Cauchy Sequences and Completeness

Suppose a sequence of rational numbers gets tighter together, term after term. Yet you cannot name the limit. It happens to be irrational. Does the sequence still "converge"? This question exposes a subtlety hiding inside the very idea of a limit. The definition of convergence requires knowing the limit in advance. But often we only know that terms are clustering. Cauchy sequences solve this problem. They define a notion of "clustering" that refers only to the terms themselves. It never refers to a limit. The remarkable fact — and the centerpiece of this article — is this. In the real numbers, this internal clustering condition is equivalent to convergence. This equivalence is not free. It is precisely where the completeness of R\mathbb{R} enters analysis. This is what distinguishes R\mathbb{R} from Q\mathbb{Q}. A sequence of rationals can be Cauchy without converging to a rational number. But every Cauchy sequence of reals converges to a real number. This single theorem — the Cauchy Criterion — lets us prove a sequence converges. We never need to exhibit its limit. The idea reaches far beyond a technical curiosity. It underlies the construction of R\mathbb{R} itself from Q\mathbb{Q} (as equivalence classes of Cauchy sequences). It is the standard tool for proving convergence of series and iterative algorithms. It also generalizes to the notion of a complete metric space. This concept is indispensable in functional analysis, differential equations, and numerical methods. This article gives the precise definition of a Cauchy sequence. It proves the Cauchy Criterion for R\mathbb{R} in full rigor. This uses the Bolzano-Weierstrass Theorem. It also shows, with an explicit counterexample, why completeness of the ambient space cannot be dropped. It previews how this machinery feeds directly into the Cauchy Criterion for Series. It also connects to the theory of Limit Superior and Limit Inferior.

Key takeaways
  • A sequence is Cauchy if its terms eventually become arbitrarily close to each other, not just to some fixed limit.
  • Every convergent sequence (in any metric space) is automatically Cauchy — this direction is easy and needs no completeness.
  • In R\mathbb{R}, the converse also holds: every Cauchy sequence converges. This is the Cauchy Criterion, and it fails in Q\mathbb{Q}.
  • The proof of the converse combines boundedness of Cauchy sequences with Bolzano-Weierstrass Theorem to extract a convergent subsequence, then shows the whole sequence must converge to the same limit.
  • "Cauchy" is a completeness-free, self-contained test for convergence — useful exactly when you cannot guess the limit.
  • The definition generalizes verbatim to any metric space, where completeness becomes the defining property of spaces (like Rn\mathbb{R}^n, or Banach spaces) where the equivalence still holds.
  • Key idea: Convergence and the Cauchy property coincide in R\mathbb{R}, and this equivalence is what "completeness" means.

Where this lives today

The Cauchy criterion is the working definition of convergence whenever an explicit limit is unavailable. In numerical analysis, iterative root-finding and fixed-point methods (Newton's method, as in Example 1) are proven to converge by showing the iterates form a Cauchy sequence, which is often far easier than guessing the exact limit. In the theory of series, the Cauchy Criterion for partial sums underlies tests like the Cauchy Criterion for Series and the ratio/root tests. In functional analysis, Banach spaces are defined exactly as normed vector spaces that are complete with respect to their norm — Cauchy sequences of functions converge to functions in the same space — which is the bedrock of the existence theory for differential equations (Picard iteration) and of Fourier analysis. In probability theory, "Cauchy in probability" or "Cauchy almost surely" versions of this concept underlie almost-sure convergence theorems for martingales and random series. In computer science and cryptography, Cauchy-sequence-style arguments justify convergence of iterative algorithms (gradient descent, fixed-point iteration in cryptographic protocols) without needing closed-form limits, and pp-adic number systems in cryptography are literally built, like R\mathbb{R} itself, as completions of Q\mathbb{Q} with respect to a different (non-Archimedean) metric — the same completion machinery pioneered by the real-number construction from Cauchy sequences.

