False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-sonnet-5)

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.

FALSE: The union of two linear subspaces is a linear subspace

Statement

False claim: if $U$ and $W$ are linear subspaces of a vector space $V$ over a field $F$ (Linear subspace of a vector space), then $U \cup W$ is a linear subspace of $V$ .

The corresponding statement for intersections is true and is The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$ . For unions it fails, and it already fails in the plane over any field: with $e_{0}, e_{1} \in F^{2}$ the vectors with coordinates $(1_{F}, 0_{F})$ and $(0_{F}, 1_{F})$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ ), take $U : = span {e_{0}}$ and $W : = span {e_{1}}$ (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ). Then $e_{0} \in U$ and $e_{1} \in W$ , so both lie in $U \cup W$ , while their sum, with coordinates $(1_{F}, 1_{F})$ , lies in neither.

Facts & Assumptions

Given: A field $F$ , the vector space $F^{2}$ over $F$ , the vectors $e_{0}$ and $e_{1}$ , and the linear subspaces $U = span {e_{0}}$ and $W = span {e_{1}}$ .

[L1]

$F^{2}$ is the vector space of functions $2 \to F$ with $(x + y)_{i} = x_{i} + y_{i}$ and $(λ x)_{i} = λ x_{i}$ , where $2 = {0, 1}$ ; two elements are equal exactly when both coordinates agree (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ , Vector space over a field, The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ).

[L2]

$span {v} = {λ v : λ \in F}$ , and a span is a linear subspace ( $span {v} = {λ v : λ \in F}$ , which is ${0_{V}}$ when $v = 0_{V}$ , and when $v \neq 0_{V}$ contains $0_{V}$ only as the multiple $0_{F} v$ , Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

[L3]

A linear subspace is closed under addition, by condition (W2) (Linear subspace of a vector space, One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $λ u + v \in W$ for all $λ \in F$ and $u, v \in W$ ).

[L4]

In a field: $1_{F} \neq 0_{F}$ ; $λ 1_{F} = λ$ ; $0_{F} λ = 0_{F}$ (Multiplication by zero: $0 \cdot a = 0$ ) and multiplication is commutative, so $λ 0_{F} = 0_{F}$ ; and $0_{F}$ is the additive identity (Field).

[L5]

The refuted claim: the union of two linear subspaces of a vector space is a linear subspace of it.

Refutation

technique · direct

1.1

$U$ and $W$ are linear subspaces of $F^{2}$ , being spans of one-element subsets.

L1L2

1.2

The elements of $U$ are the vectors $(λ, 0_{F})$ and the elements of $W$ are the vectors $(0_{F}, λ)$ , for $λ \in F$ : indeed $(λ e_{0})_{0} = λ 1_{F} = λ$ , $(λ e_{0})_{1} = λ 0_{F} = 0_{F}$ , and symmetrically for $e_{1}$ .

L1L2L4

1.3

$e_{0} + e_{1}$ has coordinates $(1_{F} + 0_{F}, 0_{F} + 1_{F}) = (1_{F}, 1_{F})$ .

L1L4

2.1

$e_{0} \in U$ and $e_{1} \in W$ , since $e_{j} = 1_{F} e_{j}$ ; so both lie in $U \cup W$ .

step 1.2L2L4

2.2

$e_{0} + e_{1} \notin U \cup W$ : if it lay in $U$ its coordinate at index $1$ would be $0_{F}$ , and if it lay in $W$ its coordinate at index $0$ would be $0_{F}$ , whereas both coordinates are $1_{F}$ and $1_{F} \neq 0_{F}$ .

step 1.2step 1.3L1L4

3.1

So $U \cup W$ contains $e_{0}$ and $e_{1}$ but not $e_{0} + e_{1}$ : it is not closed under addition, so condition (W2) fails and it is not a linear subspace of $F^{2}$ , although $U$ and $W$ both are. The claim of [L5] is false.

step 1.1step 2.1step 2.2L3L5∎

Remarks

Intersections behave, unions do not. An intersection of linear subspaces is always a linear subspace (The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$ ), which is what makes $span (S)$ definable as the smallest linear subspace containing $S$ (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ). On the union side there is no corresponding construction, and the repair is to take the sum rather than the union: $U + W$ is a linear subspace, and it is exactly $span (U \cup W)$ ( $\sum_{i < n} U_{i} = span (⋃_{i < n} U_{i})$ , so the sum is the smallest linear subspace containing every $U_{i}$ ).
The exact condition. For linear subspaces $U, W$ of a vector space $V$ , the union $U \cup W$ is a linear subspace if and only if $U \subseteq W$ or $W \subseteq U$ . One direction is immediate, the union then being the larger of the two. For the other, suppose neither inclusion holds and choose $u \in U ∖ W$ and $w \in W ∖ U$ . If $u + w$ were in $U$ then $w = (u + w) + (- u) \in U$ , and if $u + w$ were in $W$ then $u = (u + w) + (- w) \in W$ ; both contradict the choice, since a linear subspace is closed under addition and under additive inverses. So $u + w \notin U \cup W$ and the union is not closed under addition.
The witness above is an instance of that criterion, since neither of $U$ and $W$ contains the other: $e_{0} \in U$ has coordinate $1_{F} \neq 0_{F}$ at index $0$ and so is not of the form $(0_{F}, λ)$ , and symmetrically for $e_{1}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 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

Linear subspace (Wikipedia) (standard reference, not scraped)
Union (set theory) (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed. (free PDF, CC BY-NC) (standard reference, not scraped)
The union of vector subspaces (Andrea Minini) (standard reference, not scraped)