CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)

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On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ have the same topology and are not uniformly equivalent

Statement refuted

Refuted claim: topologically equivalent metrics are uniformly equivalent; equivalently, the implication "uniformly equivalent implies topologically equivalent" of Lipschitz equivalence implies uniform equivalence implies topological equivalence reverses.

Let $X : = (0, \infty) = {t \in R : t > 0}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) carry the metric $d (x, y) : = ∣ x - y ∣$ inherited from the real line (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and put

$ρ (x, y) : = ∣ \frac{1}{x} - \frac{1}{y} ∣ (x, y \in X) .$

Then $ρ$ is a metric on $X$ , the two metrics are topologically equivalent, and they are not uniformly equivalent (Topologically, uniformly and Lipschitz equivalent metrics on a set). So the second implication of the hierarchy is strict.

Facts & Assumptions

Given: The set $X = (0, \infty)$ , the metrics $d$ and $ρ$ above, the map $ι : X \to X$ with $ι (x) : = x^{- 1}$ , a point $a \in X$ , a real $ε > 0$ , and a real $δ > 0$ .

[L1]

$d_{R} (u, v) = ∣ u - v ∣$ is a metric on $R$ and restricts to a metric on any subset (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

Inverses: $u > 0$ gives $u^{- 1} > 0$ and $(u^{- 1})^{- 1} = u$ ; and $0 < u < v$ gives $0 < v^{- 1} < u^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order, Field).

[L3]

Absolute value: $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ u ∣ \geq 0$ , $∣ u ∣ = 0$ exactly when $u = 0$ , $∣ - u ∣ = ∣ u ∣$ , and $∣ u ∣ < c$ is equivalent to $- c < u < c$ for $c > 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L4]

Order arithmetic: scaling a strict inequality by a positive element (Sign rules for products and monotonicity of multiplication), adding a constant to an inequality (Order is preserved by adding a constant and by adding inequalities), transitivity and trichotomy (Ordered field, Complete ordered field (least-upper-bound property)); $0 < 1$ (The multiplicative identity is positive); halving a positive real; and the minimum of a two-element set of reals, which is one of the two (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L5]

Reciprocal Archimedean property: for every real $t > 0$ there is a natural $n \geq 1$ with $1 / n < t$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean); and $n \cdot 1_{R} > 0$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing, The natural numbers $N$ (von Neumann)).

[L6]

Open sets, balls, and the two notions of equivalence (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Topologically, uniformly and Lipschitz equivalent metrics on a set, Injection, surjection, bijection).

Counterexample

technique · direct

1.1

The map $ι$ is a well defined bijection of $X$ onto itself with $ι \circ ι = {i d}_{X}$ , since $x > 0$ gives $x^{- 1} > 0$ and $(x^{- 1})^{- 1} = x$ ; and for $x, y \in X$ one has $ρ (x, y) = d (ι x, ι y)$ together with the identity $ρ (x, y) = ∣ x - y ∣ (x y)^{- 1}$ , because $x^{- 1} - y^{- 1} = (y - x) (x y)^{- 1}$ and $∣ y - x ∣ = ∣ x - y ∣$ .

L2L3L4

2.1

$ρ$ is a metric on $X$ : it inherits symmetry and the triangle inequality from $d$ through $ι$ , and $ρ (x, y) = 0$ gives $d (ι x, ι y) = 0$ , hence $ι x = ι y$ and, $ι$ being injective, $x = y$ ; conversely $ρ (x, x) = 0$ .

step 1.1L1L2L3

2.2

Estimate A: put $δ_{A} : = \min {a / 2, ε a^{2} / 2} > 0$ . If $x \in X$ and $∣ x - a ∣ < δ_{A}$ then $x > a - δ_{A} \geq a - a / 2 = a / 2 > 0$ , so $x a > a^{2} / 2 > 0$ and $(x a)^{- 1} < 2 a^{- 2}$ ; hence $ρ (x, a) = ∣ x - a ∣ (x a)^{- 1} < δ_{A} (x a)^{- 1} < 2 δ_{A} a^{- 2} \leq ε$ when $x \neq a$ , and $ρ (a, a) = 0 < ε$ otherwise. So $B_{d} (a, δ_{A}) \subseteq B_{ρ} (a, ε)$ .

