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The post-office metric d(x,y)=x+yd(x,y) = \|x\| + \|y\| for xyx \ne y on Rn\mathbb{R}^n, and its isolated points

Example

Let n1n \ge 1 and let Rn\mathbb{R}^n carry the Euclidean metric d2d_2 of Rn\mathbb{R}^n as the set of functions nRn \to \mathbb{R}, and d1d_1, d2d_2, dd_\infty are metrics on it. Write 0\mathbf{0} for the element of Rn\mathbb{R}^n with all coordinates 00 and

x:=d2(x,0)=k<nxk2.\|x\| := d_2(x, \mathbf{0}) = \sqrt{\textstyle\sum_{k<n} x_k^2} .

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

ρ(x,y):={0x=y,x+yxy.\rho(x,y) := \begin{cases} 0 & x = y, \\ \|x\| + \|y\| & x \ne y. \end{cases}

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

  1. ρ\rho is a metric on Rn\mathbb{R}^n (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric).
  2. Every x0x \ne \mathbf{0} is an isolated point of (Rn,ρ)(\mathbb{R}^n, \rho) (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space), because Bρ(x,x)={x}B_\rho\big(x, \|x\|\big) = \{x\}.
  3. 0\mathbf{0} is not an isolated point: every ball Bρ(0,r)B_\rho(\mathbf{0}, r) contains points other than 0\mathbf{0}.

So (Rn,ρ)(\mathbb{R}^n,\rho) has exactly one non-isolated point, which no metric equivalent to d2d_2 could achieve: in the Euclidean topology no point of Rn\mathbb{R}^n is isolated.

Facts & Assumptions

Given: A natural n1n \ge 1; elements x,y,zRnx, y, z \in \mathbb{R}^n; a real r>0r > 0; and the element wRnw \in \mathbb{R}^n with w0:=r/2w_0 := r/2 and wk:=0w_k := 0 for 1k<n1 \le 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 mnm \le n, and k<pλ=pλ\sum_{k<p}\lambda = p\lambda, so a sum all of whose terms are 00 is 00; and N\mathbb{N} contains 00, so the index 00 exists as soon as n1n \ge 1 (The natural numbers N\mathbb{N} (von Neumann)).

[L4]

Halving: for a real r>0r > 0 the element r/2:=r(1+1)1r/2 := r(1+1)^{-1} is positive with r/2+r/2=rr/2 + r/2 = r, hence r/2<rr/2 < r; this uses 0<10 < 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 aba \ge b rules out a<ba < 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}B_\rho(u,t) = \{v : \rho(u,v) < t\}, and uu is isolated in a set AuA \ni u when some ball meets AA only in uu (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\rho(u,v) \ge 0 for all u,vu,v, being 00 or a sum of two nonnegative numbers; ρ\rho is symmetric, since both defining clauses are; and ρ(u,v)=0\rho(u,v) = 0 exactly when u=vu = v, because for uvu \ne v the value u+v\|u\| + \|v\| vanishes only if u=v=0\|u\| = \|v\| = 0, that is only if u=v=0u = v = \mathbf{0}, contradicting uvu \ne v.

L1L5
1.2

Triangle inequality: if x=zx = z then ρ(x,z)=0ρ(x,y)+ρ(y,z)\rho(x,z) = 0 \le \rho(x,y) + \rho(y,z); if xzx \ne z and y=xy = x, then yzy \ne z and ρ(x,y)+ρ(y,z)=0+x+z=ρ(x,z)\rho(x,y) + \rho(y,z) = 0 + \|x\| + \|z\| = \rho(x,z); if xzx \ne z and y=zy = z, the same computation applies with the roles exchanged; and if xzx \ne z with yxy \ne x and yzy \ne z, then ρ(x,y)+ρ(y,z)=x+y+y+zx+z=ρ(x,z)\rho(x,y) + \rho(y,z) = \|x\| + \|y\| + \|y\| + \|z\| \ge \|x\| + \|z\| = \rho(x,z), since y0\|y\| \ge 0.

L1L5
1.3

The element ww satisfies k<nwk2=w02+0=(r/2)2\sum_{k<n} w_k^2 = w_0^2 + 0 = (r/2)^2, splitting the sum at index 1n1 \le n and using that all remaining terms are 00; hence w=(r/2)2=r/2<r\|w\| = \sqrt{(r/2)^2} = r/2 < r, and w0w \ne \mathbf{0} because w0=r/20w_0 = r/2 \ne 0.

L2L3L4
2.1

Claim 1: ρ\rho satisfies (M1) and (M2) by step 1.1 and (M3) by step 1.2, so it is a metric on Rn\mathbb{R}^n.

step 1.1step 1.2
3.1

Claim 2: let x0x \ne \mathbf{0}, so x>0\|x\| > 0 by [L1] and trichotomy. For yxy \ne x we get ρ(x,y)=x+yx\rho(x,y) = \|x\| + \|y\| \ge \|x\|, so yBρ(x,x)y \notin B_\rho(x,\|x\|); and xBρ(x,x)x \in B_\rho(x,\|x\|) since ρ(x,x)=0<x\rho(x,x) = 0 < \|x\|. Hence Bρ(x,x)={x}B_\rho(x,\|x\|) = \{x\} and xx is isolated in Rn\mathbb{R}^n.

step 1.1step 2.1L1L5L6
3.2

Claim 3: let r>0r > 0. For y0y \ne \mathbf{0} one has ρ(0,y)=0+y=y\rho(\mathbf{0},y) = \|\mathbf{0}\| + \|y\| = \|y\|, so the element ww of step 1.3 satisfies ρ(0,w)=w<r\rho(\mathbf{0},w) = \|w\| < r and w0w \ne \mathbf{0}; thus Bρ(0,r)B_\rho(\mathbf{0},r) contains a point other than 0\mathbf{0}, for every r>0r > 0, and 0\mathbf{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

  • The topology is almost discrete. Every singleton {x}\{x\} with x0x \ne \mathbf{0} is open by claim 2, so every subset of Rn{0}\mathbb{R}^n \setminus \{\mathbf{0}\} is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed); the only points a set has to be careful about are those near 0\mathbf{0}.
  • This is not equivalent to any of d1d_1, d2d_2, dd_\infty, not even topologically. Those three share a topology (The metrics d1d_1, d2d_2 and dd_\infty on Rn\mathbb{R}^n are metrics and are Lipschitz equivalent, with explicit constants) in which no point is isolated: for xRnx \in \mathbb{R}^n and r>0r > 0 the element yy with y0:=x0+r/2y_0 := x_0 + r/2 and yk:=xky_k := x_k for 1k<n1 \le k < n satisfies d2(x,y)=r/2<rd_2(x, y) = r/2 < r, by the computation of step 1.3, and differs from xx. In ρ\rho, by claim 2, every point but 0\mathbf{0} is isolated.
  • The same construction works over any metric space with a distinguished point, replacing x\|x\| by the distance to that point; nothing above uses more about \|\cdot\| than nonnegativity and vanishing exactly at 0\mathbf{0}.

Depends on

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