ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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The standard unit families $e_{k} \in F^{N}$ form a basis of the linear subspace of eventually zero families: an explicit infinite basis, built with no choice principle

Example

Let $F$ be a field (Field) and let $F^{N}$ be the function space of all families $x : N \to F$ with the pointwise operations (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ ); $N$ contains $0$ (The natural numbers $N$ (von Neumann)). Put

$E : = {x \in F^{N} : there is N \in N with x (n) = 0_{F} for every n \geq N},$

the eventually zero families, and for $k \in N$ let $e_{k} \in F^{N}$ be the standard unit family with $e_{k} (k) = 1_{F}$ and $e_{k} (n) = 0_{F}$ for $n \neq k$ . Write $B : = {e_{k} : k \in N}$ . Then:

$E$ is a linear subspace of $F^{N}$ (Linear subspace of a vector space);
$B \subseteq E$ and $span (B) = E$ (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ );
$B$ is linearly independent (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), hence a basis of $E$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), and $k \mapsto e_{k}$ is a bijection $N \to B$ , so $B \approx N$ (Equinumerous sets, $A \approx B$ and $A ⪯ B$ );
$E$ is infinite-dimensional over $F$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis): it has no finite basis.

No choice principle is used anywhere below: the basis is written down.

Facts & Assumptions

Given: A field $F$ , the vector space $F^{N}$ with pointwise operations, the set $E$ of eventually zero families, the families $e_{k}$ , and $B = {e_{k} : k \in N}$ .

[L1]

$F^{X}$ is a vector space over $F$ with $(x + y) (n) = x (n) + y (n)$ , $(λ x) (n) = λ x (n)$ and zero the constant family at $0_{F}$ ; two elements are equal exactly when they agree at every point (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).

[L2]

One-step test: a nonempty $W \subseteq V$ with $λ u + v \in W$ for all $λ \in F$ , $u, v \in W$ is a linear subspace; a linear subspace is a vector space in its own right, and independence and spans of its subsets agree with those computed in the ambient space (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 $u, v \in W$ , Linear subspace of a vector space, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, section on bases of a linear subspace).

[L3]

$span (T)$ is the set of linear combinations $\sum_{k < p} μ_{k} y_{k}$ of finite lists into $T$ , and it is the smallest linear subspace containing $T$ ( $span (S)$ is exactly the set of linear combinations of finite lists of elements of $S$ , and $span (\emptyset) = {0_{V}}$ , Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

[L4]

A finite sum in a function space is pointwise: $(\sum_{k < p} u_{k}) (n) = \sum_{k < p} u_{k} (n)$ (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 $0$ , claim 1). Finite sums obey $\sum_{k < 0} u_{k} = 0$ and $\sum_{k < σ (p)} u_{k} = (\sum_{k < p} u_{k}) + u_{p}$ (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).

[L5]

$F$ is a vector space over itself (A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$ -vector space is a $K$ -vector space by restricting the scalars, claim 1), so (F1) and (F3) apply to lists of scalars: an all- $0_{F}$ list sums to $0_{F}$ , and a list vanishing off a single index sums to its value at that index (The sum $U + W$ of two linear subspaces and the sum $\sum_{i < n} U_{i}$ of a finite family).

[L6]

In $F$ : $λ 1_{F} = λ$ , $λ 0_{F} = 0_{F}$ , $0_{F} + 0_{F} = 0_{F}$ , and $1_{F} \neq 0_{F}$ (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 $v = 0_{V}$ ).

[L7]

A subset $S$ is linearly independent when every injective finite list into $S$ is; a list is independent exactly when it is injective with independent image (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, Finite sums re-indexed along an injection, with a zero term deleted, and concatenated; and the closure properties of linear independence: an independent list is injective and never $0_{V}$ , its sublists are independent, a list is independent exactly when it is injective with linearly independent image, and every subset of a linearly independent set is linearly independent, claim 6).

[L8]

If a space has a spanning set with $p$ elements, then no linearly independent subset of it is equinumerous with $N$ (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 $N$ , claim 2); a basis is a spanning set, and a finite set is equinumerous with exactly one natural (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, The pigeonhole principle on $N$ , Finite, countably infinite, countable, uncountable).

[L9]

The order of $N$ is total, $m < σ (N) ⟺ m \leq N$ , and $m < n$ implies $m \neq n$ ( $\leq$ is a linear order on $N$ , Order on the natural numbers, On $N$ the order is membership: $m < n ⟺ m \in n$ ); induction (The principle of mathematical induction); injectivity and images (Injection, surjection, bijection).

Verification

technique · direct

1.1

Claim 1. $E$ is nonempty: the zero family has value $0_{F}$ everywhere, so $N = 0$ witnesses that it lies in $E$ . And $E$ is closed under the one-step expression: for $λ \in F$ and $u, v \in E$ with witnesses $N_{1}$ and $N_{2}$ , let $N$ be the larger of the two, which exists because the order of $N$ is total; then for $n \geq N$ we have $n \geq N_{1}$ and $n \geq N_{2}$ , so $(λ u + v) (n) = λ u (n) + v (n) = λ 0_{F} + 0_{F} = 0_{F}$ , and $N$ witnesses $λ u + v \in E$ . So $E$ is a linear subspace of $F^{N}$ by the one-step test.

L1L2L6L9

1.2

Each $e_{k}$ lies in $E$ , so $B \subseteq E$ : if $n \geq σ (k)$ then $k < n$ , hence $n \neq k$ and $e_{k} (n) = 0_{F}$ , so $σ (k)$ is a witness.

