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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$\{(1,0), (0,1), (1,1)\}$ spans $F^{2}$ and is linearly dependent, so a spanning set need not be a basis; each of its three two-element subsets is a basis

Statement refuted

False claim: a spanning subset of a vector space is linearly independent, and hence a basis.

Let $F$ be any field (Field) and let $F^{2}$ be the function space on $2 = \{0,1\}$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ ), whose elements we write $(y_0,y_1)$ . Put

$e_0 := (1_F, 0_F), \qquad e_1 := (0_F, 1_F), \qquad d := (1_F, 1_F) = e_0 + e_1,$

and $S := \{e_0, e_1, d\}$ , a set with exactly three elements. Then

$\operatorname{span}(S) = F^{2}$ ;
$S$ is linearly dependent (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent), so $S$ is not a basis of $F^{2}$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis);
each of the three two-element subsets $\{e_0,e_1\}$ , $\{e_0,d\}$ , $\{e_1,d\}$ is a basis of $F^{2}$ .

The field is arbitrary; over a field in which $1_F + 1_F = 0_F$ the identity $-1_F = 1_F$ holds and changes nothing below, the three displayed vectors still being distinct because $1_F \ne 0_F$ .

Facts & Assumptions

Given: A field $F$ , the vector space $F^{2}$ with pointwise operations, and the vectors $e_0$ , $e_1$ , $d = e_0+e_1$ and the set $S = \{e_0,e_1,d\}$ .

[L1]

$e : 2 \to F^{2}$ is an ordered basis of $F^{2}$ , $e[2] = \{e_0,e_1\}$ is a basis, $\bigl(\sum_{i<2}\lambda_i e_i\bigr)(j) = \lambda_j$ , and $\dim_F F^{2} = 2$ (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ , claims 2, 3 and 4).

[L2]

A list $w : n \to V$ is an ordered basis if and only if every $y \in V$ is $\sum_{i<n}\lambda_i w_i$ for exactly one $\lambda : n \to F$ ; an ordered basis is injective with a basis as its image (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i<n} \lambda_i v_i$ for exactly one $\lambda : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L3]

$\operatorname{span}$ is monotone and $\operatorname{span}(T) \subseteq V$ always (The span is monotone and idempotent, $\operatorname{span}(S) = S$ exactly when $S$ is a linear subspace, and $\operatorname{span}(S \cup \{0_V\}) = \operatorname{span}(S)$ , Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ).

[L4]

Finite sums: $\sum_{i<3}u_i = ((0_V + u_0) + u_1) + u_2$ and $\sum_{i<2}u_i = u_0 + u_1$ (The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ).

[L5]

$(V,+,0_V)$ is an abelian group; $1_F y = y$ , $(-1_F)y = -y$ , $0_F y = 0_V$ ; (V2) and (V3); and $1_F \ne 0_F$ , with $-1_F \ne 0_F$ (Vector space over a field, 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 $v = 0_V$ , Field).

[L6]

A subset $T$ is dependent when some injective finite list into $T$ is dependent; equivalently when some $t \in T$ lies in $\operatorname{span}(T\setminus\{t\})$ (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent, A subset $S \subseteq V$ is linearly dependent if and only if some $s \in S$ lies in $\operatorname{span}(S \setminus \{s\})$ ; and $\operatorname{span}(S)$ is already the set of linear combinations of INJECTIVE finite lists into $S$ ).

[L7]

Elements of $F^{2}$ are equal exactly when they agree at $0$ and at $1$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ , The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ); injectivity is as in Injection, surjection, bijection.

Counterexample

technique · direct

1.1

$S$ has exactly three elements. $e_0 \ne e_1$ because their values at $0$ are $1_F$ and $0_F$ ; $e_0 \ne d$ because their values at $1$ are $0_F$ and $1_F$ ; and $e_1 \ne d$ because their values at $0$ are $0_F$ and $1_F$ . All three inequalities use only $1_F \ne 0_F$ .

L5L7

1.2

Claim 1. From $\{e_0,e_1\} \subseteq S$ and monotonicity, $F^{2} = \operatorname{span}\{e_0,e_1\} \subseteq \operatorname{span}(S) \subseteq F^{2}$ , so $\operatorname{span}(S) = F^{2}$ .