Ask the tutor: Can you show me how the Cauchy Criterion is used to prove the Ratio Test for series convergence?

DefinitionCauchy sequence

Let (an)n=1∞(a_n)_{n=1}^\infty be a sequence of real numbers. We say (an)(a_n) is a Cauchy sequence if for every ε>0\varepsilon > 0 there exists N∈NN \in \mathbb{N} such that for all m,n≥Nm, n \geq N,

∣an−am∣<ε.|a_n - a_m| < \varepsilon.

Equivalently: for every ε>0\varepsilon>0, all but finitely many terms of the sequence lie within an interval of length ε\varepsilon of each other, i.e. the tail (aN,aN+1,aN+2,… )(a_N, a_{N+1}, a_{N+2}, \dots) has diameter at most ε\varepsilon. (The same definition applies verbatim in any metric space (X,d)(X,d), replacing ∣an−am∣|a_n - a_m| with d(an,am)d(a_n, a_m). A metric space in which every Cauchy sequence converges is called complete.)

Intuition

Ordinary convergence an→La_n \to L says: pick any tolerance ε\varepsilon, and eventually every term is within ε\varepsilon of the target LL. This definition needs LL as an input. The Cauchy condition instead compares terms to each other: eventually, any two terms (however far apart in index) are within ε\varepsilon of each other. It's a statement purely about the internal geometry of the sequence, with no reference to an external point. Picture the sequence as a cloud of points on the number line, indexed by time. Convergence says the cloud collapses onto a single point LL. The Cauchy condition says the cloud's diameter shrinks to zero as time goes on — the points bunch together, but a priori we don't know they're bunching around any particular real number. The theorem below says: on the real line, a shrinking-diameter cloud of points must be collapsing onto some point, and that point is a real number. This fails on Q\mathbb{Q} because the "hole" the cloud collapses toward might not be a rational number — the cloud shrinks in diameter, but there's no rational point at its center.

Figure
Two sequences: one collapses onto a real limit (top), one shrinks in diameter but the "hole" it targets is irrational (bottom), illustrating why completeness of the ambient space matters
ℝ L = 2 a₁ a₂ a₃ a₄ terms cluster onto L ∈ ℝ ℚ √2 ∉ ℚ (no point here) 1.4 1.41 1.414 Cauchy in ℚ, no limit in ℚ

The main results

TheoremCauchy sequences are bounded

Every Cauchy sequence of real numbers (an)(a_n) is bounded: there exists M>0M > 0 such that ∣an∣≤M|a_n| \le M for all n∈Nn \in \mathbb{N}.

Proof

Apply the definition of Cauchy with ε=1\varepsilon = 1: there exists N∈NN \in \mathbb{N} such that for all m,n≥Nm,n \ge N, ∣an−am∣<1|a_n - a_m| < 1. In particular, taking m=Nm = N, for all n≥Nn \ge N,

∣an∣=∣an−aN+aN∣≤∣an−aN∣+∣aN∣<1+∣aN∣.|a_n| = |a_n - a_N + a_N| \le |a_n - a_N| + |a_N| < 1 + |a_N|.

So every term with index ≥N\ge N is bounded by 1+∣aN∣1 + |a_N|. The remaining finitely many terms a1,…,aN−1a_1, \dots, a_{N-1} are each individually a fixed real number, hence bounded by max⁡{∣a1∣,…,∣aN−1∣}\max\{|a_1|, \dots, |a_{N-1}|\} (an empty max, if N=1N=1, is vacuously not needed). Set

M=max⁡{∣a1∣,…,∣aN−1∣, 1+∣aN∣}.M = \max\{ |a_1|, \dots, |a_{N-1}|,\ 1 + |a_N| \}.

Then ∣an∣≤M|a_n| \le M for every n∈Nn \in \mathbb{N}. ■\blacksquare

∎
TheoremConvergent implies Cauchy

If (an)(a_n) is a sequence of real numbers that converges to a limit L∈RL \in \mathbb{R}, then (an)(a_n) is a Cauchy sequence.

