LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The span is monotone and idempotent, $\operatorname{span}(S) = S$ exactly when $S$ is a linear subspace, and $\operatorname{span}(S \cup \{0_V\}) = \operatorname{span}(S)$

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field) and let $S, T \subseteq V$ . Then:

Extensive. $S \subseteq \operatorname{span}(S)$ .
Monotone. If $S \subseteq T$ then $\operatorname{span}(S) \subseteq \operatorname{span}(T)$ .
Idempotent. $\operatorname{span}(\operatorname{span}(S)) = \operatorname{span}(S)$ .
$\operatorname{span}(S) = S$ if and only if $S$ is a linear subspace of $V$ (Linear subspace of a vector space).
$\operatorname{span}(S \cup \{0_V\}) = \operatorname{span}(S)$ .

Facts & Assumptions

Given: A field $F$ , a vector space $V$ over $F$ , and subsets $S, T \subseteq V$ .

[L1]

$\operatorname{span}(S)$ is a linear subspace of $V$ , it contains $S$ , and it is contained in every linear subspace of $V$ that contains $S$ ; the same holds with $T$ or any other subset in place of $S$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ).

[L2]

Every linear subspace of $V$ contains $0_V$ , by condition (W1) (Linear subspace of a vector space).

Proof

technique · direct

1.1

Claim 1 is part of the defining description of the span: it contains $S$ .

1.2

Claim 2. Suppose $S \subseteq T$ . Then $\operatorname{span}(T)$ is a linear subspace of $V$ containing $T$ , hence containing $S$ ; since $\operatorname{span}(S)$ is contained in every linear subspace containing $S$ , we get $\operatorname{span}(S) \subseteq \operatorname{span}(T)$ .

1.3

If $S$ is a linear subspace of $V$ , then $S$ is itself a linear subspace containing $S$ , so $\operatorname{span}(S) \subseteq S$ ; together with $S \subseteq \operatorname{span}(S)$ this gives $\operatorname{span}(S) = S$ .

1.4

Conversely, if $\operatorname{span}(S) = S$ then $S$ is a linear subspace of $V$ , because $\operatorname{span}(S)$ is one.

1.5

$0_V \in \operatorname{span}(S)$ , since $\operatorname{span}(S)$ is a linear subspace of $V$ .

L1L2

2.1

Steps 1.3 and 1.4 together are claim 4.

step 1.3step 1.4

2.2

Claim 5. From $S \subseteq S \cup \{0_V\}$ and step 1.2 we get $\operatorname{span}(S) \subseteq \operatorname{span}(S \cup \{0_V\})$ . Conversely $S \subseteq \operatorname{span}(S)$ by step 1.1 and $0_V \in \operatorname{span}(S)$ by step 1.5, so $S \cup \{0_V\} \subseteq \operatorname{span}(S)$ , and since $\operatorname{span}(S)$ is a linear subspace containing $S \cup \{0_V\}$ , minimality gives $\operatorname{span}(S \cup \{0_V\}) \subseteq \operatorname{span}(S)$ .

step 1.1step 1.2step 1.5L1

3.1

Claim 3. The set $\operatorname{span}(S)$ is a linear subspace of $V$ , so applying claim 4 to it gives $\operatorname{span}(\operatorname{span}(S)) = \operatorname{span}(S)$ .

step 2.1L1

4.1

Claims 1, 2, 3, 4 and 5 are steps 1.1, 1.2, 3.1, 2.1 and 2.2 respectively.

step 1.1step 1.2step 2.1step 2.2step 3.1∎

Remarks

These three properties are exactly what makes the span a closure operator on the subsets of $V$ : extensive, monotone and idempotent. The closed sets of that operator are the linear subspaces, which is the content of claim 4. The subgroup generated by a subset (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups) is the closure operator of the same shape on the subsets of a group.
Claim 5 says $0_V$ is invisible to the span. Adding or deleting the zero vector changes nothing, since every linear subspace contains it. It does not follow, and is not claimed here, that deleting any other single vector changes nothing: that question is about linear independence and belongs to a later page.
Claim 2 is used constantly in the form "a bigger set spans more". It is what lets a spanning set be enlarged freely, and, with claim 3, what lets the span be computed in stages: spanning a set, then spanning the result, gains nothing.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Linear span (Wikipedia) (standard reference, not scraped)
Closure operator (Wikipedia) (standard reference, not scraped)

The span is monotone and idempotent, span⁡(S)=S\operatorname{span}(S) = Sspan(S)=S exactly when SSS is a linear subspace, and span⁡(S∪{0V})=span⁡(S)\operatorname{span}(S \cup \{0_V\}) = \operatorname{span}(S)span(S∪{0V​})=span(S)