L1L3

1.3

The pair $(e_0, d)$ is an ordered basis. For $\lambda : 2 \to F$ , $\sum_{i<2}\lambda_i v_i$ with $v_0 = e_0$ , $v_1 = d$ is $\lambda_0 e_0 + \lambda_1(e_0+e_1) = (\lambda_0 + \lambda_1)e_0 + \lambda_1 e_1$ by (V2), (V3) and the abelian group laws, which evaluates to $(\lambda_0+\lambda_1,\ \lambda_1)$ . Given $y = (y_0,y_1)$ , the equations $\lambda_0 + \lambda_1 = y_0$ and $\lambda_1 = y_1$ have the unique solution $\lambda_1 = y_1$ , $\lambda_0 = y_0 - y_1$ ; so every $y$ has exactly one representation and $(e_0,d)$ is an ordered basis, whence $\{e_0, d\}$ is a basis of $F^{2}$ .

L1L2L4L5

1.4

The pair $(e_1, d)$ is an ordered basis. Likewise $\lambda_0 e_1 + \lambda_1 d = \lambda_1 e_0 + (\lambda_0 + \lambda_1)e_1$ evaluates to $(\lambda_1,\ \lambda_0 + \lambda_1)$ , and $\lambda_1 = y_0$ , $\lambda_0 = y_1 - y_0$ is the unique solution; so $\{e_1, d\}$ is a basis of $F^{2}$ .

L1L2L4L5

2.1

Claim 2. The list $v : 3 \to S$ with $v_0 = e_0$ , $v_1 = e_1$ , $v_2 = d$ is injective by step 1.1, and with $\lambda_0 = \lambda_1 = 1_F$ and $\lambda_2 = -1_F$ we get $\sum_{i<3}\lambda_i v_i = ((0_V + 1_F e_0) + 1_F e_1) + (-1_F)d = (e_0 + e_1) + (-d) = d + (-d) = 0_V$ , while $\lambda_0 = 1_F \ne 0_F$ . So $v$ is a dependent injective list into $S$ and $S$ is linearly dependent; a basis is independent, so $S$ is not a basis of $F^{2}$ .

step 1.1L4L5L6

3.1

Claim 3, and the conclusion. $\{e_0,e_1\}$ is a basis by the standard basis lemma, and $\{e_0,d\}$ and $\{e_1,d\}$ are bases by steps 1.3 and 1.4; these are the three two-element subsets of the three-element set $S$ of step 1.1. Together with step 1.2 and step 2.1, the set $S$ spans $F^{2}$ and is dependent, refuting the false claim.

step 1.1step 1.2step 1.3step 1.4step 2.1L1∎

Remarks

This makes Every spanning subset of a vector space contains a basis concrete, and shows the contained basis is not unique. $S$ spans and is not a basis; it contains three different bases, and no argument singles one out. By For $B \subseteq V$ the following are equivalent: $B$ is a basis; $B$ is a maximal linearly independent subset of $V$ ; $B$ is a minimal spanning subset of $V$ — maximality and minimality being in the inclusion order each of them is a minimal spanning subset of $F^{2}$ , while $S$ itself is not minimal.
A spanning set may be larger than the dimension. Here $\dim_F F^{2} = 2$ and $S$ has three elements. The bound of If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $\mathbb{N}$ runs the other way: it is independent sets that cannot exceed the size of a finite spanning set, and indeed no three vectors of $F^{2}$ are independent, by If $V$ has a basis with $n$ elements and a basis with $m$ elements then $n = m$ ; and if $V$ has one finite basis then every basis of $V$ is finite and that bound.
The field is arbitrary and is named. Every step uses only the field axioms and $1_F \ne 0_F$ . Over the two-element field the dependence reads $e_0 + e_1 + d = 0_V$ , since $-1_F = 1_F$ there; the three vectors remain distinct and the three two-element subsets remain bases.

Depends on

Used by

FALSE: the union of two linearly independent subsets of a vector space is linearly independent False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 96 results over 29 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

Basis (linear algebra) (Wikipedia) (standard reference, not scraped)
Linear independence (Wikipedia) (standard reference, not scraped)
Auburn University notes: Spanning sets and linear independence (standard reference, not scraped)

{(1,0),(0,1),(1,1)}\{(1,0), (0,1), (1,1)\}{(1,0),(0,1),(1,1)} spans F2F^{2}F2 and is linearly dependent, so a spanning set need not be a basis; each of its three two-element subsets is a basis