Proof

Let ε>0\varepsilon > 0. Since an→La_n \to L, applying the definition of convergence with tolerance ε/2>0\varepsilon/2 > 0, there exists N∈NN \in \mathbb{N} such that for all n≥Nn \ge N,

∣an−L∣<ε2.|a_n - L| < \frac{\varepsilon}{2}.

Now let m,n≥Nm, n \ge N. By the triangle inequality,

∣an−am∣=∣(an−L)−(am−L)∣≤∣an−L∣+∣am−L∣<ε2+ε2=ε.|a_n - a_m| = |(a_n - L) - (a_m - L)| \le |a_n - L| + |a_m - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.

Since ε>0\varepsilon>0 was arbitrary, (an)(a_n) satisfies the Cauchy condition. ■\blacksquare

∎
TheoremThe Cauchy Criterion for R\mathbb{R}

A sequence of real numbers (an)(a_n) converges (to some real number) if and only if it is a Cauchy sequence.

Proof

(⇒\Rightarrow) This is exactly the previous theorem, Convergent implies Cauchy.

(⇐\Leftarrow) Suppose (an)(a_n) is Cauchy. We must produce a real number LL with an→La_n \to L. The proof technique is: bound the sequence, extract a convergent subsequence via Bolzano-Weierstrass Theorem, then show the whole sequence converges to that subsequential limit using the Cauchy property to control the gap. Step 1 (boundedness). By the theorem Cauchy sequences are bounded, (an)(a_n) is bounded. Step 2 (extract a convergent subsequence). Since (an)(a_n) is a bounded sequence of real numbers, the Bolzano-Weierstrass Theorem guarantees a subsequence (ank)k=1∞(a_{n_k})_{k=1}^\infty that converges to some L∈RL \in \mathbb{R}. Step 3 (the whole sequence converges to LL). Let ε>0\varepsilon > 0. We must find NN such that n≥N  ⟹  ∣an−L∣<εn \ge N \implies |a_n - L| < \varepsilon. Since (an)(a_n) is Cauchy, choose N1∈NN_1 \in \mathbb{N} such that for all m,n≥N1m, n \ge N_1,

∣an−am∣<ε2.(1)|a_n - a_m| < \frac{\varepsilon}{2}. \tag{1}

Since ank→La_{n_k} \to L, choose K∈NK \in \mathbb{N} such that for all k≥Kk \ge K,

∣ank−L∣<ε2.(2)|a_{n_k} - L| < \frac{\varepsilon}{2}. \tag{2}

Because the indices nkn_k form a strictly increasing sequence of natural numbers, nk≥kn_k \ge k for all kk (an elementary induction: n1≥1n_1 \ge 1, and if nk≥kn_k \ge k then since nk+1>nk≥kn_{k+1} > n_k \ge k and both are integers, nk+1≥k+1n_{k+1} \ge k+1). Hence we may choose a single index k0≥Kk_0 \ge K with nk0≥N1n_{k_0} \ge N_1 (possible since nk→∞n_k \to \infty as k→∞k \to \infty). Set N=N1N = N_1. Let n≥Nn \ge N. Using the fixed index nk0≥N1n_{k_0} \ge N_1 together with n≥N1n \ge N_1, inequality (1) gives

∣an−ank0∣<ε2.|a_n - a_{n_{k_0}}| < \frac{\varepsilon}{2}.

Since k0≥Kk_0 \ge K, inequality (2) gives

∣ank0−L∣<ε2.|a_{n_{k_0}} - L| < \frac{\varepsilon}{2}.

By the triangle inequality,

∣an−L∣≤∣an−ank0∣+∣ank0−L∣<ε2+ε2=ε.|a_n - L| \le |a_n - a_{n_{k_0}}| + |a_{n_{k_0}} - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.

Since ε>0\varepsilon > 0 was arbitrary and NN depends only on ε\varepsilon (not on nn), this proves an→La_n \to L. ■\blacksquare

∎
CorollaryCauchy Criterion, contrapositive form

A sequence (an)(a_n) of real numbers diverges if and only if there exists ε0>0\varepsilon_0 > 0 such that for every N∈NN \in \mathbb{N} there exist m,n≥Nm, n \ge N with ∣an−am∣≥ε0|a_n - a_m| \ge \varepsilon_0.

Proof

This is the logical negation of the Cauchy condition, combined with the Cauchy Criterion: (an)(a_n) converges iff it is Cauchy, so (an)(a_n) diverges iff it is not Cauchy, i.e. iff

¬[∀ε>0 ∃N ∀m,n≥N: ∣an−am∣<ε],\neg\big[\forall \varepsilon>0\ \exists N\ \forall m,n\ge N:\ |a_n-a_m|<\varepsilon\big],

which by moving the negation through the quantifiers (flipping ∀↔∃\forall \leftrightarrow \exists and << to ≥\ge) is exactly

∃ε0>0 ∀N ∃m,n≥N: ∣an−am∣≥ε0.■\exists \varepsilon_0 > 0\ \forall N\ \exists m,n \ge N:\ |a_n - a_m| \ge \varepsilon_0. \qquad \blacksquare
∎

Worked examples

Example 1 (proving convergence without knowing the limit). Define (an)(a_n) recursively by

a1=1,an+1=12(an+3an)(n≥1).a_1 = 1, \qquad a_{n+1} = \frac{1}{2}\left(a_n + \frac{3}{a_n}\right) \quad (n \ge 1).

This is Newton's method approximating 3\sqrt{3}. We show (an)(a_n) is Cauchy directly from the recursion, without assuming we know 3\sqrt 3 is the limit. First, a1=1>0a_1 = 1 > 0, and if an>0a_n > 0 then an+1=12(an+3/an)>0a_{n+1} = \frac12(a_n + 3/a_n) > 0; by induction an>0a_n > 0 for all nn. Next, for n≥1n \ge 1,

an+12−3=14(an+3an)2−3=14an2(an2−3)2≥0,a_{n+1}^2 - 3 = \frac{1}{4}\left(a_n + \frac{3}{a_n}\right)^2 - 3 = \frac{1}{4a_n^2}\left(a_n^2-3\right)^2 \ge 0,

so an+12≥3a_{n+1}^2 \ge 3 for every n≥1n \ge 1, i.e. an≥3a_n \ge \sqrt3 for all n≥2n \ge 2. Now for n≥2n \ge 2,

an+1−an=12(3an−an)=3−an22an≤0,a_{n+1} - a_n = \frac{1}{2}\left(\frac{3}{a_n} - a_n\right) = \frac{3 - a_n^2}{2a_n} \le 0,

using an2≥3a_n^2 \ge 3. So (an)n≥2(a_n)_{n\ge 2} is monotone decreasing and bounded below by 3>0\sqrt3 > 0; in particular it's bounded. We can go further and show it is Cauchy using a quantitative estimate. From an+1−3=(an−3)22ana_{n+1} - \sqrt3 = \frac{(a_n-\sqrt3)^2}{2a_n} (a direct algebraic identity: expand 12(an+3/an)−3=an2−23an+32an=(an−3)22an\frac{1}{2}(a_n+3/a_n) - \sqrt3 = \frac{a_n^2 - 2\sqrt3 a_n+3}{2a_n} = \frac{(a_n-\sqrt3)^2}{2a_n}), and since an≥3a_n \ge \sqrt3 for n≥2n\ge2,

0≤an+1−3≤(an−3)223.0 \le a_{n+1} - \sqrt3 \le \frac{(a_n - \sqrt3)^2}{2\sqrt3}.

