Completeness of the reals

Theorem

Completeness of the reals

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Statement

Let \( (x_n)_{n \in \mathbb{N}} \) be a sequence of real numbers, and equip \( \mathbb{R} \) with its standard metric \( d(x,y) = |x-y| \). Then \( (x_n) \) converges to some limit \( L \in \mathbb{R} \) if and only if \( (x_n) \) is a Cauchy sequence: for every \( \varepsilon \gt 0 \) there exists \( N \in \mathbb{N} \) such that \( |x_n - x_m| \lt \varepsilon \) for all \( m, n \ge N \). The forward implication holds in any metric space; the reverse implication uses the completeness axiom of \( \mathbb{R} \) (every non-empty set bounded above has a least upper bound), invoked here through the Bolzano–Weierstrass theorem.

Why it matters

The Cauchy criterion is an intrinsic test for convergence: it lets you prove a sequence converges without knowing, guessing, or even being able to name its limit. Almost every existence theorem in analysis — convergence of absolutely convergent series, the contraction mapping principle, existence of solutions to differential equations by Picard iteration, completeness of function spaces — ultimately funnels through this theorem. It converts “the terms eventually cluster among themselves” into “there is an actual real number they cluster around.”

It is also the property that distinguishes \( \mathbb{R} \) from \( \mathbb{Q} \). Both are ordered fields, both are dense in themselves, but only \( \mathbb{R} \) has no “holes”: Cauchy sequences of rationals can aim at a gap (such as \( \sqrt{2} \)), while Cauchy sequences of reals always land. Indeed Cantor’s 1872 construction defines the reals as equivalence classes of rational Cauchy sequences, making this theorem true by design.

Hypotheses

Hypothesis (the ambient space is \( \mathbb{R} \), a complete ordered field).In \( \mathbb{Q} \) the theorem fails: set \( x_1 = 1 \), \( x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n} \). Each \( x_n \) is rational, the sequence is Cauchy (it converges in \( \mathbb{R} \) to \( \sqrt{2} \), and any sequence convergent in \( \mathbb{R} \) is Cauchy), but it has no limit in \( \mathbb{Q} \) because \( \sqrt{2} \notin \mathbb{Q} \). The same failure occurs in any non-closed subset of \( \mathbb{R} \): in \( (0,1) \) the sequence \( x_n = \frac{1}{n+1} \) is Cauchy but its would-be limit \( 0 \) is missing from the space.

Hypothesis (the standard metric \( d(x,y)=|x-y| \)).Completeness is a metric property, not a topological one. Put the metric \( d(x,y) = |\arctan x - \arctan y| \) on \( \mathbb{R} \); it induces the usual topology, yet \( x_n = n \) becomes Cauchy (since \( \arctan n \to \frac{\pi}{2} \)) while converging to nothing. So the theorem genuinely depends on which metric you use, not merely on the underlying topological space.

Hypothesis (Cauchy condition uniform over ALL pairs \( m,n \ge N \)).The weaker condition \( |x_{n+1} - x_n| \to 0 \) does not suffice: the harmonic partial sums \( H_n = \sum_{k=1}^{n} \frac{1}{k} \) satisfy \( |H_{n+1} - H_n| = \frac{1}{n+1} \to 0 \) yet \( H_{2n} - H_n \ge \frac{1}{2} \) for every \( n \), so \( (H_n) \) is not Cauchy and diverges.

Proof

We prove the two implications separately. Write \( \varepsilon \gt 0 \) for an arbitrary tolerance throughout.

Direction 1: convergent \( \Rightarrow \) Cauchy.

Suppose \( x_n \to L \). Given \( \varepsilon \gt 0 \), choose \( N \in \mathbb{N} \) such that \[ |x_n - L| \lt \frac{\varepsilon}{2} \quad \text{for all } n \ge N. \]

Definition of the limit of a sequence, applied with tolerance \( \varepsilon/2 \). A

For all \( m, n \ge N \): \[ |x_n - x_m| \le |x_n - L| + |L - x_m| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \]

Triangle inequality for the absolute value, then the bound from Step 1 applied to each term. Hence \( (x_n) \) is Cauchy. A

Direction 2: Cauchy \( \Rightarrow \) convergent. Suppose \( (x_n) \) is Cauchy.

