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Every Cauchy sequence of reals is bounded

Statement

Every Cauchy sequence of reals is bounded: if $(x_{k})$ is a Cauchy sequence (Limits and Cauchy sequences of reals) then there is $M \in R$ with $∣ x_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

This is the real-number counterpart of the lemma proving the same statement for Cauchy sequences of rationals inside $Q$ , and the argument is the same one: the Cauchy condition at a single value of $ε$ confines all but finitely many terms, and the finitely many exceptions are handled by a maximum.

Facts & Assumptions

Given: A Cauchy sequence $(x_{k})$ of reals.

[A1]

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

[L1]

Triangle inequality: $∣ x ∣ = ∣ (x - y) + y ∣ \leq ∣ x - y ∣ + ∣ y ∣$ for all reals $x, y$ (The triangle inequality).

[L2]

Every nonempty finite list of reals has a maximum, so $\max {a_{0}, \dots, a_{n}}$ is a well-determined real that dominates each listed value (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L3]

The rational $1$ is positive, and the embedding of $Q$ in $R$ carries it to $1 \in R$ , so $ε = 1$ is an admissible test value in [A1] (The rationals embed densely in the reals).

[L4]

Order arithmetic in $R$ : translation invariance, $a < b \Rightarrow a + c < b + c$ (Order is preserved by adding a constant and by adding inequalities); and the mixed transitivity $a \leq b < c \Rightarrow a < c$ , immediate from the reading of $a \leq b$ as " $a < b$ or $a = b$ " together with transitivity of $<$ (Complete ordered field (least-upper-bound property), Ordered field).

[L5]

The order on $N$ is total, so every index satisfies $k \leq K$ or $k \geq K$ ( $\leq$ is a linear order on $N$ ).

[L6]

A sequence of reals is bounded when some $M \in R$ satisfies $∣ x_{k} ∣ \leq M$ at every index (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Proof

technique · direct

1.1A1L3choose

Apply [A1] with the rational test value $ε = 1$ : fix $K \in N$ such that $∣ x_{k} - x_{l} ∣ < 1$ for all $k, l \geq K$ .

1.2L1

For all reals $x_{k}$ and $x_{K}$ the triangle inequality gives $∣ x_{k} ∣ \leq ∣ x_{k} - x_{K} ∣ + ∣ x_{K} ∣$ .

2.1step 1.1step 1.2L4

For every $k \geq K$ : $∣ x_{k} - x_{K} ∣ < 1$ by step 1.1, and adding $∣ x_{K} ∣$ to both sides then combining with step 1.2 gives $∣ x_{k} ∣ < ∣ x_{K} ∣ + 1$ .

2.2step 1.1L2choose

Define $M : = \max {∣ x_{0} ∣, ∣ x_{1} ∣, \dots, ∣ x_{K} ∣, ∣ x_{K} ∣ + 1}$ , the maximum of a nonempty finite list of reals, which exists by [L2].

3.1step 2.2L2

For every $k \leq K$ : $∣ x_{k} ∣$ is one of the listed values, so $∣ x_{k} ∣ \leq M$ .

3.2step 2.1step 2.2L2

For every $k \geq K$ : $∣ x_{k} ∣ < ∣ x_{K} ∣ + 1 \leq M$ , since $∣ x_{K} ∣ + 1$ is one of the listed values.

4.1step 3.1step 3.2L5L6∎

Every index satisfies $k \leq K$ or $k \geq K$ , so $∣ x_{k} ∣ \leq M$ for every $k \in N$ and $(x_{k})$ is bounded.

Remarks

One value of $ε$ suffices, and $ε = 1$ is not special. Any single positive rational would do; what matters is that the Cauchy condition confines all terms from some index onward to within a fixed distance of one term, after which only finitely many terms remain, and a finite list of reals has a maximum (Every nonempty finite set of reals has a maximum and a minimum). This is the same division of labour as in Every convergent sequence is bounded.
The converse is false. A bounded sequence need not be Cauchy: the alternating sequence of FALSE: every bounded sequence converges is bounded and, being divergent, is not Cauchy (Every convergent sequence is Cauchy would otherwise make it convergent by The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges). Boundedness is strictly weaker, and what it does yield is a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).
No completeness is used. The argument runs in any ordered field, and it is used here as the first of the three steps by which the least-upper-bound property is converted into Cauchy completeness in The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges.
The rational counterpart, proved on the Cauchy-construction page, is Every Cauchy sequence of rationals is bounded. It is the house-style exemplar for this argument, and nothing here depends on it, since the two lemmas live in different fields.

Every Cauchy sequence of reals is bounded

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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