Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)
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The cube [−M,M]n in Rn is totally bounded, with an explicit finite ε-net of grid points and no appeal to the integer part

Example

Let n∈N with n≥1, let M∈R with M>0, and let (Rn,d2) carry the Euclidean metric (Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it). The cube

Q  :=  { x∈Rn:−M≤xk≤M  for every k<n }

is a totally bounded metric subspace of (Rn,d2) (Finite ε-net and totally bounded metric space, Isometry, isometric embedding, and the subspace metric on a subset), and a finite ε-net can be written down: choose a natural m≥1 with 1/m<ε/(2Mι(n)), put h:=2M/m, and take the grid

G  :=  { g∈Q:each gk=−M+jkh for some natural jk≤m }.

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

Facts & Assumptions

Given: n≥1, a real M>0, the cube Q⊆Rn with the metric d2 restricted to it, and a real ε>0.

[L1]

d2(x,y)=∑k<n(xk−yk)2 is a metric on Rn, and d∞(x,y)=max⁡{∣xk−yk∣:k<n} is another (Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, Finite sums and finite products, by recursion, Absolute value in an ordered field).

[L2]

d2(x,y)≤ι(n) d∞(x,y), since each (xk−yk)2≤d∞(x,y)2 gives d2(x,y)2≤ι(n)d∞(x,y)2≤(ι(n)d∞(x,y))2 using ι(n)≥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 a≥0 with (a)2=a; the positives are {x2:x≠0}, The canonical natural ι(n)=n⋅1F of a field).

[L3]

A finite ε-net for a metric space is a finite subset F of it with the balls B(y,ε), y∈F, covering the space, and total boundedness asks for one at every real ε>0; balls of a subspace are traces of ambient balls (Finite ε-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 {a0,…,ap}, 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 N has a least element (The well-ordering principle).

[L6]

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

Verification

technique · direct
1.1

Take a natural m≥1 with 1/m<ε/(2Mι(n)), which exists because 2Mι(n)>0, and put h:=2M/m>0, so that ι(n)h=2Mι(n)/m<ε.

L6
2.1

Let G be the set of points of Rn each of whose coordinates is −M+jh for some natural j≤m; every such point lies in Q, since 0≤jh≤mh=2M gives −M≤−M+jh≤M.

L1step 1.1
3.1

G is finite: writing a natural t<(m+1)n in base m+1 gives digits t0,…,tn−1, each a natural ≤m, and t↦ the point with k-th coordinate −M+tkh 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∈Q and k<n; the set of naturals j≤m with yk≤−M+jh is nonempty, containing m because yk≤M=−M+mh, so it has a least element jk.

L5step 2.1
4.1

Then ∣yk−(−M+jkh)∣≤h: if jk=0 then yk≤−M and also yk≥−M, so the difference is 0; and if jk≥1 then minimality gives yk>−M+(jk−1)h, so −h<yk−(−M+jkh)≤0.

step 3.2
5.1

Writing g for the point of G with k-th coordinate −M+jkh, step 4.1 gives d∞(y,g)≤h, hence d2(y,g)≤ι(n)h<ε by step 1.1, so y lies in the ball of radius ε about g in the subspace Q.

L1L2L3step 1.1step 4.1
6.1

So G is a finite ε-net for Q; as ε>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 Rn, hence compact (Heine-Borel in Rn: with the Euclidean metric a subset of Rn 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 jk is the integer part of (yk+M)/h, and this library has no integer-part function at this point in the reading order. Taking the least j≤m with yk≤−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 · two levels

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Sources