False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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FALSE: two metrics inducing the same topology have the same Cauchy sequences

Statement

The following statement is FALSE.

Let $d$ and $d^{'}$ be topologically equivalent metrics on a set $X$ , that is $T_{d} = T_{d^{'}}$ (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). Then a sequence in $X$ is Cauchy for $d$ if and only if it is Cauchy for $d^{'}$ (Cauchy sequence in a metric space).

The claim is plausible because convergence really is determined by the topology (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ); the mistake is to extend that to Cauchyness, which is not a topological notion.

Facts & Assumptions

Given: The set $X : = (0, \infty) \subseteq R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), the metrics $d (x, y) : = ∣ x - y ∣$ and $d^{'} (x, y) : = ∣ 1 / x - 1 / y ∣$ on it, a point $a \in X$ , and reals $r, s > 0$ .

[A1]

The false claim: topologically equivalent metrics have the same Cauchy sequences.

[L1]

The absolute value makes $R$ a metric space; a restriction of a metric to a subset is a metric; and $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ (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 are those in which every point has a ball inside the set; a set is open exactly when it is a union of balls around its points (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).

[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).

Refutation

technique · direct

1.1L1L2

$d$ is a metric on $X$ , being the restriction of the usual metric of $R$ ; and $d^{'}$ is a metric on $X$ , since symmetry and the triangle inequality are inherited from the absolute value applied to the reals $1 / x$ , while $d^{'} (x, y) = 0$ forces $1 / x = 1 / y$ and hence $x = y$ by injectivity of reciprocation on the positives.

1.2L1L2L5

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 $1 / x > 1 / a - 1 / (2 a) = 1 / (2 a)$ and hence $x < 2 a$ ; therefore $∣ x - a ∣ = ∣ 1 / x - 1 / a ∣ \cdot x a < s \cdot 2 a \cdot a = 2 a^{2} s \leq r / 2 < r$ . So $B_{d^{'}} (a, s) \subseteq B_{d} (a, r)$ .

1.3L1L2L5

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 $B_{d} (a, r) \subseteq B_{d^{'}} (a, s)$ .

1.4L1L4L6

Put $x_{k} : = 1 / (k + 2)$ , a sequence in $X$ since $k + 2 \geq 2 > 0$ . Given a real $ε > 0$ , [L4] gives $N \geq 1$ with $1 / N < ε / 2$ , and for $k, l \geq N$ we have $1 / (k + 2) < 1 / N$ and $1 / (l + 2) < 1 / N$ , so $d (x_{k}, x_{l}) \leq 1 / (k + 2) + 1 / (l + 2) < 2 / N < ε$ . Hence $(x_{k})$ is $d$ -Cauchy.

2.1step 1.2step 1.3L3

Hence $T_{d} = T_{d^{'}}$ : if $U$ is $d$ -open and $a \in U$ , take $r$ with $B_{d} (a, r) \subseteq U$ and then $s$ as in step 1.2, so $B_{d^{'}} (a, s) \subseteq U$ and $U$ is $d^{'}$ -open; the converse uses step 1.3 in the same way. So $d$ and $d^{'}$ are topologically equivalent.

2.2step 1.4L1L2L6

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

3.1step 2.1step 1.4step 2.2A1∎

So $d$ and $d^{'}$ are topologically equivalent metrics on $X$ with a sequence that is $d$ -Cauchy and not $d^{'}$ -Cauchy, which refutes [A1]. The displayed statement is false.

Remarks

What is true instead. Uniformly equivalent metrics do share their Cauchy sequences, and the reason is one line: uniform equivalence says both identity maps are uniformly continuous (Topologically, uniformly and Lipschitz equivalent metrics on a set, Uniform continuity of a map of metric spaces: one $δ$ serving every point), and a uniformly continuous map sends Cauchy sequences to Cauchy sequences (A uniformly continuous map sends Cauchy sequences to Cauchy sequences). So the pair above is also a witness that topological equivalence does not imply uniform equivalence.
The failure is symmetric. The sequence $y_{k} : = k + 2$ is $d^{'}$ -Cauchy, because $d^{'} (y_{k}, y_{l}) = ∣ 1 / (k + 2) - 1 / (l + 2) ∣$ is small for large indices by the same computation as step 1.4, and it is not $d$ -Cauchy, because $d (y_{K}, y_{K + 1}) = 1$ for every $K$ . So neither metric's Cauchy sequences contain the other's.
This pair does not separate completeness. Both displayed metric spaces are incomplete: $x_{k} = 1 / (k + 2)$ is $d$ -Cauchy with missing limit $0$ , while $y_{k} = k + 2$ is $d^{'}$ -Cauchy and its reciprocals again have missing limit $0$ . Topologically equivalent metrics can nevertheless differ in completeness, as FALSE: completeness of a metric space is determined by its topology records; that requires a different witness.
The worked witness, with both directions of the Cauchy comparison, is On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ share their topology and not their Cauchy sequences ↗ on the companion page.

FALSE: two metrics inducing the same topology have the same Cauchy sequences

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources