ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

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.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The cube $[-M,M]^n$ in $\mathbb{R}^n$ is totally bounded, with an explicit finite $\varepsilon$ -net of grid points and no appeal to the integer part

Example

Let $n \in \mathbb{N}$ with $n \ge 1$ , let $M \in \mathbb{R}$ with $M > 0$ , and let $(\mathbb{R}^n, d_2)$ carry the Euclidean metric ( $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it). The cube

$Q \;:=\; \{\, x \in \mathbb{R}^n : -M \le x_k \le M \ \text{ for every } k < n \,\}$

is a totally bounded metric subspace of $(\mathbb{R}^n,d_2)$ (Finite $\varepsilon$ -net and totally bounded metric space, Isometry, isometric embedding, and the subspace metric on a subset), and a finite $\varepsilon$ -net can be written down: choose a natural $m \ge 1$ with $1/m < \varepsilon/(2M\iota(n))$ , put $h := 2M/m$ , and take the grid

$G \;:=\; \{\, g \in Q : \text{each } g_k = -M + j_k h \text{ for some natural } j_k \le m \,\}.$

No integer part and no floor function is used: the index $j_k$ attached to a point of $Q$ is produced as a least natural meeting an inequality (The well-ordering principle).

Facts & Assumptions

Given: $n \ge 1$ , a real $M > 0$ , the cube $Q \subseteq \mathbb{R}^n$ with the metric $d_2$ restricted to it, and a real $\varepsilon > 0$ .

[L1]

$d_2(x,y) = \sqrt{\sum_{k<n}(x_k-y_k)^2}$ is a metric on $\mathbb{R}^n$ , and $d_\infty(x,y) = \max\{|x_k-y_k| : k<n\}$ is another ( $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, Finite sums and finite products, by recursion, Absolute value in an ordered field).

[L2]

$d_2(x,y) \le \iota(n)\, d_\infty(x,y)$ , since each $(x_k-y_k)^2 \le d_\infty(x,y)^2$ gives $d_2(x,y)^2 \le \iota(n) d_\infty(x,y)^2 \le (\iota(n)d_\infty(x,y))^2$ using $\iota(n) \ge 1$ , and squaring is monotone on the nonnegatives (Laws of finite sums and finite products, Squaring is monotone on the nonnegatives, Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L3]

A finite $\varepsilon$ -net for a metric space is a finite subset $F$ of it with the balls $B(y,\varepsilon)$ , $y \in F$ , covering the space, and total boundedness asks for one at every real $\varepsilon > 0$ ; balls of a subspace are traces of ambient balls (Finite $\varepsilon$ -net and totally bounded metric space, Open ball, closed ball and sphere in a metric space, Isometry, isometric embedding, and the subspace metric on a subset).

[L4]

A set listed as $\{a_0, \dots, a_p\}$ , that is the image of a function whose domain is a natural number, is finite (Open cover, subcover, compact metric space, and compact subset of a metric space).

[L5]

Every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[L6]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ ; reciprocals of positives are positive and reverse the order; and integer powers are those of Integer powers $a^m$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

Verification

technique · direct

1.1

Take a natural $m \ge 1$ with $1/m < \varepsilon/(2M\iota(n))$ , which exists because $2M\iota(n) > 0$ , and put $h := 2M/m > 0$ , so that $\iota(n) h = 2M\iota(n)/m < \varepsilon$ .

2.1

Let $G$ be the set of points of $\mathbb{R}^n$ each of whose coordinates is $-M + jh$ for some natural $j \le m$ ; every such point lies in $Q$ , since $0 \le jh \le mh = 2M$ gives $-M \le -M + jh \le M$ .

L1step 1.1

3.1

$G$ is finite: writing a natural $t < (m+1)^n$ in base $m+1$ gives digits $t_0, \dots, t_{n-1}$ , each a natural $\le m$ , and $t \mapsto$ the point with $k$ -th coordinate $-M + t_k h$ is a function from the natural number $(m+1)^n$ onto $G$ , so $G$ is listed by that function.

L4step 2.1

3.2

Let $y \in Q$ and $k < n$ ; the set of naturals $j \le m$ with $y_k \le -M + jh$ is nonempty, containing $m$ because $y_k \le M = -M + mh$ , so it has a least element $j_k$ .

L5step 2.1

4.1

Then $|y_k - (-M + j_k h)| \le h$ : if $j_k = 0$ then $y_k \le -M$ and also $y_k \ge -M$ , so the difference is $0$ ; and if $j_k \ge 1$ then minimality gives $y_k > -M + (j_k-1)h$ , so $-h < y_k - (-M+j_k h) \le 0$ .

step 3.2

5.1

Writing $g$ for the point of $G$ with $k$ -th coordinate $-M + j_k h$ , step 4.1 gives $d_\infty(y,g) \le h$ , hence $d_2(y,g) \le \iota(n)h < \varepsilon$ by step 1.1, so $y$ lies in the ball of radius $\varepsilon$ about $g$ in the subspace $Q$ .

L1L2L3step 1.1step 4.1

6.1

So $G$ is a finite $\varepsilon$ -net for $Q$ ; as $\varepsilon > 0$ was arbitrary, $Q$ is totally bounded.

L3step 3.1step 5.1∎

Remarks

A second proof, and why the explicit one is worth having. $Q$ is closed and bounded in $\mathbb{R}^n$ , hence compact (Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), hence totally bounded (A compact metric space is complete and totally bounded, and neither implication uses any choice principle), which proves the same statement in one line. The explicit grid is given because it exhibits the net rather than asserting that one exists, and because the count of grid points, $(m+1)^n$ , shows how the size of a net grows with the dimension — the feature that makes total boundedness a genuinely metric notion rather than a consequence of boundedness (FALSE: a bounded metric space is totally bounded).

Why the least index and not the integer part. The natural choice of $j_k$ is the integer part of $(y_k+M)/h$ , and this library has no integer-part function at this point in the reading order. Taking the least $j \le m$ with $y_k \le -M + jh$ produces the same index, using only that a nonempty set of naturals has a least element.

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: 137 results over 26 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

Totally bounded space (Wikipedia) (standard reference, not scraped)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)

The cube [−M,M]n[-M,M]^n[−M,M]n in Rn\mathbb{R}^nRn is totally bounded, with an explicit finite ε\varepsilonε-net of grid points and no appeal to the integer part