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

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FALSE: completeness of a metric space is determined by its topology

Statement

The following statement is FALSE.

Let $d$ and $d^{'}$ be metrics on the same set $X$ that are topologically equivalent, 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 $(X, d)$ is complete if and only if $(X, d^{'})$ is complete (Complete metric space: every Cauchy sequence converges in the space).

Equivalently, and this is the form in which the error is usually made: completeness is a topological property of a metrisable space, so that it makes sense to call a topological space "complete".

Facts & Assumptions

Given: The set $P : = {n \in N : n \geq 1}$ of positive naturals, regarded inside $R$ through the canonical embedding; the functions $d (m, n) : = ∣ m - n ∣$ and $d^{'} (m, n) : = ∣ 1 / m - 1 / n ∣$ on $P \times P$ ; a real $ε > 0$ .

[A1]

The false claim: topologically equivalent metrics on one set are either both complete or both incomplete.

[L1]

The absolute value makes $R$ a metric space, so $∣ x - y ∣$ satisfies (M1), (M2) and (M3), and its balls are the intervals $(x - r, x + r)$ ; a restriction of a metric to a subset 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, so $0 < a < b$ gives $0 < 1 / b < 1 / a$ ; in particular $m \mapsto 1 / m$ is injective on $P$ (Inverses of positives are positive, and reciprocation reverses order).

[L3]

Positive naturals sit in $R$ in their own order, and for naturals $m \neq n$ one has $∣ m - n ∣ \geq 1$ , since one of them is at least the successor of the other (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 in a metric space, 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).

Refutation

technique · direct

1.1L1

$d$ is a metric on $P$ , being the restriction to $P \times P$ of the usual metric of $R$ .

1.2L1L2

$d^{'}$ is a metric on $P$ : symmetry and the triangle inequality are inherited from the absolute value applied to the reals $1 / m$ , and $d^{'} (m, n) = 0$ forces $1 / m = 1 / n$ and hence $m = n$ , because $m \mapsto 1 / m$ is injective on $P$ .

2.1step 1.1L3L5

Every subset of $P$ is open for $d$ : for $m \in P$ the ball $B_{d} (m, 1 / 2)$ is ${m}$ , since $d (m, n) = ∣ m - n ∣ \geq 1 > 1 / 2$ for $n \neq m$ ; so every subset is a union of open balls.

2.2step 1.2L2L4L5

Every subset of $P$ is open for $d^{'}$ : fix $m \in P$ and put $r : = 1 / m - 1 / (m + 1)$ , a positive real. If $n > m$ then $1 / n \leq 1 / (m + 1)$ , so $d^{'} (m, n) = 1 / m - 1 / n \geq r$ ; and if $n < m$ then $m \geq 2$ and $1 / n \geq 1 / (m - 1)$ , so $d^{'} (m, n) = 1 / n - 1 / m \geq 1 / (m - 1) - 1 / m > 1 / m - 1 / (m + 1) = r$ . Hence $B_{d^{'}} (m, r) = {m}$ and every subset is a union of open balls.

2.3step 1.1L3L5

$(P, d)$ is complete: let $(x_{k})$ be $d$ -Cauchy and apply the definition with $ε = 1$ to get $K$ with $∣ x_{m} - x_{n} ∣ < 1$ for all $m, n \geq K$ ; by [L3] this forces $x_{m} = x_{n}$ for $m, n \geq K$ , so the sequence is constant from $K$ on and converges to $x_{K} \in P$ .

2.4step 1.2L2L4L5

$(P, d^{'})$ is not complete. Put $a_{k} : = k + 1$ , a sequence in $P$ . Given a real $ε > 0$ , [L4] gives $N \geq 1$ with $1 / N < ε / 2$ ; for $k, l \geq N$ we have $k + 1 > N$ and $l + 1 > N$ , hence $1 / (k + 1) < 1 / N$ and $1 / (l + 1) < 1 / N$ , so $d^{'} (a_{k}, a_{l}) = ∣ 1 / (k + 1) - 1 / (l + 1) ∣ \leq 1 / (k + 1) + 1 / (l + 1) < 2 / N < ε$ . Hence $(a_{k})$ is $d^{'}$ -Cauchy.

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

Suppose $a_{k} \to p$ in $(P, d^{'})$ for some $p \in P$ . Since $1 / p > 0$ , [L4] gives $N \geq 1$ with $1 / N < 1 / (2 p)$ , and then for every $k \geq N$ we have $1 / (k + 1) \leq 1 / N < 1 / (2 p)$ , so $d^{'} (a_{k}, p) = ∣ 1 / (k + 1) - 1 / p ∣ \geq 1 / p - 1 / (k + 1) > 1 / p - 1 / (2 p) = 1 / (2 p) > 0$ . So $d^{'} (a_{k}, p)$ never drops below $1 / (2 p)$ from any index on, contradicting convergence to $p$ ; as $p \in P$ was arbitrary, $(a_{k})$ has no limit in $(P, d^{'})$ .

4.1step 3.1step 2.3step 2.4step 3.2A1∎

So $d$ and $d^{'}$ are topologically equivalent metrics on $P$ with $(P, d)$ complete and $(P, d^{'})$ not, which refutes [A1]. The displayed statement is false.

Remarks

What is true instead. Completeness is preserved by uniform equivalence, not by topological equivalence, because uniform equivalence says exactly that both identity maps are uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point, Topologically, uniformly and Lipschitz equivalent metrics on a set) and uniformly continuous maps preserve Cauchy sequences (A uniformly continuous map sends Cauchy sequences to Cauchy sequences). The two metrics above are therefore not uniformly equivalent, even though they are topologically equivalent; that is another way to read the counterexample.
Where exactly the failure sits. Convergence of a sequence is topological (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ), but Cauchyness is not: it compares two terms of the sequence with each other rather than with a point of the space, and the comparison is metric. The two metrics above have the same convergent sequences and different Cauchy sequences (FALSE: two metrics inducing the same topology have the same Cauchy sequences).
The topological invariant that does exist is a weaker one. A topological space is completely metrisable when at least one metric inducing its topology is complete, and the class of completely metrisable spaces is genuinely a topological class. Nothing about it is proved here; the orientation is Completeness belongs to the metric; complete metrizability is characterised on a later page.
The fully worked witness, with both topologies computed and both completeness verdicts verified, is On the positive integers the metrics $∣ m - n ∣$ and $∣ 1 / m - 1 / n ∣$ both induce the discrete topology, and only the first is complete ↗ on the companion page.

FALSE: completeness of a metric space is determined by its topology

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources