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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$F^{N}$ is a vector space and the eventually zero families form a linear subspace of it that is the span of the standard unit families

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 $X = n = {0, 1, \dots, n - 1}$ ), written $x = (x_{n})_{n \in N}$ with $x_{n} : = x (n)$ ; the index runs over $N$ , which 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 set of eventually zero families, and for $k \in N$ let $e_{k} \in F^{N}$ be the standard unit family given by $e_{k} (k) = 1_{F}$ and $e_{k} (n) = 0_{F}$ for $n \neq k$ . Then:

$E$ is a linear subspace of $F^{N}$ (Linear subspace of a vector space);
$E = span {e_{k} : k \in N}$ (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ );
$E \neq F^{N}$ , the constant family at $1_{F}$ lying outside $E$ .

Facts & Assumptions

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

[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 subset closed under $λ u + v$ is a linear subspace, and a linear subspace satisfies (W1), (W2), (W3) (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).

[L3]

$span (S)$ is the set of linear combinations $\sum_{k < N} λ_{k} v_{k}$ of elements of $S$ , and it is the smallest linear subspace containing $S$ ( $span (S)$ 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]

Finite sums in a vector space: $\sum_{k < 0} u_{k} = 0$ and $\sum_{k < σ (N)} u_{k} = (\sum_{k < N} u_{k}) + u_{N}$ (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]

Induction on $N$ (The principle of mathematical induction).

[L6]

The order of $N$ is total and reflexive, $m < σ (N)$ is equivalent to $m \leq N$ , 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: $m < n ⟺ m \in n$ ).

[L7]

In a field: $0_{F} λ = 0_{F}$ (Multiplication by zero: $0 \cdot a = 0$ ) and multiplication is commutative, so $λ 0_{F} = 0_{F}$ ; $λ 1_{F} = λ$ ; $0_{F} + 0_{F} = 0_{F}$ and $0_{F}$ is the additive identity; and $1_{F} \neq 0_{F}$ (Field).

Verification

technique · direct

1.1

$F^{N}$ is a vector space over $F$ , being the function space on the index set $N$ , and its zero is the constant family at $0_{F}$ .

1.2

$E$ is nonempty: the zero family has $x_{n} = 0_{F}$ for every $n$ , so $N = 0$ witnesses that it lies in $E$ .

1.3

$E$ is closed under the one-step expression. Let $λ \in F$ and $u, v \in E$ , with witnesses $N_{1}$ for $u$ and $N_{2}$ for $v$ . The order of $N$ is total, so one of $N_{1}, N_{2}$ is at least the other; let $N$ be that one. For $n \geq N$ we have $n \geq N_{1}$ and $n \geq N_{2}$ , hence $(λ u + v)_{n} = λ u_{n} + v_{n} = λ 0_{F} + 0_{F} = 0_{F}$ . So $N$ witnesses $λ u + v \in E$ .

L1L6L7

1.4

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

L6L7

1.5

For $x \in F^{N}$ and $N \in N$ , the finite sum $s_{N} : = \sum_{k < N} x_{k} e_{k}$ satisfies $s_{N} (n) = x_{n}$ for $n < N$ and $s_{N} (n) = 0_{F}$ for $N \leq n$ . By induction on $N$ : at $N = 0$ the sum is the zero family, there is no $n < 0$ , and the second clause holds. Assuming it at $N$ , we have $s_{σ (N)} = s_{N} + x_{N} e_{N}$ , so $s_{σ (N)} (n) = s_{N} (n) + x_{N} e_{N} (n)$ ; for $n < N$ this is $x_{n} + x_{N} 0_{F} = x_{n}$ , since $n \neq N$ ; for $n = N$ it is $0_{F} + x_{N} 1_{F} = x_{N}$ ; and for $σ (N) \leq n$ we have $N \leq n$ and $n \neq N$ , so it is $0_{F} + x_{N} 0_{F} = 0_{F}$ . As $n < σ (N)$ is equivalent to $n \leq N$ , this is the claim at $σ (N)$ .

L1L4L5L6L7

1.6

The constant family $c$ with $c_{n} = 1_{F}$ for every $n$ does not lie in $E$ : for any candidate witness $N$ we have $N \geq N$ and $c_{N} = 1_{F} \neq 0_{F}$ . Hence $E \neq F^{N}$ , which is claim 3.

L1L6L7

2.1

$E$ is a linear subspace of $F^{N}$ , by the one-step test applied to steps 1.2 and 1.3; this is claim 1.

step 1.2step 1.3L2

2.2

If $x \in E$ with witness $N$ , then $x = \sum_{k < N} x_{k} e_{k}$ . Indeed the two families agree at every $n$ : at $n < N$ both are $x_{n}$ , and at $N \leq n$ the sum is $0_{F}$ while $x_{n} = 0_{F}$ by the choice of $N$ . So $x$ is a linear combination of elements of ${e_{k} : k \in N}$ .

step 1.5L1

3.1

Claim 2. By step 2.2 every element of $E$ is a linear combination of elements of ${e_{k} : k \in N}$ , hence lies in its span. Conversely each $e_{k}$ lies in $E$ by step 1.4 and $E$ is a linear subspace by step 2.1, so the span, being the smallest linear subspace containing all the $e_{k}$ , is contained in $E$ .

step 1.4step 2.1step 2.2L3

4.1

Claim 1 is step 2.1, claim 2 is step 3.1 and claim 3 is step 1.6.

step 1.6step 2.1step 3.1∎

Remarks

The spanning set is infinite and every combination is finite. The span of an infinite set consists of the vectors built from finitely many of its elements ( $span (S)$ is exactly the set of linear combinations of finite lists of elements of $S$ , and $span (\emptyset) = {0_{V}}$ ), which is exactly why the span of all the $e_{k}$ is the eventually zero families and not all of $F^{N}$ . Claim 3 is the concrete form of that distinction.
No topology and no convergence are involved. "Eventually zero" is a purely algebraic condition on a family indexed by $N$ : some tail is identically $0_{F}$ . Nothing here needs an order or a metric on $F$ , and $F$ is an arbitrary field.
The witness $N$ is not unique and nothing above assumes it is. If $N$ witnesses that $x$ is eventually zero then so does every larger natural number, which is what makes the argument in step 1.3 work: two families are handled by taking the larger of their witnesses.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Examples of vector spaces (Wikipedia) (standard reference, not scraped)
Sequence space (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed. (free PDF, CC BY-NC) (standard reference, not scraped)

FNFN is a vector space and the eventually zero families form a linear subspace of it that is the span of the standard unit families