Claim: \( (x_n) \) is bounded. Apply the Cauchy condition with \( \varepsilon = 1 \) to get \( N_0 \) with \( |x_n - x_{N_0}| \lt 1 \) for all \( n \ge N_0 \); hence \( |x_n| \lt |x_{N_0}| + 1 \) for \( n \ge N_0 \), and \[ |x_n| \le M := \max\left\{ |x_1|, \dots, |x_{N_0 - 1}|, \, |x_{N_0}| + 1 \right\} \quad \forall n \in \mathbb{N}. \]

Cauchy condition instantiated at \( \varepsilon = 1 \), reverse triangle inequality \( |x_n| - |x_{N_0}| \le |x_n - x_{N_0}| \), and the fact that a maximum of finitely many reals exists. A

Since \( (x_n) \) is bounded, it has a convergent subsequence: there exist indices \( n_1 \lt n_2 \lt n_3 \lt \cdots \) and \( L \in \mathbb{R} \) with \[ x_{n_k} \to L \quad \text{as } k \to \infty. \]

Bolzano–Weierstrass theorem: every bounded real sequence has a convergent subsequence. This is where completeness of \( \mathbb{R} \) enters: Bolzano–Weierstrass follows from the least upper bound axiom via the monotone subsequence lemma (every real sequence has a monotone subsequence) combined with the monotone convergence theorem (a bounded monotone sequence converges to its supremum or infimum, which exists by the axiom). B

Claim: the whole sequence converges to \( L \). Given \( \varepsilon \gt 0 \), pick \( N \) with \[ |x_n - x_m| \lt \frac{\varepsilon}{2} \quad \text{for all } m, n \ge N, \] then pick \( k \) large enough that both \( n_k \ge N \) and \( |x_{n_k} - L| \lt \frac{\varepsilon}{2} \).

The first choice is the Cauchy condition at tolerance \( \varepsilon/2 \). The second is possible because \( x_{n_k} \to L \) (Step 4) and because subsequence indices satisfy \( n_k \ge k \) (proved by induction on \( k \) from \( n_1 \ge 1 \) and \( n_{k+1} \gt n_k \)), so both requirements hold for all sufficiently large \( k \) simultaneously. C

For every \( n \ge N \): \[ |x_n - L| \le |x_n - x_{n_k}| + |x_{n_k} - L| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \]

Triangle inequality; the first term is controlled by Step 5’s Cauchy bound (valid since \( n \ge N \) and \( n_k \ge N \)), the second by the subsequential convergence. Since \( \varepsilon \gt 0 \) was arbitrary, \( x_n \to L \). This completes both directions. B

Result

\[ (x_n) \text{ converges in } \mathbb{R} \iff \forall \varepsilon \gt 0 \ \exists N \in \mathbb{N} \ \forall m, n \ge N : \ |x_n - x_m| \lt \varepsilon. \]

Reading. A real sequence has a limit exactly when its terms eventually get, and stay, arbitrarily close to each other. You never need to produce the limit in advance: internal clustering guarantees an actual real number is being approached.

Scope. Both directions hold for sequences in \( \mathbb{R} \) (and, coordinatewise, in \( \mathbb{R}^k \) and \( \mathbb{C} \)) with the standard metric. In a general metric space only the forward direction survives; the reverse direction is precisely the definition of the space being complete, and it fails in \( \mathbb{Q} \), in \( (0,1) \), and in \( \mathbb{R} \) under exotic metrics.

