LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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A Cauchy sequence in a metric space with a convergent subsequence converges to that subsequence’s limit

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), let $(x_k)$ be a Cauchy sequence in $X$ (Cauchy sequence in a metric space), and suppose some subsequence $(x_{n_j})$ converges to $p \in X$ , that is $p$ is a subsequential limit of $(x_k)$ in $(X,d)$ (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Subsequential limit of a real sequence, and the subsequential limit set). Then the whole sequence converges to $p$ .

In particular a Cauchy sequence has at most one subsequential limit, and it has one exactly when it converges.

Facts & Assumptions

Given: A metric space $(X,d)$ , a Cauchy sequence $(x_k)$ in $X$ , a strictly increasing $n : \mathbb{N} \to \mathbb{N}$ , a point $p \in X$ with $x_{n_j} \to p$ , and a real $\varepsilon > 0$ .

[A1]

Cauchyness: for every real $\eta > 0$ there is $K_1$ with $d(x_m,x_l) < \eta$ for all $m,l \ge K_1$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

[A2]

Metric convergence: a sequence $(y_j)$ converges to $p$ if and only if for every real $\eta > 0$ there is $K$ such that $d(y_j,p) < \eta$ for all $j \ge K$ . Applied to $y_j = x_{n_j}$ , the subsequence hypothesis gives $K_2$ with $d(x_{n_j},p) < \eta$ for all $j \ge K_2$ (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).

[L1]

A strictly increasing index map satisfies $n_j \ge j$ for every $j$ (A strictly increasing index map satisfies $n_k \ge k$ ).

[L2]

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

[L3]

Two naturals have a maximum, which is one of them and is at least each of them (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · direct

1.1

Apply [A1] with $\eta = \varepsilon/2$ to get $K_1$ with $d(x_m,x_l) < \varepsilon/2$ for all $m,l \ge K_1$ , and [A2] with $\eta = \varepsilon/2$ to get $K_2$ with $d(x_{n_j},p) < \varepsilon/2$ for all $j \ge K_2$ .

A1A2algebra

1.2

Put $K := \max\{K_1, K_2\}$ , so $K \ge K_1$ and $K \ge K_2$ .

L3choose

2.1

The index $n_K$ satisfies $n_K \ge K \ge K_1$ , and $K \ge K_2$ , so $d(x_{n_K}, p) < \varepsilon/2$ .

step 1.1step 1.2L1

3.1

Let $k \ge K$ . Then $k \ge K_1$ and $n_K \ge K_1$ , so $d(x_k, x_{n_K}) < \varepsilon/2$ , and therefore $d(x_k,p) \le d(x_k, x_{n_K}) + d(x_{n_K}, p) < \varepsilon/2 + \varepsilon/2 = \varepsilon$ .

step 1.1step 1.2step 2.1L1L2algebra

4.1

Since $\varepsilon > 0$ was an arbitrary real, $x_k \to p$ in $(X,d)$ .

step 3.1A2∎

Remarks

Where $n_j \ge j$ is used. In step 2.1, and quoted again in step 3.1, and both places matter: it puts the single chosen subsequence term $x_{n_K}$ beyond the Cauchy index $K_1$ , and it is what lets one index $K$ control the subsequence and the sequence at once. Without it the subsequence could in principle look only at small indices, and the Cauchy hypothesis would never be reached.
Only one term of the subsequence is used. The proof does not pass to the limit along the subsequence; it fixes the single index $n_K$ and routes every later term of the whole sequence through it. That is why no further hypothesis on the subsequence is needed.
This is the lemma that turns compactness into completeness. A space in which every sequence has a convergent subsequence is complete (Complete metric space: every Cauchy sequence converges in the space) by this lemma alone, applied to a Cauchy sequence. The converse is false, not merely harder: $\mathbb{R}$ is complete ( $\mathbb{R}$ and $\mathbb{R}^n$ for $n \ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$ ) and the sequence $x_k = k$ has no convergent subsequence, its terms being at pairwise distance at least $1$ so that no subsequence is Cauchy. What does hold is that a complete space in which every sequence has a convergent subsequence is exactly a complete and totally bounded one, and total boundedness is a later page of this library.

Depends on

Used by

A sequentially compact metric space is complete, with no choice principle used Lemma

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 30 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)