1.3

For $x \in F^{N}$ and $N \in N$ put $s_{N} : = \sum_{k < N} x (k) e_{k}$ . Then $s_{N} (n) = x (n)$ for $n < N$ and $s_{N} (n) = 0_{F}$ for $n \geq N$ . Indeed $s_{N} (n) = \sum_{k < N} (x (k) e_{k}) (n) = \sum_{k < N} x (k) e_{k} (n)$ by pointwise evaluation and pointwise scalar multiplication; the scalar list $k \mapsto x (k) e_{k} (n)$ has the value $x (k) 0_{F} = 0_{F}$ at every $k \neq n$ . If $n < N$ this list vanishes off the single index $n$ , where its value is $x (n) 1_{F} = x (n)$ , so the sum is $x (n)$ ; if $n \geq N$ then no $k < N$ equals $n$ , the list is all $0_{F}$ , and the sum is $0_{F}$ .

L1L4L5L6L9

1.4

Claim 3, independence. The map $k \mapsto e_{k}$ is injective, since $e_{k} (k) = 1_{F} \neq 0_{F} = e_{l} (k)$ for $l \neq k$ . Let $w : m \to B$ be an injective finite list and $λ : m \to F$ with $\sum_{j < m} λ_{j} w_{j} = 0$ in $F^{N}$ . Each $w_{j}$ is $e_{i_{j}}$ for exactly one $i_{j} \in N$ , and $j \mapsto i_{j}$ is injective because $w$ is. Fix $j < m$ and evaluate at $i_{j}$ : pointwise evaluation gives $0_{F} = \sum_{l < m} λ_{l} e_{i_{l}} (i_{j})$ , and $e_{i_{l}} (i_{j}) = 0_{F}$ unless $i_{l} = i_{j}$ , that is unless $l = j$ , where it is $1_{F}$ . So the scalar list vanishes off the single index $j$ and sums to $λ_{j} 1_{F} = λ_{j}$ , giving $λ_{j} = 0_{F}$ . Hence every injective finite list into $B$ is independent, that is $B$ is linearly independent.

L1L4L5L6L7L9

2.1

$B \approx N$ : the map $k \mapsto e_{k}$ is injective by step 1.4 and its image is $B$ by definition, so it is a bijection $N \to B$ .

step 1.4L9

2.2

Claim 2. Each $e_{k}$ lies in $E$ and $E$ is a linear subspace, so $span (B) \subseteq E$ by minimality of the span. Conversely let $x \in E$ with witness $N$ ; then $x$ and $s_{N}$ of step 1.3 agree at every $n$ , since for $n < N$ both take the value $x (n)$ and for $n \geq N$ both take the value $0_{F}$ , so $x = s_{N} = \sum_{k < N} x (k) e_{k}$ , a linear combination of elements of $B$ . Hence $E \subseteq span (B)$ .

step 1.1step 1.2step 1.3L1L3

3.1

Claim 4. Suppose $E$ had a finite basis $C$ , say with $p$ elements. Then $C$ is a spanning set of $E$ with $p$ elements, so no linearly independent subset of $E$ is equinumerous with $N$ . But $B \subseteq E$ is linearly independent by step 1.4 and $B \approx N$ by step 2.1. So no finite basis exists and $E$ is infinite-dimensional over $F$ .

step 1.4step 2.1L8

4.1

Claim 3, that $B$ is a basis of $E$ . By step 1.4 the set $B$ is linearly independent and by step 2.2 it spans $E$ ; independence and spans computed in the linear subspace $E$ agree with those computed in $F^{N}$ , so $B$ is a basis of the vector space $E$ .

step 1.4step 2.2L2∎

Remarks

Agreement with the order-69 examples page. Claims 1 and 2 above are exactly claims 1 and 2 of $F^{N}$ is a vector space and the eventually zero families form a linear subspace of it that is the span of the standard unit families, which states that $E$ is a linear subspace of $F^{N}$ and that $E = span {e_{k} : k \in N}$ , for the same $E$ and the same $e_{k}$ . They are rebuilt here from 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 $u, v \in W$ rather than quoted, because an examples page is a leaf of the library and nothing outside it may depend on the items homed there; the statements agree, and neither is stronger than the other. Claims 3 and 4 are new: that page had no notion of independence or dimension available.
This is the counterweight to Every vector space has a basis. Here an infinite basis is exhibited and every step is explicit; there a basis is produced from Zorn's lemma and none is exhibited. The two extremes are placed on the same page on purpose, and the middle case is $R$ as a vector space over $Q$ has a basis, and every such basis is infinite; the existence proof exhibits none, where a basis exists by the same Zorn argument and this page exhibits none.
Infinite-dimensional is a negation, and that is all claim 4 asserts. No number and no cardinal is attached to $E$ : Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis assigns a dimension only to a space with a finite basis, and the fact that some basis of $E$ is equinumerous with $N$ is a statement about that basis, not about a dimension.

Depends on

Used by

Dependency tree · two levels

62 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

Sequence space (Wikipedia) (standard reference, not scraped)
Basis (linear algebra) (Wikipedia) (standard reference, not scraped)
Cambridge University Press excerpt: Vector spaces and bases (standard reference, not scraped)

The standard unit families ek∈FNek​∈FN form a basis of the linear subspace of eventually zero families: an explicit infinite basis, built with no choice principle