Corollaries & converses

Cauchy criterion for series. \( \sum a_k \) converges iff \( \forall \varepsilon \gt 0 \ \exists N \ \forall m \gt n \ge N: \left| \sum_{k=n+1}^{m} a_k \right| \lt \varepsilon \) — apply the theorem to the partial sums.
Absolute convergence implies convergence. If \( \sum |a_k| \) converges then \( \left| \sum_{k=n+1}^{m} a_k \right| \le \sum_{k=n+1}^{m} |a_k| \), so the partial sums of \( \sum a_k \) are Cauchy. This underwrites the ratio, root, and comparison tests.
Contractive sequences converge. If \( |x_{n+2} - x_{n+1}| \le r\,|x_{n+1} - x_n| \) with \( 0 \le r \lt 1 \), a geometric tail estimate shows \( (x_n) \) is Cauchy, hence convergent (see Worked Example 1). This is the one-dimensional seed of the Banach fixed point theorem.
Completeness of \( \mathbb{R}^k \) and \( \mathbb{C} \). A sequence in \( \mathbb{R}^k \) is Cauchy iff each coordinate sequence is Cauchy; apply the theorem coordinatewise.
Converse. The statement is already a biconditional, so there is no separate converse for \( \mathbb{R} \). The correct general-space statement: “convergent \( \Rightarrow \) Cauchy” holds in every metric space, while “Cauchy \( \Rightarrow \) convergent” characterises complete spaces — counterexample \( x_n = \frac{1}{n} \) in the space \( (0,1) \).
Equivalence with other completeness forms. Over an Archimedean ordered field, Cauchy completeness is equivalent to the least upper bound property, to the monotone convergence theorem, to Bolzano–Weierstrass, and to the nested interval principle (Problem 5 proves one arrow).

Fails without

Drop completeness of the field (work in \( \mathbb{Q} \)): the Babylonian sequence \( x_1 = 1 \), \( x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n} \) is a Cauchy sequence of rationals with no rational limit. Its limit in \( \mathbb{R} \) satisfies \( L = \frac{L}{2} + \frac{1}{L} \), i.e. \( L^2 = 2 \), and \( \sqrt{2} \notin \mathbb{Q} \) by the classical parity argument. So “Cauchy \( \Rightarrow \) convergent” is false in \( \mathbb{Q} \).
Drop closedness of the ambient set (work in \( (0,1) \) with the usual metric): \( x_n = \frac{1}{n+1} \) is Cauchy in \( (0,1) \) — the Cauchy condition only mentions distances between terms — but converges to no point of \( (0,1) \), since its only candidate limit in \( \mathbb{R} \) is \( 0 \), which lies outside.
Change the metric: with \( d(x,y) = |\arctan x - \arctan y| \) on \( \mathbb{R} \), the sequence \( x_n = n \) satisfies \( d(x_n, x_m) = |\arctan n - \arctan m| \to 0 \), so it is Cauchy, but it converges to no real number in this metric either (its \( \arctan \)-images approach \( \frac{\pi}{2} \), which is \( \arctan \) of nothing).
Weaken “all pairs \( m,n \ge N \)” to “consecutive terms close”: \( H_n = \sum_{k=1}^{n} \frac{1}{k} \) has \( H_{n+1} - H_n = \frac{1}{n+1} \to 0 \), yet \( H_{2n} - H_n = \sum_{k=n+1}^{2n} \frac{1}{k} \ge n \cdot \frac{1}{2n} = \frac{1}{2} \), so \( (H_n) \) is not Cauchy and indeed diverges to \( +\infty \).

Common errors

The consecutive-terms fallacy. Concluding “Cauchy” from \( |x_{n+1} - x_n| \to 0 \). The Cauchy condition demands one \( N \) working for every pair \( m,n \ge N \) at once; the harmonic partial sums kill this instantly.
The fixed-gap fallacy. Verifying only \( |x_{n+k} - x_n| \to 0 \) for each fixed \( k \). Same trap in disguise: \( H_{n+k} - H_n \to 0 \) for every fixed \( k \), but \( (H_n) \) diverges. The gap \( m - n \) must be allowed to grow with \( n \).
Circular use of the limit. Proving a sequence is Cauchy by estimating \( |x_n - L| \) when the whole point is that \( L \) is not yet known to exist. Legitimate Cauchy estimates bound \( |x_n - x_m| \) using only the terms themselves.
Quantifier inversion. Writing \( \exists N \ \forall \varepsilon \gt 0 \ \forall m,n \ge N: |x_n - x_m| \lt \varepsilon \). This says the sequence is eventually constant — a far stronger property. The order is \( \forall \varepsilon \ \exists N \).
Portability error. Applying “Cauchy \( \Rightarrow \) convergent” inside a subset such as \( \mathbb{Q} \), \( (0,1) \), or \( C[0,1] \) with the \( L^1 \) norm. The theorem certifies a limit in \( \mathbb{R} \); whether the limit stays inside your subset is a separate (closedness/completeness) question.
Proving Bolzano–Weierstrass from this theorem in a foundations course. Depending on the axiom chosen for \( \mathbb{R} \), the dependency may run the other way; citing results in a circle is a real danger when several completeness formulations float around. Fix a dependency chain (here: LUB axiom \( \Rightarrow \) monotone convergence \( \Rightarrow \) Bolzano–Weierstrass \( \Rightarrow \) this theorem) and stick to it.

