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

1.1L1L2

$d$ 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 $m \mapsto 1 / m$ .

2.1step 1.1L3L5

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.

2.2step 1.1L2L5

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.

2.3step 1.1L3L5

$(P, d)$ 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 $P$ .

2.4step 1.1L2L4L5

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 $d^{'} (a_{k}, a_{l}) \leq 1 / (k + 1) + 1 / (l + 1) < 2 / N < ε$ .

3.1step 2.1step 2.2L5

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

3.2step 2.4L2L4L5

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.

3.3step 2.4L3L6

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.

4.1step 3.1step 2.3step 3.2step 3.3∎

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.

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