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 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 $-1$

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 the vector addition, then $W$ is a linear subspace of $V$ (Linear subspace of a vector space).

The first quadrant of $\mathbb{R}^{2}$ refutes it. Take $F = \mathbb{R}$ (The reals form a field) and $V = \mathbb{R}^{2}$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ ), and put

$Q \;:=\; \{\, x \in \mathbb{R}^{2} \;:\; 0 \le x_0 \text{ and } 0 \le x_1 \,\}.$

Then $0_V \in Q$ and $Q$ is closed under addition, but the vector $e_0$ with coordinates $(1,0)$ lies in $Q$ while $(-1)e_0$ , with coordinates $(-1,0)$ , does not. So $Q$ is not closed under scalar multiplication and is not a linear subspace.

Facts & Assumptions

Given: The field $\mathbb{R}$ (The reals form a field, The real numbers) with its order, the vector space $\mathbb{R}^{2}$ over $\mathbb{R}$ , and the subset $Q$ displayed above.

[L1]

$\mathbb{R}$ is an ordered field with positive cone $P$ : (O1) for each $x$ exactly one of $x \in P$ , $x = 0$ , $-x \in P$ holds; (O2) $P$ is closed under addition and multiplication; $x > 0$ means $x \in P$ , and $a \le b$ means $a < b$ or $a = b$ (The reals form a totally ordered field, Ordered field).

[L2]

Every nonzero square of an ordered field is positive (Squares of nonzero elements are positive).

[L3]

$\mathbb{R}^{2}$ is a vector space over $\mathbb{R}$ with $(x+y)_i = x_i + y_i$ and $(\lambda x)_i = \lambda x_i$ for $i \in 2 = \{0,1\}$ , and its zero vector has both coordinates $0$ (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: $m < n \iff m \in n$ ).

[L4]

A linear subspace satisfies (W1) $0_V \in W$ , (W2) closure under $+$ , and (W3) closure under scalar multiplication (Linear subspace of a vector space); the three conditions are together equivalent to the one-step test on a nonempty subset (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 $u, v \in W$ ).

[L5]

Every linear subspace is a subgroup of the additive group of the space, and a subgroup is closed under inverses (The additive group of a vector space is an abelian group and every linear subspace is a subgroup of it; conversely a subgroup closed under scalar multiplication is a linear subspace, Subgroup).

[L6]

$(-1_F)v = -v$ in any vector space (In any vector space $0_F v = 0_V$ , $\lambda 0_V = 0_V$ , $(-\lambda)v = -(\lambda v)$ , $(-1_F)v = -v$ , and $\lambda v = 0_V$ forces $\lambda = 0_F$ or $v = 0_V$ ).

[L7]

Field arithmetic in $\mathbb{R}$ : $1 \ne 0$ ; $1 \cdot a = a$ ; $a \cdot 0 = 0$ ; $0$ is the additive identity; and $-(-1) = 1$ , since $(F,+,0)$ is an abelian group (Field).

[L8]

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

Counterexample

technique · direct

1.1

$\mathbb{R}^{2}$ is the vector space of functions $2 \to \mathbb{R}$ with coordinatewise operations, $2 = \{0,1\}$ , so an element is $x = (x_0, x_1)$ and the zero vector is $(0,0)$ .

1.2

The zero vector lies in $Q$ , since $0 \le 0$ ; in particular $Q$ is nonempty.

L1L3

1.3

$0 < 1$ in $\mathbb{R}$ : $1 \ne 0$ and $1 = 1 \cdot 1$ is a square, so $1 \in P$ .

L1L2L7

1.4

If $a, b \in \mathbb{R}$ satisfy $0 \le a$ and $0 \le b$ , then $0 \le a + b$ : if $a = 0$ then $a + b = b$ ; if $b = 0$ then $a + b = a$ ; and otherwise $a, b \in P$ , so $a + b \in P$ by (O2).

L1L7

2.1

$Q$ is closed under addition: for $x, y \in Q$ and $i \in \{0,1\}$ we have $(x+y)_i = x_i + y_i$ with $0 \le x_i$ and $0 \le y_i$ , hence $0 \le (x+y)_i$ .

step 1.4L1L3

2.2

It is not the case that $0 \le -1$ : applying trichotomy to $-1$ , exactly one of $-1 \in P$ , $-1 = 0$ , $-(-1) = 1 \in P$ holds, and the last one does, so $-1 \notin P$ and $-1 \ne 0$ .

step 1.3L1L7

2.3

The vector $e_0$ with $(e_0)_0 = 1$ and $(e_0)_1 = 0$ lies in $Q$ , since $0 \le 1$ and $0 \le 0$ .

step 1.3L1L3

3.1

$(-1)e_0$ has coordinates $((-1)\cdot 1, (-1)\cdot 0) = (-1, 0)$ , and its coordinate at index $0$ fails $0 \le -1$ , so $(-1)e_0 \notin Q$ . Hence $Q$ is not closed under scalar multiplication: condition (W3) fails, and $Q$ is not a linear subspace of $\mathbb{R}^{2}$ .

step 2.2step 2.3L3L4L7

4.1

So $Q$ contains the zero vector and is closed under addition, by steps 1.2 and 2.1, and is not a linear subspace, by step 3.1; the claim of [L8] is therefore false. The failure can also be read in the additive group: $(-1)e_0 = -e_0$ lies outside $Q$ , so $Q$ is not even a subgroup of the additive group of $\mathbb{R}^{2}$ , whereas a linear subspace always is.

step 1.2step 2.1step 3.1L5L6L8∎

Remarks

Exactly one of the three conditions fails. $Q$ satisfies (W1) and (W2) and fails (W3), and it fails it at a single scalar, $-1$ . The reverse failure, a subset closed under scalar multiplication but not under addition, is recorded in 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, so neither closure condition implies the other.
What the order is doing here. The example needs a field in which some element is not the negative of a nonnegative one, so it needs an order; over an arbitrary field there is no "first quadrant" to speak of. That is why this witness is stated over $\mathbb{R}$ while its companion is stated over an arbitrary field.
$Q$ is closed under multiplication by nonnegative scalars. If $0 \le \lambda$ and $x \in Q$ then $\lambda x \in Q$ , by the closure of $P$ under multiplication together with the zero cases. So the failure is confined to the negative scalars; a subset with this weaker closure property is a cone, not a linear subspace, and the difference is exactly what the example isolates.

Depends on

Used by

The union of the two coordinate axes of F² is closed under scalar multiplication and is not closed under addition, so neither closure condition implies the other Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 74 results over 25 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 subspace (Wikipedia) (standard reference, not scraped)
Ordered field (Wikipedia) (standard reference, not scraped)
D. Margalit and J. Rabinoff, Interactive Linear Algebra, 2.6 Subspaces (standard reference, not scraped)

The first quadrant of R2\mathbb{R}^{2}R2 contains 000 and is closed under addition and is not a linear subspace, since it is not closed under multiplication by −1-1−1