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Lipschitz equivalence implies uniform equivalence implies topological equivalence
Statement
Let and be metrics on the same set (Metric space: iff , 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 and are Lipschitz equivalent, they are uniformly equivalent.
- If and 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 and two metrics on it; a real .
Lipschitz equivalence: there are reals with for all (Topologically, uniformly and Lipschitz equivalent metrics on a set).
Uniform equivalence: for every real there are such that implies and implies , for all (Topologically, uniformly and Lipschitz equivalent metrics on a set).
Inverses and products of positives: gives (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)).
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).
Continuity of a map at a point in the - form (Continuity of a map between metric spaces, at a point and globally, in the - form); a map continuous at every point has open preimages of open sets (For a map of metric spaces the following agree: - continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and ).
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
Claim 1: assume [A1] and let . Put and , both positive since are. If then ; and if then , so after multiplying by . Hence and are uniformly equivalent.
Assume [A2]. Then the identity map is continuous at every point : given , the of [A2] satisfies , which is the - condition at ; symmetrically is continuous at every point, using .
By [L3] applied to the two continuous identity maps of step 1.2: the preimage under of a -open set is itself and is -open, so every -open set is -open; and symmetrically every -open set is -open. Hence , which is claim 2.
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 the metrics and have the same topology and are not uniformly equivalent (On the metrics and have the same topology and are not uniformly equivalent ↗); on the metrics and are uniformly equivalent and not Lipschitz equivalent (On the metrics and 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 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.
Depends on
- Topologically, uniformly and Lipschitz equivalent metrics on a set
- 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
- For a map of metric spaces the following agree: $\varepsilon$-$\delta$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f(\overline{A}) \subseteq \overline{f(A)}$
- Open ball, closed ball and sphere in a metric space
- Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$-$\delta$ form
- Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric
- Inverses of positives are positive, and reciprocation reverses order
- Sign rules for products and monotonicity of multiplication
- Order is preserved by adding a constant and by adding inequalities
- Ordered field
- Complete ordered field (least-upper-bound property)
Used by
- On (0,∞) the metrics |x-y| and |1/x - 1/y| have the same topology and are not uniformly equivalent Counterexample
- On ℝ the metrics |x-y| and min(|x-y|,1) are uniformly but not Lipschitz equivalent Counterexample
- ℝ carries both an unbounded and a bounded metric inducing the same topology Counterexample
- Equivalent norms, and the dictionary with equivalent metrics Definition
- C([0,1], ℝ) is complete, and on it the uniform metric and the supremum metric induce the same topology Example
- min(|x-y|, 1) on ℝ has the usual topology and diameter at most 1 Example
- The metrics d₁, d₂ and d_∞ on ℝⁿ are metrics and are Lipschitz equivalent, with explicit constants Example
- FALSE: boundedness of a metric space is determined by its topology False statement
- For n ≥ 1 the product topology on n copies of the usual topology of ℝ is the metric topology of d_∞ on ℝⁿ, and hence also of d₁ and d₂, so ℝⁿ as a product and ℝⁿ as a metric space are one space Lemma
- 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 Lemma
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.
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)