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

$d(m,n) := |m - n|, \qquad d'(m,n) := \Big|\tfrac{1}{m} - \tfrac{1}{n}\Big| \qquad (m,n \in P).$

Then both are metrics on $P$ , both induce the discrete topology, so $\mathcal{T}_d = \mathcal{T}_{d'}$ ; $(P,d)$ 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 $\mathbb{R}$ ; the functions $d$ and $d'$ above; the sequence $a_k := k+1$ in $P$ ; a real $\varepsilon > 0$ .

[L1]

The absolute value makes $\mathbb{R}$ 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 $\mathbb{R}$ 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]

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

[L3]

For distinct naturals $m \ne n$ one has $|m-n| \ge 1$ (Canonical naturals are positive and strictly increasing, The natural numbers $\mathbb{N}$ (von Neumann), The unique embedding of ℚ into an ordered field).

[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]

Open sets, balls, Cauchyness and convergence, tested with real $\varepsilon > 0$ (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 $\mathbb{R}$ , 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 $\delta$ 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 $\mathbb{R}$ ; 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 $d(m,n) = |m-n| \ge 1 > 1/2$ for $n \ne m$ ; 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 $1/n \le 1/(m+1)$ , so $d'(m,n) = 1/m - 1/n \ge r$ ; if $n < m$ then $m \ge 2$ and $1/n \ge 1/(m-1)$ , so $d'(m,n) = 1/n - 1/m \ge 1/(m-1) - 1/m = 1/(m(m-1)) > r$ . 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 $\varepsilon = 1$ has an index $K$ with $|x_k - x_l| < 1$ for all $k,l \ge K$ , 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 $\varepsilon > 0$ , take $N \ge 1$ with $1/N < \varepsilon/2$ ; for $k,l \ge N$ we have $k+1 > N$ and $l+1 > N$ , so $d'(a_k,a_l) \le 1/(k+1) + 1/(l+1) < 2/N < \varepsilon$ .

step 1.1L2L4L5

3.1

Therefore $\mathcal{T}_d = \mathcal{T}_{d'}$ : 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 $N \ge 1$ with $1/N < 1/(2p)$ gives, for every $k \ge N$ , $1/(k+1) < 1/N < 1/(2p)$ and hence $d'(a_k,p) = |1/(k+1) - 1/p| \ge 1/p - 1/(k+1) > 1/(2p) > 0$ ; so from no index on is $d'(a_k,p)$ below $1/(2p)$ , and $a_k \not\to p$ . 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 $\mathbb{R}$ , which is not closed there: the point $0$ is missing (A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed). The sequence $(a_k)$ of step 4.1 is exactly the sequence heading for that missing point.
Reciprocal metrics on $\mathbb{N}$ would not work. $0 \in \mathbb{N}$ 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|∣m−n∣ and ∣1/m−1/n∣|1/m - 1/n|∣1/m−1/n∣ both induce the discrete topology, and only the first is complete