CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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On the positive integers the metrics $∣ m - n ∣$ and $∣ 1 / m - 1 / n ∣$ both induce the discrete topology, and only the first is complete

Statement refuted

Refuted claim: completeness is determined by the topology, so topologically equivalent metrics are either both complete or both incomplete (FALSE: completeness of a metric space is determined by its topology, Topologically, uniformly and Lipschitz equivalent metrics on a set).

Let $P : = {n \in N : n \geq 1}$ be the positive integers, regarded inside $R$ through the canonical embedding, and put

$d (m, n) : = ∣ m - n ∣, d^{'} (m, n) : = ∣ \frac{1}{m} - \frac{1}{n} ∣ (m, n \in P) .$

Then both are metrics on $P$ , both induce the discrete topology, so $T_{d} = T_{d^{'}}$ ; $(P, d)$ is complete and $(P, d^{'})$ is not; and consequently $d$ and $d^{'}$ are not uniformly equivalent either (Topologically, uniformly and Lipschitz equivalent metrics on a set).

Facts & Assumptions

Given: The set $P$ of positive integers inside $R$ ; the functions $d$ and $d^{'}$ above; the sequence $a_{k} : = k + 1$ in $P$ ; a real $ε > 0$ .

[L1]

The absolute value makes $R$ a metric space, a restriction of a metric to a subset 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]

Reciprocation is strictly decreasing on the positive reals, hence injective there (Inverses of positives are positive, and reciprocation reverses order).

[L3]

For distinct naturals $m \neq n$ one has $∣ m - n ∣ \geq 1$ (Canonical naturals are positive and strictly increasing, The natural numbers $N$ (von Neumann), The unique embedding of ℚ into an ordered field).

[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]

Open sets, balls, Cauchyness and convergence, tested with real $ε > 0$ (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, Cauchy sequence in a metric space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The rationals embed densely in the reals).

[L6]

Uniform equivalence of two metrics says exactly that 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 $P$ , being the restriction of the usual metric of $R$ ; and $d^{'}$ is a metric on $P$ , since it is the pullback of that metric along the injective map $m \mapsto 1 / m$ .

L1L2

2.1

The $d$ -topology is discrete: $B_{d} (m, 1 / 2) = {m}$ for every $m \in P$ , because $d (m, n) = ∣ m - n ∣ \geq 1 > 1 / 2$ for $n \neq m$ ; so every subset of $P$ is a union of open balls, hence open.

step 1.1L3L5

2.2

The $d^{'}$ -topology is discrete as well. Fix $m \in P$ and put $r : = 1 / m - 1 / (m + 1) = 1 / (m (m + 1)) > 0$ . If $n > m$ then $1 / n \leq 1 / (m + 1)$ , so $d^{'} (m, n) = 1 / m - 1 / n \geq r$ ; if $n < m$ then $m \geq 2$ and $1 / n \geq 1 / (m - 1)$ , so $d^{'} (m, n) = 1 / n - 1 / m \geq 1 / (m - 1) - 1 / m = 1 / (m (m - 1)) > r$ . Hence $B_{d^{'}} (m, r) = {m}$ and every subset of $P$ is $d^{'}$ -open.

step 1.1L2L5

2.3

$(P, d)$ is complete: a $d$ -Cauchy sequence tested at $ε = 1$ has an index $K$ with $∣ x_{k} - x_{l} ∣ < 1$ for all $k, l \geq K$ , which by [L3] forces $x_{k} = x_{l}$ ; so the sequence is constant from $K$ on and converges in $P$ .

step 1.1L3L5

2.4

The sequence $a_{k} = k + 1$ is $d^{'}$ -Cauchy: given a real $ε > 0$ , take $N \geq 1$ with $1 / N < ε / 2$ ; for $k, l \geq N$ we have $k + 1 > N$ and $l + 1 > N$ , so $d^{'} (a_{k}, a_{l}) \leq 1 / (k + 1) + 1 / (l + 1) < 2 / N < ε$ .

step 1.1L2L4L5

3.1

Therefore $T_{d} = T_{d^{'}}$ : both are the collection of all subsets of $P$ , so $d$ and $d^{'}$ are topologically equivalent.

step 2.1step 2.2L5

3.2

It has no $d^{'}$ -limit in $P$ : if $p \in P$ then $1 / p > 0$ , and taking $N \geq 1$ with $1 / N < 1 / (2 p)$ gives, for every $k \geq N$ , $1 / (k + 1) < 1 / N < 1 / (2 p)$ and hence $d^{'} (a_{k}, p) = ∣ 1 / (k + 1) - 1 / p ∣ \geq 1 / p - 1 / (k + 1) > 1 / (2 p) > 0$ ; so from no index on is $d^{'} (a_{k}, p)$ below $1 / (2 p)$ , and $a_{k} \to̸ p$ . Hence $(P, d^{'})$ is not complete.

step 2.4L2L4L5

3.3

The same sequence is not $d$ -Cauchy, since $d (a_{K}, a_{K + 1}) = 1$ for every $K$ ; so if $d$ and $d^{'}$ were uniformly equivalent, the identity map $(P, d^{'}) \to (P, d)$ would be uniformly continuous and would carry the $d^{'}$ -Cauchy sequence $(a_{k})$ to a $d$ -Cauchy sequence, which it is not. Hence $d$ and $d^{'}$ are not uniformly equivalent.

step 2.4L3L6

4.1

So $d$ and $d^{'}$ are topologically equivalent metrics on one set of which exactly one is complete, which refutes the claim above; and step 3.3 locates the reason, namely that topological equivalence is strictly weaker than uniform equivalence.

step 3.1step 2.3step 3.2step 3.3∎

Remarks

The two spaces are the same set with the same open sets and different geometry. In $(P, d)$ the points are uniformly spaced, at distance at least $1$ ; in $(P, d^{'})$ they crowd together, the distance from $n$ to $n + 1$ being $1 / (n (n + 1))$ , which tends to $0$ . A discrete topology cannot see that difference, and completeness can.
What $(P, d^{'})$ is missing. Under $m \mapsto 1 / m$ it is isometric to the set ${1, 1 / 2, 1 / 3, \dots}$ inside $R$ , which is not closed there: the point $0$ is missing (Closed subspaces of complete metric spaces are complete; the converse under countable choice). The sequence $(a_{k})$ of step 4.1 is exactly the sequence heading for that missing point.
Reciprocal metrics on $N$ would not work. $0 \in N$ in this library, and $1 / 0$ does not exist, which is why the underlying set here is the positive integers and the sequence is $a_{k} = k + 1$ rather than $a_{k} = k$ .
The same pair of phenomena, one level down. That these two metrics do not share their Cauchy sequences is the statement refuted by FALSE: two metrics inducing the same topology have the same Cauchy sequences, witnessed on the half-line by On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ share their topology and not their Cauchy sequences; completeness fails here for precisely that reason.

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Sources

Complete metric space (Wikipedia) (standard reference, not scraped)
Equivalence of metrics (Wikipedia) (standard reference, not scraped)
Discrete space (Wikipedia) (standard reference, not scraped)

On the positive integers the metrics ∣m−n∣∣m−n∣ and ∣1/m−1/n∣∣1/m−1/n∣ both induce the discrete topology, and only the first is complete