Discussion

The theorem is the precise statement that \( \mathbb{R} \) has no holes. In \( \mathbb{Q} \), a sequence can satisfy every finitary closeness condition and still fail to converge because its target was never in the space; the decimal truncations \( 1, 1.4, 1.41, 1.414, \dots \) of \( \sqrt{2} \) are the canonical example. Completeness says this cannot happen in \( \mathbb{R} \): internal coherence of a sequence is sufficient evidence for the existence of a limit. This is a genuinely non-constructive existence principle — the proof via Bolzano–Weierstrass tells you a limit exists without giving a formula for it — and that is exactly what makes it powerful in applications where limits have no closed form.

Historically, the criterion appears in Bolzano’s 1817 pamphlet on the intermediate value theorem and in Cauchy’s 1821 Cours d’analyse, both of whom treated the “Cauchy \( \Rightarrow \) convergent” direction as evident — understandably, since no rigorous definition of the real numbers yet existed to prove it from. The gap was closed in 1872 when Cantor (following Heine and Méray) constructed \( \mathbb{R} \) as equivalence classes of rational Cauchy sequences, and Dedekind independently via cuts. On Cantor’s construction the theorem is nearly definitional; on Dedekind’s (or from the LUB axiom, as here) it is a theorem requiring the Bolzano–Weierstrass argument above.

The concept generalises far beyond \( \mathbb{R} \): a metric space in which every Cauchy sequence converges is called complete, and completeness is the working hypothesis of the deepest tools of analysis — the Banach fixed point theorem, the Baire category theorem, the completeness of \( L^p \) spaces (Riesz–Fischer). Every metric space embeds densely in an essentially unique completion, constructed exactly as Cantor constructed \( \mathbb{R} \): the completion of \( \mathbb{Q} \) under \( |\cdot| \) is \( \mathbb{R} \), and the same machine applied to the \( p \)-adic absolute value yields \( \mathbb{Q}_p \).

A subtler point: Cauchy completeness alone does not pin down \( \mathbb{R} \) among ordered fields. There exist non-Archimedean ordered fields that are Cauchy complete (with respect to sequences, or even nets), such as fields of formal Laurent series with real coefficients; in such fields the least upper bound property still fails — the set of infinitesimals is bounded above by \( 1 \) but has no supremum. What is true is that Cauchy completeness + the Archimedean property \( \iff \) the least upper bound property (Problem 5 proves the forward arrow; the reverse is Steps 3–6 plus the standard proof that LUB implies Archimedes). So when this theorem is advertised as “equivalent to completeness,” the Archimedean property is being smuggled in, and honest axiomatics must say so.

Common misconceptions. “Cauchy” is not the same as “consecutive terms getting close” (see Common errors); “complete” is not the same as “closed” (completeness is intrinsic to the space, closedness is relative to an ambient space — though a subset of a complete space is complete iff it is closed); and the theorem does not say Cauchy sequences converge quickly — no rate is implied, and none can be, since the criterion is purely qualitative.

Worked examples

Example 1 (a recursively defined sequence with no obvious limit). Define \( x_1 = 1 \) and \( x_{n+1} = 1 + \dfrac{1}{1 + x_n} \). Show \( (x_n) \) converges and find its limit.

Every term satisfies \( x_n \ge 1 \): true for \( x_1 = 1 \), and if \( x_n \ge 1 \) then \( x_{n+1} = 1 + \frac{1}{1+x_n} \gt 1 \).