With a2=12(1+3)=2a_2 = \frac12(1+3) = 2, so a2−3≤2−1=1a_2 - \sqrt3 \le 2 - 1 = 1 (since 3>1\sqrt3 > 1), a quick induction shows an−3→0a_n - \sqrt 3 \to 0 doubly-exponentially fast, so certainly an−am→0a_n - a_m \to 0 as m,n→∞m,n\to\infty: the sequence is Cauchy. By the Cauchy Criterion, (an)(a_n) converges to some real LL; passing to the limit in the recursion L=12(L+3/L)L = \frac12(L + 3/L) gives L2=3L^2 = 3, and since an≥3>0a_n \ge \sqrt3>0, L=3L = \sqrt3. The point of this example: we established convergence (via the Cauchy property, monotonicity and boundedness together) before pinning down that the limit is 3\sqrt3 — exactly the situation where the Cauchy criterion is essential, since 3\sqrt3 was not given to us in advance as a number to test against.

Example 2 (a divergence proof via the contrapositive). Let an=∑k=1n1k=1+12+⋯+1na_n = \sum_{k=1}^n \frac{1}{k} = 1 + \frac12 + \cdots + \frac1n (the harmonic partial sums). We show (an)(a_n) is not Cauchy, hence diverges, using the corollary above. Take ε0=12\varepsilon_0 = \tfrac12. Given any NN, set m=Nm = N and n=2Nn = 2N. Then

an−am=∑k=N+12N1k.a_n - a_m = \sum_{k=N+1}^{2N} \frac{1}{k}.

This sum has exactly NN terms, each term is ≥12N\ge \frac{1}{2N} (since k≤2Nk \le 2N), so

a2N−aN≥N⋅12N=12=ε0.a_{2N} - a_N \ge N \cdot \frac{1}{2N} = \frac12 = \varepsilon_0.

Thus for every NN we exhibited m=N,n=2N≥Nm=N, n=2N \ge N with ∣an−am∣≥ε0=12|a_n - a_m| \ge \varepsilon_0 = \tfrac12. By the contrapositive corollary, (an)(a_n) is not Cauchy, and therefore (by the Cauchy Criterion) (an)(a_n) diverges. This reproves the classical fact that the harmonic series diverges, and directly foreshadows the Cauchy Criterion for Series, which restates exactly this test in terms of partial sums.

Counterexamplewhy completeness of the ambient space is needed

The forward direction (convergent ⇒\Rightarrow Cauchy) used no special property of R\mathbb{R}; it holds in any metric space. The reverse direction — Cauchy ⇒\Rightarrow convergent — genuinely used completeness of R\mathbb{R} (via Bolzano–Weierstrass). To see this hypothesis cannot be dropped, work inside the metric space Q\mathbb{Q} (with the usual absolute-value metric), which is not complete. Define (qn)(q_n) by truncating the decimal expansion of 2\sqrt2:

q1=1.4,q2=1.41,q3=1.414,q4=1.4142, …q_1 = 1.4,\quad q_2 = 1.41,\quad q_3 = 1.414,\quad q_4 = 1.4142,\ \dots

