LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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$S \subseteq V$ is linearly independent if and only if every finite subset of $S$ is; consequently the union of a nonempty chain of linearly independent subsets of $V$ , ordered by inclusion, is linearly independent

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field).

Finite character. A subset $S \subseteq V$ is linearly independent (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent) if and only if every finite subset of $S$ (Finite, countably infinite, countable, uncountable) is linearly independent.
Chains. Let $C$ be a nonempty chain (Chain in a poset) in the poset of subsets of $V$ ordered by inclusion (Partial order and partially ordered set), every member of which is a linearly independent subset of $V$ . Then $⋃ C$ is linearly independent.

Facts & Assumptions

Given: A field $F$ and a vector space $V$ over $F$ .

[L1]

A subset $S \subseteq V$ is linearly independent when every injective finite list $v : n \to S$ is linearly independent (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[L2]

Every subset of a linearly independent subset of $V$ is linearly independent; and a linearly independent list $v : n \to V$ is injective with $v [n] \approx n$ (Finite sums re-indexed along an injection, with a zero term deleted, and concatenated; and the closure properties of linear independence: an independent list is injective and never $0_{V}$ , its sublists are independent, a list is independent exactly when it is injective with linearly independent image, and every subset of a linearly independent set is linearly independent, claims 6 and 7).

[L3]

An injective $v : n \to A$ is a bijection onto its image, so $v [n] \approx n$ and $v [n]$ is finite (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Finite, countably infinite, countable, uncountable).

[L4]

Inclusion is a partial order on the subsets of $V$ , and a chain is a subset of a poset any two of whose elements are comparable (Partial order and partially ordered set, Chain in a poset).

[L5]

Induction on $N$ , whose elements are the von Neumann naturals with $σ (n) = n \cup {n}$ (The principle of mathematical induction, The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ).

[L6]

Finite sums of vectors, and hence the vanishing condition defining independence, are computed in $(V, +, 0_{V})$ and do not depend on which subset of $V$ a list is read as landing in (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , Field).

Proof

technique · direct

1.1

Claim 1, from left to right. If $S$ is independent then every subset of $S$ is independent, and in particular every finite subset of $S$ is.

1.2

Claim 1, from right to left. Suppose every finite subset of $S$ is independent and let $v : n \to S$ be an injective finite list. Its image $v [n]$ is a subset of $S$ with $v [n] \approx n$ , hence a finite subset of $S$ , so $v [n]$ is independent by hypothesis; and $v$ , read as a function $n \to v [n]$ , is an injective finite list into $v [n]$ , hence independent. As $v$ was an arbitrary injective finite list into $S$ , the set $S$ is independent.

L1L3L6

1.3

In claim 2, for every $n \in N$ and every list $v : n \to ⋃ C$ there is $A \in C$ with $v [n] \subseteq A$ . By induction on $n$ . At $n = 0$ the image is empty and any member of $C$ will do, $C$ being nonempty; this is the only place the nonemptiness hypothesis is used. Assume the statement at $n$ and let $v : σ (n) \to ⋃ C$ ; the restriction of $v$ to $n$ gives some $A \in C$ with $v [n] \subseteq A$ , and $v_{n}$ lies in some $B \in C$ by the definition of the union. Since $C$ is a chain, any two of its members are comparable under inclusion, so either $A \subseteq B$ or $B \subseteq A$ ; in the first case $B$ contains $v [n] \cup {v_{n}} = v [σ (n)]$ , and in the second case $A$ does.

L4L5

2.1

Claim 2. Let $v : m \to ⋃ C$ be an injective finite list. By step 1.3 there is $A \in C$ with $v [m] \subseteq A$ , so $v$ is an injective finite list into $A$ ; since $A$ is independent, $v$ is independent. As $v$ was arbitrary, $⋃ C$ is independent.

step 1.3L1L6

3.1

Claim 1 is steps 1.1 and 1.2 together, and claim 2 is step 2.1.

step 1.1step 1.2step 2.1∎

Remarks

Where the chain hypothesis is spent. Only in step 1.3, and only through comparability of two members at a time. That is exactly what an arbitrary family of independent sets does not give: a union of two independent sets is in general dependent, as the companion page records as a false statement. A chain is precisely a family for which the finite-character argument goes through.
Finite character is what "finite" is doing here. Independence is by definition a condition on finite lists, so no condition on $S$ can be violated without being violated inside a finite subset. Claim 1 makes that observation formal, and claim 2 is its standard consequence; the same two-step shape proves that any property of finite character satisfies the hypothesis of Zorn's lemma on the poset of sets having it.
Nonemptiness of the chain is not removable from claim 2 as stated. The union of the empty chain is $\emptyset$ , which is independent, so the conclusion happens to survive; what fails is the inductive argument above, which has no member of $C$ to name at $n = 0$ . Claim 2 is what makes Zorn's lemma applicable to a poset of linearly independent subsets, and the empty chain is handled separately where that matters, in Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ , because Zorn's lemma as proved here quantifies over every chain, the empty one included.

Depends on

Used by

Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if L ⊆ S ⊆ V with L independent and span(S) = V, there is a basis B of V with L ⊆ B ⊆ S Theorem

Dependency tree · two levels

46 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Matroid (Wikipedia) (standard reference, not scraped)
Linear independence (Wikipedia) (standard reference, not scraped)
Carnegie Mellon University linear algebra notes: Bases (standard reference, not scraped)

S⊆VS⊆V is linearly independent if and only if every finite subset of SS is; consequently the union of a nonempty chain of linearly independent subsets of VV, ordered by inclusion, is linearly independent