Induction on \( n \); the fraction \( \frac{1}{1+x_n} \) is positive because \( 1 + x_n \ge 2 \gt 0 \). A

Consecutive gaps contract: \[ |x_{n+1} - x_n| = \left| \frac{1}{1+x_n} - \frac{1}{1+x_{n-1}} \right| = \frac{|x_n - x_{n-1}|}{(1+x_n)(1+x_{n-1})} \le \frac{1}{4}\,|x_n - x_{n-1}|. \]

Common denominator, then the bound \( 1 + x_k \ge 2 \) from Step 1 applied to both factors, giving denominator \( \ge 4 \). B

Iterating: \( |x_{n+1} - x_n| \le \left(\frac{1}{4}\right)^{n-1} |x_2 - x_1| = \frac{1}{2}\left(\frac{1}{4}\right)^{n-1} \), and for \( m \gt n \), \[ |x_m - x_n| \le \sum_{j=n}^{m-1} |x_{j+1} - x_j| \le \frac{1}{2} \sum_{j=n}^{\infty} \left(\frac{1}{4}\right)^{j-1} = \frac{2}{3}\left(\frac{1}{4}\right)^{n-1}. \]

Induction using Step 2 (with \( x_2 = \frac{3}{2} \), so \( |x_2 - x_1| = \frac{1}{2} \)); telescoping and the triangle inequality; then the geometric series formula \( \sum_{j \ge n} r^{j-1} = \frac{r^{n-1}}{1-r} \) with \( r = \frac{1}{4} \). B

Given \( \varepsilon \gt 0 \), choose \( N \) with \( \frac{2}{3}\left(\frac{1}{4}\right)^{N-1} \lt \varepsilon \); then \( |x_m - x_n| \lt \varepsilon \) for all \( m, n \ge N \). So \( (x_n) \) is Cauchy, and by the theorem it converges to some \( L \in \mathbb{R} \).

Such \( N \) exists because \( \left(\frac{1}{4}\right)^{n} \to 0 \) (geometric decay); then this theorem supplies the limit \( L \) — note we could not have named \( L \) in advance. B

Passing to the limit in the recursion: \( L = 1 + \frac{1}{1+L} \), so \( (L-1)(L+1) = 1 \), i.e. \( L^2 = 2 \). Since \( x_n \ge 1 \) forces \( L \ge 1 \), we get \( L = \sqrt{2} \).

Algebra of limits (the map \( t \mapsto 1 + \frac{1}{1+t} \) is continuous at \( L \ge 1 \), and \( x_{n+1} \to L \) because a tail of a convergent sequence converges to the same limit); order limit theorem for \( L \ge 1 \). A

\[ x_1 = 1, \quad x_{n+1} = 1 + \frac{1}{1+x_n} \ \Longrightarrow \ x_n \to \sqrt{2}. \]

Reading. Contraction of successive gaps gave a geometric tail bound, the Cauchy criterion converted that into existence of a limit, and only then could we solve for the limit. Existence first, identification second — the standard division of labour this theorem enables.

Scope. The same three-step pattern (contract, sum the geometric tail, invoke completeness) proves the Banach fixed point theorem in any complete metric space.

Example 2 (convergence of a series with no closed-form sum). Show that \( \displaystyle\sum_{k=1}^{\infty} \frac{\cos k}{k^2} \) converges.

Let \( s_n = \sum_{k=1}^{n} \frac{\cos k}{k^2} \). For \( m \gt n \ge 1 \): \[ |s_m - s_n| = \left| \sum_{k=n+1}^{m} \frac{\cos k}{k^2} \right| \le \sum_{k=n+1}^{m} \frac{|\cos k|}{k^2} \le \sum_{k=n+1}^{m} \frac{1}{k^2}. \]

Triangle inequality on the finite sum, then \( |\cos k| \le 1 \). A

For \( k \ge 2 \), \( \frac{1}{k^2} \le \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k} \), so the tail telescopes: \[ \sum_{k=n+1}^{m} \frac{1}{k^2} \le \sum_{k=n+1}^{m} \left( \frac{1}{k-1} - \frac{1}{k} \right) = \frac{1}{n} - \frac{1}{m} \lt \frac{1}{n}. \]

Termwise comparison \( k^2 \ge k(k-1) \) for \( k \ge 2 \) (valid since \( n + 1 \ge 2 \)), partial fractions, telescoping cancellation. B

Given \( \varepsilon \gt 0 \), choose \( N \in \mathbb{N} \) with \( N \gt \frac{1}{\varepsilon} \). Then for all \( m \gt n \ge N \), \( |s_m - s_n| \lt \frac{1}{n} \le \frac{1}{N} \lt \varepsilon \). Hence \( (s_n) \) is Cauchy, and by the theorem it converges.