so qnq_n is the decimal expansion of 2\sqrt2 truncated after nn digits past the decimal point. Each qn∈Qq_n \in \mathbb{Q}. This sequence is Cauchy in Q\mathbb{Q}. For n≥mn \ge m, qnq_n and qmq_m agree in their first mm decimal digits by construction, so ∣qn−qm∣<10−m|q_n - q_m| < 10^{-m} (truncating further digits changes the value by less than one unit in the mm-th decimal place). Given ε>0\varepsilon > 0, choose NN with 10−N<ε10^{-N} < \varepsilon (possible by the Archimedean property); then for m,n≥Nm,n \ge N, ∣qn−qm∣<10−N<ε|q_n - q_m| < 10^{-N} < \varepsilon. So (qn)(q_n) satisfies the Cauchy condition, using only the metric on Q\mathbb{Q}. This sequence does not converge in Q\mathbb{Q}. Suppose toward contradiction qn→rq_n \to r for some r∈Qr \in \mathbb{Q}. Viewing (qn)(q_n) as a sequence of real numbers, the same estimate ∣qn−qm∣<10−m|q_n - q_m|<10^{-m} shows (qn)(q_n) converges in R\mathbb{R} (it is Cauchy in R\mathbb{R}, and R\mathbb{R} is complete by the theorem above) to the unique real number whose decimal expansion is 1.41421356…1.41421356\ldots, namely 2\sqrt2. Limits in R\mathbb{R} are unique, so if qn→rq_n \to r with r∈Q⊂Rr \in \mathbb{Q} \subset \mathbb{R}, uniqueness forces r=2r = \sqrt2. But 2\sqrt2 is irrational (the standard proof: if 2=p/q\sqrt2 = p/q in lowest terms then p2=2q2p^2 = 2q^2 forces both p,qp,q even, contradicting lowest terms) — contradiction, since rr was assumed rational. Hence (qn)(q_n) is a Cauchy sequence in the metric space Q\mathbb{Q} that has no limit in Q\mathbb{Q}. This shows the reverse implication of the Cauchy Criterion fails when the ambient space (Q\mathbb{Q}) is not complete — precisely the hypothesis "completeness of R\mathbb{R}" (encoded in our proof through the Bolzano–Weierstrass theorem, which itself rests on the least upper bound property of R\mathbb{R}) that was used essentially in Step 2 of the proof above.

Scope, edge cases & common traps

SituationWhat holdsWhat can fail
Sequence in R\mathbb{R}Cauchy   ⟺  \iff convergent—
Sequence in Q\mathbb{Q}Convergent ⇒\Rightarrow Cauchy alwaysCauchy ⇏\not\Rightarrow convergent (in Q\mathbb{Q})
Sequence in a general metric spaceConvergent ⇒\Rightarrow Cauchy alwaysCauchy ⇒\Rightarrow convergent only if space is complete
Checking finitely many m,nm,nIrrelevantDefinition needs all m,n≥Nm,n \ge N simultaneously, not just consecutive terms
∣an+1−an∣→0|a_{n+1}-a_n|\to 0Necessary for CauchyNot sufficient — does not imply Cauchy (see below)

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Early 19th-century France, in the aftermath of the French Revolution and during the rise of rigorous analysis at the École Polytechnique

Cauchy was one of the principal architects of modern rigorous calculus, formalizing limits, continuity, and convergence with the ε\varepsilon-style arguments used throughout analysis today. His 1821 Cours d'Analyse introduced what we now call the Cauchy criterion for convergence of sequences and series, replacing informal 18th-century reasoning about infinitesimals with precise inequalities. He also made foundational contributions to complex analysis (Cauchy's integral theorem), differential equations, and elasticity theory, publishing an extraordinary volume of work — over 800 papers — across nearly every branch of mathematics and mathematical physics of his time.

History
- 1789 — Cauchy born in Paris, shortly after the storming of the Bastille.
- 1821 — Publishes Cours d'Analyse, stating the Cauchy convergence criterion for sequences and series in rigorous form.
- 1829 — Develops further foundations of complex analysis and real analysis while teaching at the École Polytechnique.
- 1872 — Eduard Heine and Georg Cantor, building on Cauchy's criterion, use equivalence classes of Cauchy sequences of rationals to construct the real numbers rigorously.
- 1920s — The abstract notion of a complete metric space, modeled directly on the Cauchy criterion, becomes central to the emerging field of functional analysis (Banach, Fréchet).
What exactly is the difference between "convergent" and "Cauchy"?

Convergent means there is a specific real number LL such that terms get arbitrarily close to LL; the definition explicitly references LL. Cauchy means terms get arbitrarily close to each other, with no reference to any particular target value. In R\mathbb{R}, the Cauchy Criterion proves these are logically equivalent notions, so in practice "Cauchy" is often used precisely when you want to establish convergence without first knowing (or being able to compute) what the limit is. In general metric spaces they are not equivalent: convergent always implies Cauchy, but Cauchy implies convergent only if the space is complete.

