ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Horizontal translations of Z on the Euclidean plane are proper but not cobounded

Example

Let $Z$ act on $R^{2}$ by horizontal translations $n \cdot (x, y) : = (x + n, y) .$ This action is isometric and proper, but it is not cobounded.

Facts & Assumptions

Given: The Euclidean metric on $R^{2}$ and the translation action $n \cdot (x, y) : = (x + n, y)$ of $Z$ .

[L1]

The Euclidean distance is a metric on $R^{2}$ ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

[L2]

Properness and coboundedness are the conditions of Isometric, proper, and cobounded actions on metric spaces.

Verification

technique · direct

1.1

Horizontal translation preserves Euclidean distance, so the action is isometric. If bounded sets $B, C \subseteq R^{2}$ meet after translation by $n \in Z$ , then the $x$ -coordinates of points in $B$ and $C$ differ by $n$ , so only finitely many integers occur. Thus the action is proper in the sense of [L2].

L1L2algebra

1.2

Every orbit is a horizontal line $Z + x$ at fixed $y$ -coordinate. So the distance from $(0, m)$ to every orbit point of $(0, 0)$ is at least $∣ m ∣$ , and these distances are unbounded as $m \to \infty$ . Hence no bounded set of translates covers $R^{2}$ , so the action is not cobounded.

L2algebra

2.1

Therefore the action is proper but not cobounded.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

C. Löh, Geometric Group Theory, Sections 4.4 and 5.1 (standard reference, not scraped)