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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 dd and dd' be metrics on the same set XX (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = 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:

  1. If dd and dd' are Lipschitz equivalent, they are uniformly equivalent.
  2. If dd and dd' 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 XX and two metrics d,dd, d' on it; a real ε>0\varepsilon > 0.

[A1]

Lipschitz equivalence: there are reals α,β>0\alpha, \beta > 0 with αd(x,y)d(x,y)βd(x,y)\alpha\, d(x,y) \le d'(x,y) \le \beta\, d(x,y) for all x,yXx, y \in X (Topologically, uniformly and Lipschitz equivalent metrics on a set).

[A2]

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

[L1]

Inverses and products of positives: γ>0\gamma > 0 gives γ1>0\gamma^{-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).

Proof

technique · direct
1.1

Claim 1: assume [A1] and let ε>0\varepsilon > 0. Put δ:=εβ1\delta := \varepsilon \beta^{-1} and δ:=αε\delta' := \alpha \varepsilon, both positive since α,β,ε\alpha, \beta, \varepsilon are. If d(x,y)<δd(x,y) < \delta then d(x,y)βd(x,y)<βδ=εd'(x,y) \le \beta\, d(x,y) < \beta\delta = \varepsilon; and if d(x,y)<δd'(x,y) < \delta' then αd(x,y)d(x,y)<αε\alpha\, d(x,y) \le d'(x,y) < \alpha\varepsilon, so d(x,y)<εd(x,y) < \varepsilon after multiplying by α1>0\alpha^{-1} > 0. Hence dd and dd' are uniformly equivalent.

A1L1L2
1.2

Assume [A2]. Then the identity map id:(X,d)(X,d)\mathrm{id} : (X,d) \to (X,d') is continuous at every point aXa \in X: given ε>0\varepsilon > 0, the δ\delta of [A2] satisfies d(x,a)<δd(x,a)<εd(x,a) < \delta \Rightarrow d'(x,a) < \varepsilon, which is the ε\varepsilon-δ\delta condition at aa; symmetrically id:(X,d)(X,d)\mathrm{id} : (X,d') \to (X,d) is continuous at every point, using δ\delta'.

A2L3
2.1

By [L3] applied to the two continuous identity maps of step 1.2: the preimage under id:(X,d)(X,d)\mathrm{id} : (X,d) \to (X,d') of a dd'-open set VV is VV itself and is dd-open, so every dd'-open set is dd-open; and symmetrically every dd-open set is dd'-open. Hence Td=Td\mathcal{T}_d = \mathcal{T}_{d'}, which is claim 2.

step 1.2L3L4
3.1

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.

step 1.1step 2.1

Remarks

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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