step 1.1L2L3L4

2.3

Uniform equivalence fails: suppose some $δ > 0$ satisfied $d (x, y) < δ \Rightarrow ρ (x, y) < 1$ for all $x, y \in X$ . Choose a natural $n \geq 1$ with $1 / n < δ$ and put $x : = 1 / n$ , $y : = 1 / (n + 1)$ , both in $X$ . Then $d (x, y) = ∣ 1 / n - 1 / (n + 1) ∣ < 1 / n < δ$ , because $0 < 1 / (n + 1) < 1 / n$ ; but $ρ (x, y) = ∣ n - (n + 1) ∣ = ∣ - 1 ∣ = 1$ , so $ρ (x, y) < 1$ fails. Hence no such $δ$ exists, and the pair $d, ρ$ is not uniformly equivalent.

step 1.1L2L3L4L5L6

3.1

Estimate B: apply estimate A at the point $ι (a) = a^{- 1} \in X$ to get $δ_{B} > 0$ with $∣ u - a^{- 1} ∣ < δ_{B} \Rightarrow ρ (u, a^{- 1}) < ε$ for $u \in X$ ; substituting $u : = ι (x) = x^{- 1}$ , which runs over $X$ as $x$ does, and using $ι \circ ι = {i d}_{X}$ , this reads $ρ (x, a) < δ_{B} \Rightarrow d (x, a) < ε$ , that is $B_{ρ} (a, δ_{B}) \subseteq B_{d} (a, ε)$ .

step 1.1step 2.2L2L3

4.1

Topological equivalence: if $U$ is $ρ$ -open and $a \in U$ , take $ε > 0$ with $B_{ρ} (a, ε) \subseteq U$ and then $δ_{A}$ from step 2.2, so $B_{d} (a, δ_{A}) \subseteq U$ and $U$ is $d$ -open; if $U$ is $d$ -open and $a \in U$ , take $ε > 0$ with $B_{d} (a, ε) \subseteq U$ and then $δ_{B}$ from step 3.1, so $B_{ρ} (a, δ_{B}) \subseteq U$ and $U$ is $ρ$ -open. Hence the two metric topologies coincide.

step 2.2step 3.1L6

5.1

So $d$ and $ρ$ are topologically equivalent metrics on $X$ that are not uniformly equivalent, which refutes the claim and shows that the implication from uniform to topological equivalence in Lipschitz equivalence implies uniform equivalence implies topological equivalence does not reverse.

step 2.1step 2.3step 4.1∎

Remarks

Why the failure is at the origin end. The pairs $1 / n$ and $1 / (n + 1)$ get arbitrarily close in $d$ while their images $n$ and $n + 1$ under $ι$ stay a fixed distance apart. Uniform equivalence would have to control this with a single $δ$ , and no single $δ$ can, because $ι$ stretches by the unbounded factor $(x y)^{- 1}$ near $0$ .
The two metrics are isometric copies of each other, via $ι$ : the map $ι : (X, ρ) \to (X, d)$ satisfies $d (ι x, ι y) = ρ (x, y)$ by step 1.1, so the two spaces are isometric (Isometry, isometric embedding, and the subspace metric on a subset). Being isometric as spaces says nothing about the identity map being uniformly bicontinuous, and that is exactly the distinction this example draws.
Completeness is the usual casualty. Uniform equivalence preserves Cauchy sequences and topological equivalence does not; here the sequence $1 / n$ is Cauchy for $d$ and not for $ρ$ , since $ρ (1 / n, 1 / m) = ∣ n - m ∣$ . Cauchy sequences in a metric space are defined on a later page and are not available here, so that comparison is orientation only and belongs to the completeness page.

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Sources

Equivalence of metrics (Wikipedia) (standard reference, not scraped)
Uniform continuity (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §20 (standard reference, not scraped)

On (0,∞)(0,∞) the metrics ∣x−y∣∣x−y∣ and ∣1/x−1/y∣∣1/x−1/y∣ have the same topology and are not uniformly equivalent