LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Every convergent sequence in a metric space is Cauchy

Statement

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 sequence in $X$ converging to $p \in X$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ). Then $(x_{k})$ is Cauchy in $(X, d)$ (Cauchy sequence in a metric space).

The converse fails, and that failure is the subject of this page (FALSE: every Cauchy sequence in a metric space converges).

Facts & Assumptions

Given: A metric space $(X, d)$ , a sequence $(x_{k})$ in $X$ , a point $p \in X$ with $x_{k} \to p$ , and a real $ε > 0$ .

[A1]

Convergence: for every real $η > 0$ there is $K \in N$ with $d (x_{k}, p) < η$ for all $k \geq K$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Limits and Cauchy sequences of reals, The rationals embed densely in the reals).

[A2]

Cauchyness is established by producing, for every real $ε > 0$ , an index $K$ with $d (x_{m}, x_{n}) < ε$ for all $m, n \geq K$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L1]

Triangle inequality (M3) and symmetry (M2) of a metric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

Proof

technique · direct

1.1

The real $ε / 2$ is positive, so [A1] applied with $η = ε / 2$ supplies $K \in N$ with $d (x_{k}, p) < ε / 2$ for every $k \geq K$ .

A1algebra

1.2

For all $m, n \in N$ one has $d (x_{m}, x_{n}) \leq d (x_{m}, p) + d (p, x_{n})$ .

2.1

Hence for all $m, n \geq K$ : $d (x_{m}, x_{n}) \leq d (x_{m}, p) + d (p, x_{n}) < ε / 2 + ε / 2 = ε$ .

step 1.1step 1.2algebra

3.1

Since $ε > 0$ was an arbitrary real, $(x_{k})$ is Cauchy in $(X, d)$ .

step 2.1A2∎

Remarks

The proof spends the triangle inequality and symmetry, but not separation. Symmetry rewrites $d (p, x_{n})$ as the bounded quantity $d (x_{n}, p)$ in step 2.1. The separation axiom (M1) is not used, so the same argument shows that a sequence converging in a pseudometric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) is Cauchy for the pseudometric.
Halving is the whole idea. The Cauchy condition compares two terms of the sequence, and a limit compares one term with the limit; routing $m$ and $n$ through $p$ costs two applications of the convergence hypothesis, so each is run at half the target. Every proof on this page that produces a Cauchy sequence out of a convergent one repeats this step.

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Cauchy sequence (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)