The Archimedean property of \( \mathbb{R} \) supplies \( N \); then this theorem applied to the partial-sum sequence. No candidate for the sum was ever needed — indeed this series has no elementary closed form. B

\[ \sum_{k=1}^{\infty} \frac{\cos k}{k^2} \ \text{converges in } \mathbb{R}. \]

Reading. We bounded the clustering of partial sums by a telescoping tail and let completeness manufacture the limit. This is the Cauchy criterion for series, and the identical estimate proves the general comparison test.

Scope. Works verbatim for any series dominated termwise in absolute value by a convergent positive series — the engine behind “absolute convergence implies convergence.”

Problems

Show directly from the definitions that \( x_n = (-1)^n \) is not Cauchy, and deduce (using only the easy direction of the theorem) that it does not converge.
Solution
For every \( n \), \( |x_{n+1} - x_n| = |(-1)^{n+1} - (-1)^n| = 2 \). Take \( \varepsilon = 1 \): for any proposed \( N \), the pair \( m = N+1 \), \( n = N \) satisfies \( m, n \ge N \) but \( |x_m - x_n| = 2 \ge 1 \). So no \( N \) works and the sequence is not Cauchy. By the theorem’s forward direction (convergent \( \Rightarrow \) Cauchy), a convergent sequence must be Cauchy; contrapositively, \( ((-1)^n) \) does not converge.
Prove that the harmonic partial sums \( H_n = \sum_{k=1}^{n} \frac{1}{k} \) satisfy \( |H_{n+1} - H_n| \to 0 \) but that \( (H_n) \) diverges. What does this show about the Cauchy condition?
Solution
First, \( H_{n+1} - H_n = \frac{1}{n+1} \to 0 \). But for every \( n \), \[ H_{2n} - H_n = \sum_{k=n+1}^{2n} \frac{1}{k} \ge \sum_{k=n+1}^{2n} \frac{1}{2n} = n \cdot \frac{1}{2n} = \frac{1}{2}, \] since each of the \( n \) terms is at least \( \frac{1}{2n} \). Take \( \varepsilon = \frac{1}{2} \): for any \( N \), choosing \( n = N \) and \( m = 2N \) gives \( m, n \ge N \) with \( |H_m - H_n| \ge \frac{1}{2} \). So \( (H_n) \) is not Cauchy, hence by the theorem it does not converge; being increasing and non-convergent, it diverges to \( +\infty \). Moral: the Cauchy condition is a uniform condition over all pairs of large indices, strictly stronger than closeness of consecutive (or fixed-gap) terms.
(Contraction lemma.) Suppose there exist \( C \ge 0 \) and \( 0 \le r \lt 1 \) with \( |x_{n+1} - x_n| \le C r^n \) for all \( n \). Prove \( (x_n) \) converges.
Solution
For \( m \gt n \), telescope and apply the triangle inequality: \[ |x_m - x_n| \le \sum_{j=n}^{m-1} |x_{j+1} - x_j| \le C \sum_{j=n}^{m-1} r^j \le C \sum_{j=n}^{\infty} r^j = \frac{C r^n}{1-r}, \] using the geometric series formula, valid since \( 0 \le r \lt 1 \). Because \( r^n \to 0 \) (for \( r = 0 \) trivially; for \( 0 \lt r \lt 1 \) write \( r = \frac{1}{1+h} \) with \( h \gt 0 \), so Bernoulli’s inequality gives \( r^n \le \frac{1}{1+nh} \to 0 \) by the Archimedean property), for any \( \varepsilon \gt 0 \) there is \( N \) with \( \frac{C r^N}{1-r} \lt \varepsilon \), and then \( |x_m - x_n| \lt \varepsilon \) for all \( m, n \ge N \) (the case \( m = n \) is trivial, and the bound is monotone in \( n \)). Thus \( (x_n) \) is Cauchy, and by the completeness theorem it converges.
Let \( x_1 = 0 \), \( x_2 = 1 \), and \( x_{n+2} = \frac{x_{n+1} + x_n}{2} \). Prove that \( (x_n) \) converges and compute its limit.
Solution
Let \( d_n = x_{n+1} - x_n \). Then \[ d_{n+1} = x_{n+2} - x_{n+1} = \frac{x_{n+1} + x_n}{2} - x_{n+1} = -\frac{x_{n+1} - x_n}{2} = -\frac{d_n}{2}, \] so by induction \( d_n = \left(-\frac{1}{2}\right)^{n-1} d_1 = \left(-\frac{1}{2}\right)^{n-1} \) (since \( d_1 = 1 \)). Hence \( |x_{n+1} - x_n| = \left(\frac{1}{2}\right)^{n-1} \le 2 \cdot \left(\frac{1}{2}\right)^{n} \), so Problem 3 with \( C = 2 \), \( r = \frac{1}{2} \) shows \( (x_n) \) converges; call the limit \( L \). To identify \( L \): the partial sums of \( \sum d_n \) are \( x_{n+1} - x_1 = x_{n+1} \), and \( \sum_{n=1}^{\infty} d_n \) is a geometric series with first term \( 1 \) and ratio \( -\frac{1}{2} \), so \[ L = \lim_{n \to \infty} x_{n+1} = \sum_{n=1}^{\infty} \left(-\frac{1}{2}\right)^{n-1} = \frac{1}{1 - \left(-\frac{1}{2}\right)} = \frac{2}{3}. \] (Check: the limit lies between consecutive terms \( x_2 = 1 \), \( x_3 = \frac{1}{2} \), \( x_4 = \frac{3}{4} \), consistent with \( \frac{2}{3} \).)
(Cauchy completeness + Archimedean property \( \Rightarrow \) least upper bound property.) Assume every Cauchy sequence of reals converges and that \( \mathbb{R} \) is Archimedean (for every \( x \in \mathbb{R} \) there is \( n \in \mathbb{N} \) with \( n \gt x \)). Prove that every non-empty set \( S \subseteq \mathbb{R} \) bounded above has a least upper bound.
Solution
Pick \( s_0 \in S \) and an upper bound \( b_1 \) of \( S \); set \( a_1 = s_0 - 1 \), which is not an upper bound (it is less than \( s_0 \in S \)). Construct \( [a_n, b_n] \) by bisection: let \( m_n = \frac{a_n + b_n}{2} \); if \( m_n \) is an upper bound of \( S \), set \( a_{n+1} = a_n \), \( b_{n+1} = m_n \); otherwise set \( a_{n+1} = m_n \), \( b_{n+1} = b_n \). By induction: every \( b_n \) is an upper bound, no \( a_n \) is an upper bound, \( a_n \le a_{n+1} \le b_{n+1} \le b_n \), and \( b_n - a_n = 2^{1-n}(b_1 - a_1) \).

