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On $\mathbb{R}$ the metrics $|x-y|$ and $\min(|x-y|,1)$ are uniformly but not Lipschitz equivalent

Statement refuted

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

On $\mathbb{R}$ take the usual metric $d(x,y) := |x-y|$ (The absolute value makes $\mathbb{R}$ 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) and

$\rho(x,y) := \min\{\, |x-y|,\ 1 \,\},$

the minimum being that of a two-element set of reals (Maximum and minimum of a set).

These are uniformly equivalent ( $\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology) and are not Lipschitz equivalent (Topologically, uniformly and Lipschitz equivalent metrics on a set), because a Lipschitz bound $\alpha\, d \le \rho$ with $\alpha > 0$ would force $d$ to be bounded by $1/\alpha$ , and $d$ is unbounded on the real line.

Facts & Assumptions

Given: The real line with $d(x,y) = |x-y|$ and $\rho = \min\{d, 1\}$ .

[L1]

$d$ is a metric on $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ 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, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

For any metric $d$ , the function $\min\{d,1\}$ is a metric, satisfies $\min\{d,1\} \le 1$ everywhere, and is uniformly equivalent to $d$ ( $\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, Topologically, uniformly and Lipschitz equivalent metrics on a set, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L3]

Lipschitz equivalence of $d$ and $\rho$ means there are reals $\alpha, \beta > 0$ with $\alpha\, d(x,y) \le \rho(x,y) \le \beta\, d(x,y)$ for all $x,y$ (Topologically, uniformly and Lipschitz equivalent metrics on a set).

[L4]

Archimedean property: for every real $w$ there is a natural $n \ge 1$ with $w < n \cdot 1_{\mathbb{R}}$ (Every complete ordered field is Archimedean); and $n \cdot 1_{\mathbb{R}} > 0$ for $n \ge 1$ (Canonical naturals are positive and strictly increasing).

[L5]

Inverses and scaling: $\alpha > 0$ gives $\alpha^{-1} > 0$ (Inverses of positives are positive, and reciprocation reverses order), and multiplying an inequality by a positive preserves it (Sign rules for products and monotonicity of multiplication); $0 < 1$ (The multiplicative identity is positive); trichotomy and transitivity (Ordered field, Complete ordered field (least-upper-bound property)); $|t| = t$ for $t \ge 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

Counterexample

technique · direct

1.1

$\rho$ is a metric on $\mathbb{R}$ and is uniformly equivalent to $d$ .

L1L2

1.2

$\rho(x,y) \le 1$ for all $x, y \in \mathbb{R}$ .

2.1

Suppose $d$ and $\rho$ were Lipschitz equivalent, with constants $\alpha, \beta > 0$ as in [L3]. Then for all $x, y$ we would have $\alpha\, |x - y| \le \rho(x,y) \le 1$ .

step 1.2L3L5

3.1

Apply the Archimedean property to $w := \alpha^{-1}$ , which is a positive real: there is a natural $n \ge 1$ with $\alpha^{-1} < n \cdot 1_{\mathbb{R}}$ , and multiplying by $\alpha > 0$ gives $1 < \alpha\, (n \cdot 1_{\mathbb{R}})$ . Taking $x := n \cdot 1_{\mathbb{R}}$ and $y := 0$ , so that $|x - y| = n \cdot 1_{\mathbb{R}}$ since $n \cdot 1_{\mathbb{R}} > 0$ , step 2.1 gives $\alpha\,(n \cdot 1_{\mathbb{R}}) \le 1$ , contradicting $1 < \alpha\,(n \cdot 1_{\mathbb{R}})$ by trichotomy.

step 2.1L4L5

4.1

No such constants exist, so $d$ and $\rho$ are uniformly equivalent metrics on $\mathbb{R}$ that are not Lipschitz equivalent; the implication from Lipschitz to uniform equivalence in Lipschitz equivalence implies uniform equivalence implies topological equivalence therefore does not reverse.

step 1.1step 3.1∎

Remarks

Only the lower Lipschitz bound fails. The upper one holds with $\beta = 1$ , since $\rho \le d$ always; it is the requirement that $\rho$ dominate a positive multiple of $d$ that a bounded metric can never meet against an unbounded one.
The same pair also shows boundedness is not topological (FALSE: boundedness of a metric space is determined by its topology, $\mathbb{R}$ carries both an unbounded and a bounded metric inducing the same topology), which is no accident: Lipschitz equivalence is exactly the level of the hierarchy that preserves boundedness, and this is the pair that sits just below it.
Every metric space furnishes such a pair unless it is already bounded, by $\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology; the real line is chosen only because its unboundedness is already recorded (The absolute value makes $\mathbb{R}$ 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).

Depends on

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Sources

Equivalence of metrics (Wikipedia) (standard reference, not scraped)
Lipschitz continuity (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 20: The Metric Topology (East Tennessee State University) (standard reference, not scraped)

On R\mathbb{R}R the metrics ∣x−y∣|x-y|∣x−y∣ and min⁡(∣x−y∣,1)\min(|x-y|,1)min(∣x−y∣,1) are uniformly but not Lipschitz equivalent