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

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FALSE: every Cauchy sequence in a metric space converges

Statement

The following statement is FALSE.

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $(x_{k})$ be a Cauchy sequence in it (Cauchy sequence in a metric space). Then $(x_{k})$ converges to a point of $X$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

Equivalently: every metric space is complete (Complete metric space: every Cauchy sequence converges in the space), so that the word complete is redundant.

This is the error that the whole page exists to guard against. It is encouraged by the Cauchy criterion on the real line (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges), which is a theorem about $R$ and not about metric spaces.

Facts & Assumptions

Given: The open interval $X : = (0, 1) \subseteq R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) with the metric $d (x, y) : = ∣ x - y ∣$ inherited from $R$ ; the sequence $x_{k} : = 1 / (k + 2)$ for $k \in N$ ; a real $ε > 0$ .

[A1]

The false claim: every Cauchy sequence in every metric space converges in that space.

[L1]

The absolute value makes $R$ a metric space, and the restriction of a metric to a subset is a metric on that subset, with the same distances (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).

[L2]

For every real $η > 0$ there is a natural $N \geq 1$ with $1 / N < η$ ; and $0 < a < b$ gives $0 < 1 / b < 1 / a$ (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, Inverses of positives are positive, and reciprocation reverses order).

[L3]

A convergent sequence in a metric space is Cauchy (Every convergent sequence in a metric space is Cauchy), and Cauchyness and convergence may be tested with real $ε > 0$ (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).

[L4]

Limits in a metric space are unique (A sequence in a metric space has at most one limit).

Refutation

technique · direct

1.1L1L2

Every term lies in $X$ : $k + 2 \geq 2 > 0$ gives $x_{k} = 1 / (k + 2) > 0$ , and $k + 2 \geq 2$ gives $x_{k} \leq 1 / 2 < 1$ . So $(x_{k})$ is a sequence in $X$ , and $d$ is a metric on $X$ .

1.2L1L2L3

$x_{k} \to 0$ in $(R, ∣ \cdot ∣)$ : given a real $ε > 0$ , [L2] supplies $N \geq 1$ with $1 / N < ε$ , and for $k \geq N$ we have $k + 2 > N$ , hence $∣ x_{k} - 0 ∣ = 1 / (k + 2) < 1 / N < ε$ .

2.1step 1.1step 1.2L1L3

Hence $(x_{k})$ is Cauchy in $(R, ∣ \cdot ∣)$ , and since $d$ is the restriction of the metric of $R$ and all terms lie in $X$ , the same indices witness that $(x_{k})$ is Cauchy in $(X, d)$ .

2.2step 1.2L1L4

Suppose $(x_{k})$ converged in $(X, d)$ to some $p \in X$ . Distances in $(X, d)$ are distances in $R$ , so $x_{k} \to p$ in $(R, ∣ \cdot ∣)$ as well; with step 1.2 and uniqueness of limits in $R$ this forces $p = 0$ .

3.1step 1.1step 2.2

But $0 \notin X$ , since $X = (0, 1)$ contains only reals $> 0$ . So $(x_{k})$ has no limit in $(X, d)$ .

4.1step 2.1step 3.1A1∎

Therefore $(X, d)$ is a metric space carrying a Cauchy sequence that does not converge in it, which refutes [A1]; the displayed statement is false, and $(0, 1)$ is not complete.

Remarks

The sequence starts at $k = 0$ and the index is $k + 2$ , not $k$ . In this library $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so $1 / k$ is undefined at $k = 0$ and $1 / (k + 1)$ equals $1$ at $k = 0$ , which is not in $(0, 1)$ . The shift to $k + 2$ is what puts every term inside the interval, and a version of this example copied from a text that indexes from $1$ has to be reindexed.
Nothing is wrong with the sequence; the space is missing a point. The same sequence converges perfectly well in $R$ , and in $[0, 1)$ , and in $[0, 1]$ . Cauchyness is a property of the sequence alone (Cauchy sequence in a metric space); whether the destination exists is a property of the space, and that is exactly the asymmetry that Complete metric space: every Cauchy sequence converges in the space names.
The witness is not exotic. $(0, 1)$ is an open interval of the real line with its ordinary metric, and what it lacks is not structure but the endpoint $0$ that its own Cauchy sequence was heading for. The systematic version of this observation is Closed subspaces of complete metric spaces are complete; the converse under countable choice: $(0, 1)$ is not closed in $R$ , so it cannot be complete.
The remedy is the completion. Under Countable Choice, every metric space, this one included, sits densely and isometrically inside a complete one (Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences); for $(0, 1)$ that completion is $[0, 1]$ , which is complete as a closed subset of the complete space $R$ (Closed subspaces of complete metric spaces are complete; the converse under countable choice, $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ) and contains $(0, 1)$ densely, since every point of $[0, 1]$ is at distance less than any given positive real from a point of $(0, 1)$ .

FALSE: every Cauchy sequence in a metric space converges

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources