TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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Lipschitz equivalence implies uniform equivalence implies topological equivalence

Statement

Let $d$ and $d^{'}$ be metrics on the same set $X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), with the three equivalences as in Topologically, uniformly and Lipschitz equivalent metrics on a set. Then:

If $d$ and $d^{'}$ are Lipschitz equivalent, they are uniformly equivalent.
If $d$ and $d^{'}$ are uniformly equivalent, they are topologically equivalent.

Strictness is not claimed here. The theorem asserts the two implications and nothing more; that neither reverses is witnessed by explicit pairs of metrics on the companion page, and those witnesses are not prerequisites of this theorem. See the first remark below.

Facts & Assumptions

Given: A set $X$ and two metrics $d, d^{'}$ on it; a real $ε > 0$ .

[A1]

Lipschitz equivalence: there are reals $α, β > 0$ with $α d (x, y) \leq d^{'} (x, y) \leq β d (x, y)$ for all $x, y \in X$ (Topologically, uniformly and Lipschitz equivalent metrics on a set).

[A2]

Uniform equivalence: for every real $ε > 0$ there are $δ, δ^{'} > 0$ such that $d (x, y) < δ$ implies $d^{'} (x, y) < ε$ and $d^{'} (x, y) < δ^{'}$ implies $d (x, y) < ε$ , for all $x, y \in X$ (Topologically, uniformly and Lipschitz equivalent metrics on a set).

[L1]

Inverses and products of positives: $γ > 0$ gives $γ^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order), and a product of positives is positive; multiplying an inequality by a positive preserves it, in the strict form of Sign rules for products and monotonicity of multiplication and, with the case of equality settled by totality, in the nonstrict form (Ordered field, Complete ordered field (least-upper-bound property)).

[L2]

Transitivity of the order, and addition of a constant to an inequality (Order is preserved by adding a constant and by adding inequalities, Ordered field).

[L3]

Continuity of a map at a point in the $ε$ - $δ$ form (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form); a map continuous at every point has open preimages of open sets (Metric continuity characterisations, with countable choice for the sequential converse).

[L4]

Open sets and the metric topology (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); membership in a ball (Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

1.1A1L1L2

Claim 1: assume [A1] and let $ε > 0$ . Put $δ : = ε β^{- 1}$ and $δ^{'} : = α ε$ , both positive since $α, β, ε$ are. If $d (x, y) < δ$ then $d^{'} (x, y) \leq β d (x, y) < β δ = ε$ ; and if $d^{'} (x, y) < δ^{'}$ then $α d (x, y) \leq d^{'} (x, y) < α ε$ , so $d (x, y) < ε$ after multiplying by $α^{- 1} > 0$ . Hence $d$ and $d^{'}$ are uniformly equivalent.

1.2A2L3

Assume [A2]. Then the identity map $i d : (X, d) \to (X, d^{'})$ is continuous at every point $a \in X$ : given $ε > 0$ , the $δ$ of [A2] satisfies $d (x, a) < δ \Rightarrow d^{'} (x, a) < ε$ , which is the $ε$ - $δ$ condition at $a$ ; symmetrically $i d : (X, d^{'}) \to (X, d)$ is continuous at every point, using $δ^{'}$ .

2.1step 1.2L3L4

By [L3] applied to the two continuous identity maps of step 1.2: the preimage under $i d : (X, d) \to (X, d^{'})$ of a $d^{'}$ -open set $V$ is $V$ itself and is $d$ -open, so every $d^{'}$ -open set is $d$ -open; and symmetrically every $d$ -open set is $d^{'}$ -open. Hence $T_{d} = T_{d^{'}}$ , which is claim 2.

3.1step 1.1step 2.1∎

Claims 1 and 2 are established by steps 1.1 and 2.1, so Lipschitz equivalence implies uniform equivalence and uniform equivalence implies topological equivalence.

Remarks

Neither implication reverses, and the witnesses are on the companion page. On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ have the same topology and are not uniformly equivalent (On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ have the same topology and are not uniformly equivalent ↗); on $R$ the metrics $∣ x - y ∣$ and $\min {∣ x - y ∣, 1}$ are uniformly equivalent and not Lipschitz equivalent (On $R$ the metrics $∣ x - y ∣$ and $\min (∣ x - y ∣, 1)$ are uniformly but not Lipschitz equivalent ↗). Those two items are read here as orientation only: this theorem does not depend on them, and its statement claims nothing about strictness.
Uniform equivalence is strictly more than "both identities are continuous". Continuity of both identity maps is exactly topological equivalence, by the argument of step 2.1 read in reverse; uniform equivalence additionally demands that one $δ$ serve at every point at once, and that is the whole difference between the two conditions.
What each level preserves. Lipschitz equivalence preserves boundedness and changes diameters by at most a constant factor; uniform equivalence preserves Cauchy sequences and uniform continuity, notions taken up on a later page and not defined here; topological equivalence preserves the open sets and everything defined from them, and nothing else.

Lipschitz equivalence implies uniform equivalence implies topological equivalence

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources