CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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 $L_{j} : = span {e_{j}}$ (Linear combination of a finite list, and the span $span (S)$ 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 $R^{2}$ 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 $(λ x)_{i} = λ x_{i}$ , 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 $N$ (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 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 $λ u + v \in W$ 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, 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 $R^{2}$ 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 $L_{j} = {λ e_{j} : λ \in F}$ is a linear subspace of $F^{2}$ for $j \in {0, 1}$ .

L1L2

1.2

$L_{j} = {x \in F^{2} : x_{i} = 0_{F} for the index i \neq j}$ . Indeed $(λ e_{j})_{j} = λ 1_{F} = λ$ and $(λ e_{j})_{i} = λ 0_{F} = 0_{F}$ for $i \neq j$ ; 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 $λ x \in L_{j} \subseteq A$ for every $λ \in F$ .

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 $1_{F} \neq 0_{F}$ .

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 $1_{F} \neq 0_{F}$ , 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 $span {e_{j}}$ 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.

Dependency tree · two levels

42 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)
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 F2F2 is closed under scalar multiplication and is not closed under addition, so neither closure condition implies the other