ExampleConstruction: 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 chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$F^{\mathbb{N}}$ 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 $F^{\mathbb{N}}$ be the function space of all families $x : \mathbb{N} \to F$ 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 $x = (x_n)_{n \in \mathbb{N}}$ with $x_n := x(n)$ ; the index runs over $\mathbb{N}$ , which contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)). Put

$E \;:=\; \{\, x \in F^{\mathbb{N}} \;:\; \text{there is } N \in \mathbb{N} \text{ with } x_n = 0_F \text{ for every } n \ge N \,\},$

the set of eventually zero families, and for $k \in \mathbb{N}$ let $e_k \in F^{\mathbb{N}}$ be the standard unit family given by $e_k(k) = 1_F$ and $e_k(n) = 0_F$ for $n \ne k$ . Then:

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

Facts & Assumptions

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

[L1]

$F^{X}$ is a vector space over $F$ with $(x+y)(n) = x(n)+y(n)$ , $(\lambda x)(n) = \lambda x(n)$ , 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 $\lambda u + v$ 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 $\lambda u + v \in W$ for all $\lambda \in F$ and $u, v \in W$ , Linear subspace of a vector space).

[L3]

$\operatorname{span}(S)$ is the set of linear combinations $\sum_{k<N} \lambda_k v_k$ of elements of $S$ , and it is the smallest linear subspace containing $S$ ( $\operatorname{span}(S)$ is exactly the set of linear combinations of finite lists of elements of $S$ , and $\operatorname{span}(\varnothing) = \{0_V\}$ , Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ).

[L4]

Finite sums in a vector space: $\sum_{k<0} u_k = 0$ and $\sum_{k<\sigma(N)} u_k = \bigl(\sum_{k<N} u_k\bigr) + u_N$ (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]

Induction on $\mathbb{N}$ (The principle of mathematical induction).

[L6]

The order of $\mathbb{N}$ is total and reflexive, $m < \sigma(N)$ is equivalent to $m \le N$ , and $m < n$ implies $m \ne n$ ( $\le$ is a linear order on $\mathbb{N}$ , Order on the natural numbers, On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ).

[L7]

In a field: $0_F \lambda = 0_F$ (Multiplication by zero: $0 \cdot a = 0$ ) and multiplication is commutative, so $\lambda 0_F = 0_F$ ; $\lambda 1_F = \lambda$ ; $0_F + 0_F = 0_F$ and $0_F$ is the additive identity; and $1_F \ne 0_F$ (Field).

Verification

technique · direct

1.1

$F^{\mathbb{N}}$ is a vector space over $F$ , being the function space on the index set $\mathbb{N}$ , 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 $\lambda \in F$ and $u, v \in E$ , with witnesses $N_1$ for $u$ and $N_2$ for $v$ . The order of $\mathbb{N}$ is total, so one of $N_1, N_2$ is at least the other; let $N$ be that one. For $n \ge N$ we have $n \ge N_1$ and $n \ge N_2$ , hence $(\lambda u + v)_n = \lambda u_n + v_n = \lambda 0_F + 0_F = 0_F$ . So $N$ witnesses $\lambda u + v \in E$ .

L1L6L7

1.4

Each $e_k$ lies in $E$ : if $n \ge \sigma(k)$ then $k < n$ , so $n \ne k$ and $e_k(n) = 0_F$ ; thus $\sigma(k)$ is a witness.

L6L7

1.5

For $x \in F^{\mathbb{N}}$ and $N \in \mathbb{N}$ , 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 $N \le n$ . 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 $s_{\sigma(N)} = s_N + x_N e_N$ , so $s_{\sigma(N)}(n) = s_N(n) + x_N e_N(n)$ ; for $n < N$ this is $x_n + x_N 0_F = x_n$ , since $n \ne N$ ; for $n = N$ it is $0_F + x_N 1_F = x_N$ ; and for $\sigma(N) \le n$ we have $N \le n$ and $n \ne N$ , so it is $0_F + x_N 0_F = 0_F$ . As $n < \sigma(N)$ is equivalent to $n \le N$ , this is the claim at $\sigma(N)$ .

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 $N \ge N$ and $c_N = 1_F \ne 0_F$ . Hence $E \ne F^{\mathbb{N}}$ , which is claim 3.

L1L6L7

2.1

$E$ is a linear subspace of $F^{\mathbb{N}}$ , 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 $N \le n$ the sum is $0_F$ while $x_n = 0_F$ by the choice of $N$ . So $x$ is a linear combination of elements of $\{\, e_k : k \in \mathbb{N} \,\}$ .

step 1.5L1

3.1

Claim 2. By step 2.2 every element of $E$ is a linear combination of elements of $\{\, e_k : k \in \mathbb{N} \,\}$ , 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 ( $\operatorname{span}(S)$ is exactly the set of linear combinations of finite lists of elements of $S$ , and $\operatorname{span}(\varnothing) = \{0_V\}$ ), which is exactly why the span of all the $e_k$ is the eventually zero families and not all of $F^{\mathbb{N}}$ . 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 $\mathbb{N}$ : 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 19 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

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)

FNF^{\mathbb{N}}FN is a vector space and the eventually zero families form a linear subspace of it that is the span of the standard unit families