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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 $L \in R$ , 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 $n : N \to N$ , and $L \in R$ with $x_{n_{j}} \to L$ .

[A1]

Cauchy condition: for every rational $ε > 0$ there is $K$ with $∣ x_{k} - x_{l} ∣ < ε$ for all $k, l \geq K$ (Limits and Cauchy sequences of reals).

[A2]

Convergence of the subsequence: for every rational $ε > 0$ there is $J$ with $∣ x_{n_{j}} - L ∣ < ε$ for all $j \geq J$ (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: $∣ x - L ∣ = ∣ (x - y) + (y - L) ∣ \leq ∣ x - y ∣ + ∣ y - L ∣$ (The triangle inequality).

[L2]

Growth of an index map: a strictly increasing $n$ satisfies $n_{j} \geq j$ for every $j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L3]

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

[L4]

The order on $N$ is total and transitive, so two indices $J, K$ admit an index $j$ with $j \geq J$ and $j \geq K$ ( $\leq$ is a linear order on $N$ ).

[L5]

Convergence: $(x_{k})$ converges to $L$ when for every rational $ε > 0$ there is $K$ with $∣ x_{k} - L ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

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

givenL3

2.1

By [A1] applied to $ε / 2$ , fix $K \in N$ with $∣ x_{k} - x_{l} ∣ < ε / 2$ for all $k, l \geq K$ .

step 1.1A1choose

2.2

By [A2] applied to $ε / 2$ , fix $J \in N$ with $∣ x_{n_{j}} - L ∣ < ε / 2$ for all $j \geq J$ .

step 1.1A2choose

3.1

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

step 2.1step 2.2L2L4choose

4.1

For every $k \geq K$ : $∣ x_{k} - L ∣ \leq ∣ x_{k} - x_{n_{j}} ∣ + ∣ x_{n_{j}} - L ∣ < ε / 2 + ε / 2 = ε$ .

step 2.1step 2.2step 3.1L1

5.1

The rational $ε > 0$ 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 $n_{j} \geq j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ) 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 $R$ .

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 · two levels

31 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 (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)