CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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

$e_{0} : = (1_{F}, 0_{F}), e_{1} : = (0_{F}, 1_{F}), d : = (1_{F}, 1_{F}) = e_{0} + e_{1},$

and $S : = {e_{0}, e_{1}, d}$ , a set with exactly three elements. Then

$span (S) = F^{2}$ ;
$S$ is linearly dependent (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , 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 $1_{F} \neq 0_{F}$ .

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, $(\sum_{i < 2} λ_{i} e_{i}) (j) = λ_{j}$ , 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 $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ 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 $\sum_{i < n} λ_{i} w_{i}$ for exactly one $λ : n \to F$ ; 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 $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; 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]

$span$ is monotone and $span (T) \subseteq V$ always (The span is monotone and idempotent, $span (S) = S$ exactly when $S$ is a linear subspace, and $span (S \cup {0_{V}}) = span (S)$ , Linear combination of a finite list, and the span $span (S)$ 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 $g_{0} g_{1} \dots g_{n - 1}$ 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 $span (S)$ 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 $1_{F} \neq 0_{F}$ , with $- 1_{F} \neq 0_{F}$ (Vector space over a field, In any vector space $0_{F} v = 0_{V}$ , $λ 0_{V} = 0_{V}$ , $(- λ) v = - (λ v)$ , $(- 1_{F}) v = - v$ , and $λ v = 0_{V}$ forces $λ = 0_{F}$ 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 $span (T ∖ {t})$ (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , 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 $span (S ∖ {s})$ ; and $span (S)$ 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 $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ); injectivity is as in Injection, surjection, bijection.

Counterexample

technique · direct

1.1

$S$ has exactly three elements. $e_{0} \neq e_{1}$ because their values at $0$ are $1_{F}$ and $0_{F}$ ; $e_{0} \neq d$ because their values at $1$ are $0_{F}$ and $1_{F}$ ; and $e_{1} \neq d$ because their values at $0$ are $0_{F}$ and $1_{F}$ . All three inequalities use only $1_{F} \neq 0_{F}$ .

L5L7

1.2

Claim 1. From ${e_{0}, e_{1}} \subseteq S$ and monotonicity, $F^{2} = span {e_{0}, e_{1}} \subseteq span (S) \subseteq F^{2}$ , so $span (S) = F^{2}$ .

L1L3

1.3

The pair $(e_{0}, d)$ is an ordered basis. For $λ : 2 \to F$ , $\sum_{i < 2} λ_{i} v_{i}$ with $v_{0} = e_{0}$ , $v_{1} = d$ is $λ_{0} e_{0} + λ_{1} (e_{0} + e_{1}) = (λ_{0} + λ_{1}) e_{0} + λ_{1} e_{1}$ by (V2), (V3) and the abelian group laws, which evaluates to $(λ_{0} + λ_{1}, λ_{1})$ . Given $y = (y_{0}, y_{1})$ , the equations $λ_{0} + λ_{1} = y_{0}$ and $λ_{1} = y_{1}$ have the unique solution $λ_{1} = y_{1}$ , $λ_{0} = y_{0} - y_{1}$ ; 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 $λ_{0} e_{1} + λ_{1} d = λ_{1} e_{0} + (λ_{0} + λ_{1}) e_{1}$ evaluates to $(λ_{1}, λ_{0} + λ_{1})$ , and $λ_{1} = y_{0}$ , $λ_{0} = y_{1} - y_{0}$ 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 $λ_{0} = λ_{1} = 1_{F}$ and $λ_{2} = - 1_{F}$ we get $\sum_{i < 3} λ_{i} v_{i} = ((0_{V} + 1_{F} e_{0}) + 1_{F} e_{1}) + (- 1_{F}) d = (e_{0} + e_{1}) + (- d) = d + (- d) = 0_{V}$ , while $λ_{0} = 1_{F} \neq 0_{F}$ . 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 $N$ 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 $1_{F} \neq 0_{F}$ . 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 · two levels

63 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

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)} spans F2F2 and is linearly dependent, so a spanning set need not be a basis; each of its three two-element subsets is a basis