\( (b_n) \) is Cauchy. For \( m \ge n \), both \( b_m \) and \( b_n \) lie in \( [a_n, b_n] \), so \( |b_m - b_n| \le 2^{1-n}(b_1 - a_1) \). Given \( \varepsilon \gt 0 \), Bernoulli’s inequality gives \( 2^{n} \ge 1 + n \), and the Archimedean property supplies \( N \) with \( (1 + N)\varepsilon \gt 2(b_1 - a_1) \); then \( n \ge N \) forces \( 2^{1-n}(b_1 - a_1) \lt \varepsilon \). (This is exactly where non-Archimedean fields escape the argument: there \( 2^{-n} \) need not become smaller than a fixed infinitesimal \( \varepsilon \).) By hypothesis \( b_n \to L \) for some \( L \in \mathbb{R} \); since \( |a_n - b_n| \to 0 \), also \( a_n \to L \).

\( L \) is an upper bound. Suppose \( x \in S \) with \( x \gt L \). Since \( b_n \to L \), there is \( n \) with \( b_n \lt x \) (take the limit tolerance \( \varepsilon = x - L \), noting \( b_n \le b_1 \) and \( b_n \ge a_n \) with \( b_n \to L \) gives \( b_n \lt L + (x - L) = x \)). This contradicts \( b_n \) being an upper bound of \( S \).

No smaller upper bound. Let \( u \lt L \). Since \( a_n \to L \), there is \( n \) with \( a_n \gt u \). Since \( a_n \) is not an upper bound of \( S \), there is \( s \in S \) with \( s \gt a_n \gt u \), so \( u \) is not an upper bound. Therefore \( L = \sup S \). Combined with Steps 3–6 of the main proof (which derive Cauchy completeness from the LUB axiom via Bolzano–Weierstrass) and the standard fact that LUB implies Archimedes, this shows: over an ordered field, LUB \( \iff \) Cauchy completeness + Archimedean property.