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

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The union of the two coordinate axes of $F^{2}$ is closed under scalar multiplication and is not closed under addition, so neither closure condition implies the other

Statement refuted

False claim: if $V$ is a vector space over a field $F$ and $W \subseteq V$ contains $0_V$ and is closed under scalar multiplication, then $W$ is a linear subspace of $V$ (Linear subspace of a vector space).

The union of the two coordinate axes of $F^{2}$ refutes it, over any field $F$ . Let $e_0, e_1 \in F^{2}$ be 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 $X = n = \{0, 1, \dots, n-1\}$ ), let $L_j := \operatorname{span}\{e_j\}$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ), and put $A := L_0 \cup L_1$ . Then $0_V \in A$ and $A$ is closed under scalar multiplication, while $e_0, e_1 \in A$ and $e_0 + e_1 \notin A$ .

Together with The first quadrant of $\mathbb{R}^{2}$ contains $0$ and is closed under addition and is not a linear subspace, since it is not closed under multiplication by $-1$ , which exhibits a subset closed under addition and not under scalar multiplication, this shows that neither of the two closure conditions in Linear subspace of a vector space implies the other.

Facts & Assumptions

Given: A field $F$ , the vector space $F^{2}$ over $F$ , the vectors $e_0, e_1$ , the sets $L_0, L_1$ and their union $A$ , as displayed.

[L1]

$F^{2}$ is the vector space of functions $2 \to F$ with $(x+y)_i = x_i + y_i$ and $(\lambda x)_i = \lambda x_i$ , where $2 = \{0,1\}$ , and its zero vector has both coordinates $0_F$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ , Vector space over a field, The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ).

[L2]

$\operatorname{span}\{v\} = \{\, \lambda v : \lambda \in F \,\}$ , and a span is a linear subspace ( $\operatorname{span}\{v\} = \{\, \lambda v : \lambda \in F \,\}$ , which is $\{0_V\}$ when $v = 0_V$ , and when $v \ne 0_V$ contains $0_V$ only as the multiple $0_F v$ , Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ).

[L3]

A linear subspace satisfies (W1) $0_V \in W$ , (W2) closure under $+$ , and (W3) closure under scalar multiplication (Linear subspace of a vector space); on a nonempty subset the three are equivalent to the one-step test (One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $\lambda u + v \in W$ for all $\lambda \in F$ and $u, v \in W$ ).

[L4]

In a field: $1_F \ne 0_F$ ; $\lambda 1_F = \lambda$ ; $0_F \lambda = 0_F$ (Multiplication by zero: $0 \cdot a = 0$ ) and multiplication is commutative, so $\lambda 0_F = 0_F$ ; and $0_F$ is the additive identity, so $1_F + 0_F = 1_F = 0_F + 1_F$ (Field).

[L5]

There is a subset of a vector space that contains the zero vector and is closed under addition and is not closed under scalar multiplication (The first quadrant of $\mathbb{R}^{2}$ contains $0$ and is closed under addition and is not a linear subspace, since it is not closed under multiplication by $-1$ ).

[L6]

The refuted claim: a subset of a vector space containing the zero vector and closed under scalar multiplication is a linear subspace.

Counterexample

technique · direct

1.1

$F^{2}$ is the set of functions $2 \to F$ with coordinatewise operations and $2 = \{0,1\}$ , so an element is $x = (x_0, x_1)$ ; and $L_j = \{\, \lambda e_j : \lambda \in F \,\}$ is a linear subspace of $F^{2}$ for $j \in \{0,1\}$ .

L1L2

1.2

$L_j = \{\, x \in F^{2} : x_i = 0_F \text{ for the index } i \ne j \,\}$ . Indeed $(\lambda e_j)_j = \lambda 1_F = \lambda$ and $(\lambda e_j)_i = \lambda 0_F = 0_F$ for $i \ne j$ ; conversely a vector $x$ whose other coordinate is $0_F$ agrees with $x_j e_j$ at both indices, so $x = x_j e_j$ .

L1L2L4

1.3

$0_V \in A$ : the vector $0_F e_0$ has both coordinates $0_F \cdot (e_0)_i = 0_F$ , so it is the zero vector, and it lies in $L_0 \subseteq A$ .

L1L2L4

1.4

$A$ is closed under scalar multiplication: if $x \in A$ then $x \in L_j$ for some $j \in \{0,1\}$ , and $L_j$ is a linear subspace, so $\lambda x \in L_j \subseteq A$ for every $\lambda \in F$ .

L2L3

1.5

$e_0 + e_1$ has coordinates $(1_F + 0_F,\; 0_F + 1_F) = (1_F, 1_F)$ .

L1L4

2.1

$e_0 + e_1 \notin A$ : membership in $L_0$ requires the coordinate at index $1$ to be $0_F$ and membership in $L_1$ requires the coordinate at index $0$ to be $0_F$ , and both coordinates of $e_0 + e_1$ are $1_F \ne 0_F$ .

step 1.2step 1.5L4

2.2

$e_0 \in A$ and $e_1 \in A$ , since $e_j = 1_F e_j \in L_j$ .

step 1.2L2L4

3.1

So $A$ contains the zero vector and is closed under scalar multiplication, while $e_0$ and $e_1$ lie in $A$ and their sum does not; condition (W2) therefore fails and $A$ is not a linear subspace of $F^{2}$ . The claim of [L6] is false.

step 1.3step 1.4step 2.1step 2.2L3L6

4.1

Combining with [L5]: closure under addition does not imply closure under scalar multiplication, and closure under scalar multiplication does not imply closure under addition, so neither of the two conditions implies the other, even for subsets containing the zero vector.

step 3.1L5∎

Remarks

The witness works over every field, including $F$ with two elements: the argument uses only $1_F \ne 0_F$ , and never counts the elements of $F$ or the linear subspaces of $F^{2}$ . No claim is made here about how many subsets of $F^{2}$ of this kind there are.
The union of two linear subspaces is the general phenomenon. $A$ is a union of two linear subspaces, neither of which contains the other, and such a union is never a linear subspace; that is recorded separately as FALSE: The union of two linear subspaces is a linear subspace, of which this item is the concrete instance closed under scalar multiplication.
"Axis" is informal here, as "line" is elsewhere on this page: it names the set $\operatorname{span}\{e_j\}$ and carries no claim about dimension, which is not available at this point in the library.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

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

The union of the two coordinate axes of F2F^{2}F2 is closed under scalar multiplication and is not closed under addition, so neither closure condition implies the other