LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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A Cauchy sequence with a convergent subsequence converges, to that subsequence’s limit

Statement

Let $(x_k)$ be a Cauchy sequence of reals (Limits and Cauchy sequences of reals) and suppose some subsequence $(x_{n_j})$ converges to $L \in \mathbb{R}$ , that is, $L$ is a subsequential limit of $(x_k)$ (Subsequential limit of a real sequence, and the subsequential limit set). Then the whole sequence $(x_k)$ converges, and its limit is $L$ .

So for a Cauchy sequence a single convergent subsequence already determines the behaviour of the sequence. This is exactly the step that upgrades Bolzano-Weierstrass into Cauchy completeness in the Cauchy criterion later on this page, and it is false without the Cauchy hypothesis.

Facts & Assumptions

Given: A Cauchy sequence $(x_k)$ of reals, a strictly increasing $n : \mathbb{N} \to \mathbb{N}$ , and $L \in \mathbb{R}$ with $x_{n_j} \to L$ .

[A1]

Cauchy condition: for every rational $\varepsilon > 0$ there is $K$ with $|x_k - x_l| < \varepsilon$ for all $k, l \ge K$ (Limits and Cauchy sequences of reals).

[A2]

Convergence of the subsequence: for every rational $\varepsilon > 0$ there is $J$ with $|x_{n_j} - L| < \varepsilon$ for all $j \ge J$ (Limits and Cauchy sequences of reals, Subsequential limit of a real sequence, and the subsequential limit set, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

Triangle inequality: $|x - L| = |(x - y) + (y - L)| \le |x - y| + |y - L|$ (The triangle inequality).

[L2]

Growth of an index map: a strictly increasing $n$ satisfies $n_j \ge j$ for every $j$ (A strictly increasing index map satisfies $n_k \ge k$ ).

[L3]

Halving a rational: if $\varepsilon$ is a positive rational then so is $\varepsilon/2$ , and the embedding of $\mathbb{Q}$ in $\mathbb{R}$ is a field embedding, so the image of $\varepsilon/2$ is half the image of $\varepsilon$ and the two halves sum to $\varepsilon$ (The rationals embed densely in the reals).

[L4]

The order on $\mathbb{N}$ is total and transitive, so two indices $J, K$ admit an index $j$ with $j \ge J$ and $j \ge K$ ( $\le$ is a linear order on $\mathbb{N}$ ).

[L5]

Convergence: $(x_k)$ converges to $L$ when for every rational $\varepsilon > 0$ there is $K$ with $|x_k - L| < \varepsilon$ for all $k \ge K$ (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Let $\varepsilon > 0$ be an arbitrary rational; then $\varepsilon/2$ is again a positive rational, and $\varepsilon/2 + \varepsilon/2 = \varepsilon$ .

givenL3

2.1

By [A1] applied to $\varepsilon/2$ , fix $K \in \mathbb{N}$ with $|x_k - x_l| < \varepsilon/2$ for all $k, l \ge K$ .

step 1.1A1choose

2.2

By [A2] applied to $\varepsilon/2$ , fix $J \in \mathbb{N}$ with $|x_{n_j} - L| < \varepsilon/2$ for all $j \ge J$ .

step 1.1A2choose

3.1

Fix a single index $j$ with $j \ge J$ and $j \ge K$ ; then $n_j \ge j \ge K$ , so the term $x_{n_j}$ is simultaneously within $\varepsilon/2$ of $L$ and within $\varepsilon/2$ of every $x_k$ with $k \ge K$ .

step 2.1step 2.2L2L4choose

4.1

For every $k \ge K$ : $|x_k - L| \le |x_k - x_{n_j}| + |x_{n_j} - L| < \varepsilon/2 + \varepsilon/2 = \varepsilon$ .

step 2.1step 2.2step 3.1L1

5.1

The rational $\varepsilon > 0$ was arbitrary and an index $K$ was produced for it, so $(x_k)$ converges to $L$ .

step 4.1L5∎

Remarks

The Cauchy hypothesis is doing all the work. Without it a convergent subsequence says nothing at all about the sequence, which is FALSE: a convergent subsequence forces the sequence to converge; the alternating sequence has a constant, hence convergent, subsequence and does not converge. What the Cauchy condition adds is that the terms are eventually close to each other, so being close to $L$ at one late index propagates to all late indices.
The single index $j$ chosen in step 3.1 is the whole trick. It is used once, as a bridge, and is not required to grow with $k$ ; this is why $n_j \ge j$ (A strictly increasing index map satisfies $n_k \ge k$ ) is needed only to know that some subsequence index lies beyond $K$ .
The limit is forced to be $L$ , not merely to exist. Combined with uniqueness of limits (A sequence has at most one limit), this says that a Cauchy sequence has at most one subsequential limit, so for Cauchy sequences the subsequential limit set (Subsequential limit of a real sequence, and the subsequential limit set) is empty or a single point, and The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges rules out the empty case in $\mathbb{R}$ .

Depends on

Used by

The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 29 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 (Thm 3.11(b)) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.4 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §2.4 (standard reference, not scraped)