False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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FALSE: boundedness of a metric space is determined by its topology

Statement

False claim: boundedness is a topological property of a metric space; that is, if $d$ and $d'$ are topologically equivalent metrics on a set $X$ (Topologically, uniformly and Lipschitz equivalent metrics on a set) and $(X,d)$ is a bounded metric space (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), then $(X,d')$ is bounded as well.

Equivalently, the false claim says that 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) determines whether the space is bounded. It does not: every metric space carries a bounded metric with exactly the same topology, so as soon as one unbounded metric space exists the claim collapses.

Facts & Assumptions

Given: The real line $\mathbb{R}$ with its usual metric $d_{\mathbb{R}}(u,v) = |u-v|$ , and the metric $\rho(u,v) := \min\{\, d_{\mathbb{R}}(u,v),\ 1 \,\}$ (Maximum and minimum of a set).

[L1]

$d_{\mathbb{R}}$ is a metric on $\mathbb{R}$ and $(\mathbb{R}, d_{\mathbb{R}})$ is not a bounded metric space: no ball contains $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

For any metric $d$ the function $\min\{d,1\}$ is a metric, is bounded with diameter at most $1$ on a nonempty space, and is uniformly equivalent to $d$ ( $\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

Uniform equivalence implies topological equivalence (Lipschitz equivalence implies uniform equivalence implies topological equivalence, Topologically, uniformly and Lipschitz equivalent metrics on a set).

Refutation

technique · direct

1.1

By [L1] the metric $d_{\mathbb{R}}$ makes $\mathbb{R}$ a metric space that is not bounded.

1.2

By [L2] the function $\rho = \min\{d_{\mathbb{R}}, 1\}$ is a metric on $\mathbb{R}$ , the space $(\mathbb{R}, \rho)$ is bounded with $\operatorname{diam}(\mathbb{R}) \le 1$ , and $\rho$ is uniformly equivalent to $d_{\mathbb{R}}$ .

2.1

By [L3] the two metrics are therefore topologically equivalent: $\mathcal{T}_\rho = \mathcal{T}_{d_{\mathbb{R}}}$ .

step 1.2L3

3.1

So $d_{\mathbb{R}}$ and $\rho$ are topologically equivalent metrics on the same set, $(\mathbb{R},\rho)$ is bounded and $(\mathbb{R},d_{\mathbb{R}})$ is not; the claim fails, and boundedness is a property of the metric and not of the topology.

step 1.1step 2.1∎

Remarks

What is true instead. Boundedness is preserved by Lipschitz equivalence, since $d' \le \beta d$ turns a ball for $d$ into a ball for $d'$ (Topologically, uniformly and Lipschitz equivalent metrics on a set). It is the two weaker equivalences that lose it, and the witness above sits precisely in the gap between Lipschitz equivalence and uniform equivalence (Lipschitz equivalence implies uniform equivalence implies topological equivalence).
The diameter is even less topological than boundedness. Rescaling a metric by a positive constant is a Lipschitz equivalence and multiplies every diameter by that constant, so no numerical value of the diameter is determined even by the Lipschitz class.
This is why "bounded" is never used as a topological adjective in this library. The bounded subsets of $(X,d)$ are defined from $d$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and every statement about them names the metric.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 60 results over 16 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

Bounded set (Wikipedia) (standard reference, not scraped)
Equivalence of metrics (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §20 (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 20: The Metric Topology (East Tennessee State University) (standard reference, not scraped)