Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)
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The post-office metric d(x,y)=∥x∥+∥y∥ for x≠y on Rn, and its isolated points

Example

Let n≥1 and let Rn carry the Euclidean metric d2 of Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it. Write 0 for the element of Rn with all coordinates 0 and

∥x∥:=d2(x,0)=∑k<nxk2.

Define the post-office metric (also called the SNCF metric) by

ρ(x,y):={0x=y,∥x∥+∥y∥x≠y.

The name is the picture: to travel between two places you must first go to the central post office at 0. Then:

  1. ρ is a metric on Rn (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric).
  2. Every x≠0 is an isolated point of (Rn,ρ) (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space), because Bρ(x,∥x∥)={x}.
  3. 0 is not an isolated point: every ball Bρ(0,r) contains points other than 0.

So (Rn,ρ) has exactly one non-isolated point, which no metric equivalent to d2 could achieve: in the Euclidean topology no point of Rn is isolated.

Facts & Assumptions

Given: A natural n≥1; elements x,y,z∈Rn; a real r>0; and the element w∈Rn with w0:=r/2 and wk:=0 for 1≤k<n.

[L2]

Finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): splitting a sum at an index m≤n, and ∑k<pλ=pλ, so a sum all of whose terms are 0 is 0; and N contains 0, so the index 0 exists as soon as n≥1 (The natural numbers N (von Neumann)).

[L4]

Halving: for a real r>0 the element r/2:=r(1+1)−1 is positive with r/2+r/2=r, hence r/2<r; this uses 0<1, positivity of inverses of positives, and positivity of products of positives (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication, Field, Ordered field).

[L5]

Order arithmetic: inequalities may be added, a nonnegative term may be dropped from the larger side, and by trichotomy a≥b rules out a<b (Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)).

[L6]

Balls, isolated points: Bρ(u,t)={v:ρ(u,v)<t}, and u is isolated in a set A∋u when some ball meets A only in u (Open ball, closed ball and sphere in a metric space, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

Verification

technique · direct
1.1

Basic values: ρ(u,v)≥0 for all u,v, being 0 or a sum of two nonnegative numbers; ρ is symmetric, since both defining clauses are; and ρ(u,v)=0 exactly when u=v, because for u≠v the value ∥u∥+∥v∥ vanishes only if ∥u∥=∥v∥=0, that is only if u=v=0, contradicting u≠v.

L1L5
1.2

Triangle inequality: if x=z then ρ(x,z)=0≤ρ(x,y)+ρ(y,z); if x≠z and y=x, then y≠z and ρ(x,y)+ρ(y,z)=0+∥x∥+∥z∥=ρ(x,z); if x≠z and y=z, the same computation applies with the roles exchanged; and if x≠z with y≠x and y≠z, then ρ(x,y)+ρ(y,z)=∥x∥+∥y∥+∥y∥+∥z∥≥∥x∥+∥z∥=ρ(x,z), since ∥y∥≥0.

L1L5
1.3

The element w satisfies ∑k<nwk2=w02+0=(r/2)2, splitting the sum at index 1≤n and using that all remaining terms are 0; hence ∥w∥=(r/2)2=r/2<r, and w≠0 because w0=r/2≠0.

L2L3L4
2.1

Claim 1: ρ satisfies (M1) and (M2) by step 1.1 and (M3) by step 1.2, so it is a metric on Rn.

step 1.1step 1.2
3.1

Claim 2: let x≠0, so ∥x∥>0 by [L1] and trichotomy. For y≠x we get ρ(x,y)=∥x∥+∥y∥≥∥x∥, so y∉Bρ(x,∥x∥); and x∈Bρ(x,∥x∥) since ρ(x,x)=0<∥x∥. Hence Bρ(x,∥x∥)={x} and x is isolated in Rn.

step 1.1step 2.1L1L5L6
3.2

Claim 3: let r>0. For y≠0 one has ρ(0,y)=∥0∥+∥y∥=∥y∥, so the element w of step 1.3 satisfies ρ(0,w)=∥w∥<r and w≠0; thus Bρ(0,r) contains a point other than 0, for every r>0, and 0 is not isolated.

step 1.3step 2.1L1L6
4.1

Claims 1, 2 and 3 hold by steps 2.1, 3.1 and 3.2.

step 2.1step 3.1step 3.2∎

Remarks

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources