ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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The map $x \mapsto x / (1 + ∣ x ∣)$ is a uniformly continuous homeomorphism from $R$ to $(- 1, 1)$ whose inverse is not uniformly continuous

Example

The function $h (x) = x / (1 + ∣ x ∣)$ maps $R$ onto $(- 1, 1)$ with inverse $h^{- 1} (t) = t / (1 - ∣ t ∣)$ . It is uniformly continuous, but its inverse is not.

Facts & Assumptions

Given: The usual metric uniformities on $R$ and $(- 1, 1)$ .

[L1]

The metric dictionary translates metric uniform continuity into uniform continuity (A metric on a nonempty set generates an entourage uniformity whose induced topology and uniformly continuous maps are the usual metric notions, and this uniformity is separated).

[L2]

Absolute value is nonnegative (Basic properties of the absolute value) and satisfies the triangle inequality (The triangle inequality).

[L3]

The reciprocal form of the Archimedean property says that $1 / n \to 0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Verification

technique · direct

1.1

Direct algebra gives $∣ h (x) - h (y) ∣ \leq 2 ∣ x - y ∣$ , so $h$ is uniformly continuous; its displayed inverse and the usual open-interval formulas make it a homeomorphism.

L1L2

1.2

Put $a_{n} = n / (n + 1)$ and $b_{n} = (n + 1) / (n + 2)$ . Then $∣ a_{n} - b_{n} ∣ \to 0$ , while $∣ h^{- 1} (a_{n}) - h^{- 1} (b_{n}) ∣ = 1$ .

L2L3

2.1

Thus $h^{- 1}$ is not uniformly continuous, so this homeomorphism is not a uniform isomorphism (Uniform embedding and uniform isomorphism, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

step 1.1step 1.2∎

Depends on

Used by

FALSE: every uniformizable topology has a unique compatible uniformity False statement

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The map x↦x/(1+∣x∣)x↦x/(1+∣x∣) is a uniformly continuous homeomorphism from RR to (−1,1)(−1,1) whose inverse is not uniformly continuous