Why does the proof need Bolzano–Weierstrass rather than something simpler?

The Cauchy condition alone only tells you the terms cluster together; it says nothing directly about where they cluster. Bolzano–Weierstrass guarantees that any bounded sequence in R\mathbb{R} has a convergent subsequence, which supplies a candidate limit LL. The rest of the proof (Step 3) is the genuinely Cauchy-specific argument: it uses the Cauchy property to show that because all terms eventually bunch within ε/2\varepsilon/2 of each other, and infinitely many of them (the subsequence) bunch within ε/2\varepsilon/2 of LL, the entire tail must be within ε\varepsilon of LL. Both ingredients are necessary; neither alone suffices.

Is every bounded sequence Cauchy?

No. Boundedness is necessary for a Cauchy sequence (we proved this), but it is far from sufficient. The sequence an=(−1)na_n = (-1)^n is bounded (between −1-1 and 11) but not Cauchy: taking ε0=1\varepsilon_0 = 1, for any NN we can pick consecutive m=N,n=N+1m=N, n=N+1 with ∣an−am∣=2≥1|a_n - a_m| = 2 \ge 1. This sequence oscillates forever and never settles down, so it is bounded but neither Cauchy nor convergent — it has two different subsequential limits, 11 and −1-1, which by the theorem's Step 3 argument (uniqueness of the limit target) already rules out convergence, hence rules out being Cauchy.

Does the Cauchy Criterion give a way to compute the limit?

No — and that is precisely its value. The Cauchy Criterion is an existence theorem: it certifies that a real number LL exists such that an→La_n \to L, without constructing or naming LL. This is exactly what makes it useful for cases like Example 1, where establishing convergence first (via Cauchy-ness, or via monotone + bounded arguments) is a separate step from later identifying the limit (e.g., by passing to the limit inside a recursive formula, or via other structural facts).

How does this relate to the construction of the real numbers themselves?

One standard construction of R\mathbb{R} (due to Cantor, Heine, and Méray) builds real numbers directly as Cauchy sequences of rationals, where two Cauchy sequences (qn)(q_n) and (rn)(r_n) are declared equivalent if qn−rn→0q_n - r_n \to 0. Each equivalence class is defined to be a "real number." Under this construction, the fact that R\mathbb{R} (built this way) is complete becomes a theorem to prove, and the Cauchy Criterion in this article can be viewed as the statement, once R\mathbb{R} is constructed via the least-upper-bound property instead, that both constructions agree.

Does the Cauchy criterion generalize to sequences of functions or vectors?

Yes. In Rn\mathbb{R}^n (with the Euclidean metric) the identical statement holds: Cauchy sequences of vectors converge, because Rn\mathbb{R}^n is complete (this follows by applying the one-dimensional Cauchy Criterion coordinatewise). In spaces of functions, "Cauchy in the sup-norm" sequences of continuous functions converge uniformly to a continuous function precisely because the space of bounded continuous functions with the sup norm is a complete metric space (a Banach space). This generalization is the entire basis for existence proofs via Picard iteration and for foundational results in Fourier series and PDE theory.

What is the relationship to the Cauchy Criterion for Series?

A series ∑ak\sum a_k converges exactly when its sequence of partial sums sn=∑k=1naks_n = \sum_{k=1}^n a_k converges, so applying the Cauchy Criterion to (sn)(s_n) directly yields the Cauchy Criterion for Series: ∑ak\sum a_k converges iff for every ε>0\varepsilon>0 there is NN such that for all n>m≥Nn>m\ge N, ∣∑k=m+1nak∣<ε|\sum_{k=m+1}^n a_k| < \varepsilon. Example 2's divergence proof for the harmonic series is a direct instance of this specialization.

Before this

Where this leads