
This fixes an arbitrary finite linear relation among elements of \(U\).
3.3. For each \(j \in \{1, \dots, k\}\), there exists \(i_j \in I\) such that \(v_j \in S_{i_j}\).Since \(\mathcal{C}\) is totally ordered, the finite collection \(\{S_{i_1}, \dots, S_{i_k}\}\) has a maximum element \(S_{i_{\max}} \in \mathcal{C}\) under inclusion.Thus \(\{v_1, \dots, v_k\} \subseteq S_{i_{\max}}\).
A nonempty finite subset of a totally ordered set has a largest element, by induction on its size.
3.4. Since \(S_{i_{\max}} \in \mathcal{P}\), it is linearly independent, which forces \(c_1 = \cdots = c_k = 0\).Thus \(U\) is linearly independent, so \(U \in \mathcal{P}\).
The relation of step 2 is a linear relation among distinct elements of \(S_{i_{\max}}\), which is linearly independent.
3.5. By construction, \(S_i \subseteq U\) for all \(i \in I\), so \(U\) is an upper bound for the chain \(\mathcal{C}\) in \(\mathcal{P}\).
Each \(S_i\) is contained in the union \(U\), and \(U\in\mathcal P\) by step 4.
4. Existence of a basis via maximality:
4.1. By Zorn’s Lemma applied to \(\mathcal{P}\), there exists a maximal element \(B \in \mathcal{P}\).
Step 2 shows that \(\mathcal P\) is a nonempty partially ordered set, and step 3 shows that every nonempty chain in \(\mathcal P\) has an upper bound; Zorn’s lemma applies.
4.2. Since \(B \in \mathcal{P}\), \(B\) is linearly independent.
Elements of \(\mathcal P\) are linearly independent sets by definition.
4.3. Show that \(\operatorname{span}_F(B) = V\):Suppose for contradiction that there exists \(v \in V \setminus \operatorname{span}_F(B)\).Consider \(B' = B \cup \{v\}\).
This is the hypothesis of the argument by contradiction in steps 4 and 5.
4.4. Show that \(B'\) is linearly independent:Suppose \(c v + \sum_{b \in B} c_b b = 0\) for \(c, c_b \in F\).If \(c \neq 0\), then \(v = -\sum_{b \in B} (c_b / c) b \in \operatorname{span}_F(B)\), contradicting \(v \notin \operatorname{span}_F(B)\).Thus \(c = 0\), which implies \(\sum_{b \in B} c_b b = 0 \implies c_b = 0\) for all \(b\) since \(B\) is linearly independent.Thus \(B' \in \mathcal{P}\).
Only finitely many \(c_b\) are nonzero, so the displayed relation is a finite linear relation, and the argument in the claim of this step shows that all its coefficients vanish.
4.5. Since \(v \notin B\), \(B \subsetneq B'\), which contradicts the maximality of \(B\) in \(\mathcal{P}\).Thus no such \(v\) exists, so \(\operatorname{span}_F(B) = V\).
Step 4 gives \(B'\in\mathcal P\) with \(B\subsetneq B'\), which is impossible for a maximal element.
5. Conclusion:\(B\) is linearly independent and spans \(V\), so \(B\) is a basis for \(V\). Q.E.D.
Steps <1>4.<2>2 and <1>4.<2>5 give linear independence and spanning.