False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $\mathcal{T}_d = \mathcal{T}_{d'}$ (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 $\mathbb{R}$ ); the mistake is to extend that to Cauchyness, which is not a topological notion.

Facts & Assumptions

Given: The set $X := (0,\infty) \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$ : 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 $\mathbb{R}$ a metric space; a restriction of a metric to a subset is a metric; and $|uv| = |u|\,|v|$ (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, 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 $x, a > 0$ : $1/x - 1/a = (a-x)/(xa)$ , so $|x - a| = |1/x - 1/a| \cdot x a$ and $|1/x - 1/a| = |x-a|/(xa)$ ; 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 $\eta > 0$ there is a natural $N \ge 1$ with $1/N < \eta$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , 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 $\varepsilon > 0$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

Refutation

technique · direct

1.1

$d$ is a metric on $X$ , being the restriction of the usual metric of $\mathbb{R}$ ; 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.

L1L2

1.2

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

L1L2L5

1.3

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 $|1/x - 1/a| = |x-a|/(xa) < r/((a/2)\,a) = 2r/a^2 \le s/2 < s$ . So $B_d(a,r) \subseteq B_{d'}(a,s)$ .

L1L2L5

1.4

Put $x_k := 1/(k+2)$ , a sequence in $X$ since $k+2 \ge 2 > 0$ . Given a real $\varepsilon > 0$ , [L4] gives $N \ge 1$ with $1/N < \varepsilon/2$ , and for $k,l \ge N$ we have $1/(k+2) < 1/N$ and $1/(l+2) < 1/N$ , so $d(x_k,x_l) \le 1/(k+2) + 1/(l+2) < 2/N < \varepsilon$ . Hence $(x_k)$ is $d$ -Cauchy.

L1L4L6

2.1

Hence $\mathcal{T}_d = \mathcal{T}_{d'}$ : 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.

step 1.2step 1.3L3

2.2

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 $\varepsilon = 1$ .

step 1.4L1L2L6

3.1

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.

step 2.1step 1.4step 2.2A1∎

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 $\delta$ 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.

Depends on

Used by

On (0,∞) the metrics |x-y| and |1/x - 1/y| share their topology and not their Cauchy sequences Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 78 results over 30 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

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