CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ share their topology and not their Cauchy sequences

Statement refuted

Refuted claim: topologically equivalent metrics have the same Cauchy sequences (FALSE: two metrics inducing the same topology have the same Cauchy sequences, Topologically, uniformly and Lipschitz equivalent metrics on a set, Cauchy sequence in a metric space).

Let $X : = (0, \infty) \subseteq R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) and put

$d (x, y) : = ∣ x - y ∣, d^{'} (x, y) : = ∣ \frac{1}{x} - \frac{1}{y} ∣ (x, y \in X) .$

Both are metrics on $X$ and $T_{d} = T_{d^{'}}$ . The sequence $x_{k} : = 1 / (k + 2)$ is $d$ -Cauchy and not $d^{'}$ -Cauchy; the sequence $y_{k} : = k + 2$ is $d^{'}$ -Cauchy and not $d$ -Cauchy. So neither metric's Cauchy sequences are contained in the other's, and topological equivalence controls neither direction.

Facts & Assumptions

Given: The set $X = (0, \infty)$ with the metrics $d$ and $d^{'}$ above; the sequences $x_{k} = 1 / (k + 2)$ and $y_{k} = k + 2$ ; a point $a \in X$ ; reals $r, s, ε > 0$ .

[L1]

The absolute value makes $R$ a metric space, a restriction of a metric is a metric, and the pullback of a metric along an injection is a metric (The absolute value makes $R$ 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, Isometry, isometric embedding, and the subspace metric on a subset, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Basic properties of the absolute value).

[L2]

For $x, a > 0$ : $1 / x - 1 / a = (a - x) / (x a)$ , so $∣ x - a ∣ = ∣ 1 / x - 1 / a ∣ \cdot x a$ and $∣ 1 / x - 1 / a ∣ = ∣ x - a ∣ / (x a)$ ; reciprocation is strictly decreasing on the positives, hence injective there (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

[L3]

Open sets and balls (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, Open ball, closed ball and sphere in a metric space); a set is open exactly when every point of it has a ball around it inside it.

[L4]

For every real $η > 0$ there is a natural $N \geq 1$ with $1 / N < η$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L5]

Two reals have a minimum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L6]

Cauchyness may be tested with real $ε > 0$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L7]

Uniform equivalence says both identity maps are uniformly continuous, and a uniformly continuous map sends Cauchy sequences to Cauchy sequences (Topologically, uniformly and Lipschitz equivalent metrics on a set, Uniform continuity of a map of metric spaces: one $δ$ serving every point, A uniformly continuous map sends Cauchy sequences to Cauchy sequences).

Counterexample

technique · direct

1.1

$d$ is a metric on $X$ , being the restriction of the usual metric of $R$ ; and $d^{'}$ is a metric on $X$ , being the pullback of that metric along the injective map $x \mapsto 1 / x$ .

L1L2

2.1

Given $a \in X$ and a real $r > 0$ , put $s : = \min {1 / (2 a), r / (4 a^{2})} > 0$ . If $d^{'} (x, a) < s$ then $∣ 1 / x - 1 / a ∣ < 1 / (2 a)$ , so $1 / x > 1 / (2 a)$ and hence $x < 2 a$ ; therefore $∣ x - a ∣ = ∣ 1 / x - 1 / a ∣ \cdot x a < s \cdot 2 a^{2} \leq r / 2 < r$ . So $B_{d^{'}} (a, s) \subseteq B_{d} (a, r)$ .

step 1.1L2L3L5

2.2

Given $a \in X$ and a real $s > 0$ , put $r : = \min {a / 2, s a^{2} / 4} > 0$ . If $d (x, a) < r$ then $∣ x - a ∣ < a / 2$ , so $x > a / 2$ and hence $∣ 1 / x - 1 / a ∣ = ∣ x - a ∣ / (x a) < r / ((a / 2) a) = 2 r / a^{2} \leq s / 2 < s$ . So $B_{d} (a, r) \subseteq B_{d^{'}} (a, s)$ .

step 1.1L2L3L5

2.3

The sequence $x_{k} = 1 / (k + 2)$ lies in $X$ ; it is $d$ -Cauchy, since for a real $ε > 0$ and $N \geq 1$ with $1 / N < ε / 2$ every $k, l \geq N$ gives $d (x_{k}, x_{l}) \leq 1 / (k + 2) + 1 / (l + 2) < 2 / N < ε$ .

step 1.1L4L6

3.1

Hence $T_{d} = T_{d^{'}}$ : a $d$ -open $U$ is $d^{'}$ -open by step 2.1 applied at each of its points, and conversely by step 2.2.

step 2.1step 2.2L3

3.2

It is not $d^{'}$ -Cauchy: $d^{'} (x_{k}, x_{l}) = ∣ (k + 2) - (l + 2) ∣ = ∣ k - l ∣$ , so $d^{'} (x_{K}, x_{K + 1}) = 1$ for every $K$ and the Cauchy condition fails at $ε = 1$ .

step 2.3L1L2L6

3.3

The sequence $y_{k} = k + 2$ lies in $X$ ; it is $d^{'}$ -Cauchy, since $d^{'} (y_{k}, y_{l}) = ∣ 1 / (k + 2) - 1 / (l + 2) ∣ \leq 1 / (k + 2) + 1 / (l + 2)$ , which is below $ε$ for $k, l \geq N$ by the computation of step 2.3.

step 2.3L1L2L4L6

4.1

It is not $d$ -Cauchy: $d (y_{K}, y_{K + 1}) = 1$ for every $K$ , so the Cauchy condition fails at $ε = 1$ .

step 3.3L1L6

5.1

So $d$ and $d^{'}$ are topologically equivalent metrics on $X$ whose classes of Cauchy sequences are incomparable, which refutes the claim above.

step 3.1step 2.3step 3.2step 3.3step 4.1

6.1

In particular $d$ and $d^{'}$ are not uniformly equivalent, since uniform equivalence would make both identity maps uniformly continuous and hence would preserve Cauchy sequences in both directions.

step 5.1L7∎

Remarks

The map $x \mapsto 1 / x$ is what is being tested. It is a bijection of $(0, \infty)$ onto itself and a homeomorphism, by steps 2.1 and 2.2, and $d^{'}$ is the metric it pulls back from $d$ . Homeomorphisms preserve open sets and convergence; they do not preserve Cauchyness, and this is that failure written out.
Both failures come from a missing endpoint, at opposite ends. The sequence $(x_{k})$ heads for $0$ , which $(0, \infty)$ does not contain, so it is $d$ -Cauchy without converging; its image under $x \mapsto 1 / x$ runs off to the right and is not Cauchy at all. Reading the same picture through $d^{'}$ exchanges the two ends, which is why the failure is symmetric.
Indexing. The terms are $1 / (k + 2)$ and $k + 2$ rather than $1 / k$ and $k$ because $N$ contains $0$ in this library; $1 / 0$ does not exist and $0 \notin (0, \infty)$ , so both sequences are shifted to start safely inside the space.
The completeness version of the same phenomenon is On the positive integers the metrics $∣ m - n ∣$ and $∣ 1 / m - 1 / n ∣$ both induce the discrete topology, and only the first is complete, and the general statement being refuted there is FALSE: completeness of a metric space is determined by its topology.

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Sources

Equivalence of metrics (Wikipedia) (standard reference, not scraped)
Cauchy sequence (Wikipedia) (standard reference, not scraped)

On (0,∞)(0,∞) the metrics ∣x−y∣∣x−y∣ and ∣1/x−1/y∣∣1/x−1/y∣ share their topology and not their Cauchy sequences