Cauchy sequence: a complete certificate for the reciprocal sequence, a space in which it has no limit, an alternating counterexample and an ultimately constant model

On this page
  1. Goal and scope
  2. Givens and symbols
  3. Rules and sources
  4. Main derivation
  5. Substeps and lemmas
  6. Reverse audits
  7. Error controls and negative controls
  8. Result and limitations
  9. Verification and source binding

← Guide index · Original MKO

derivation-mko-cauchy-sequence-01 · production-20261019-adopted-96

Goal and scope

(1) Prove that x_n=1/n (n≥1) is a Cauchy sequence in (ℝ, |s−t|): for every ε>0 give N such that |1/m−1/n|<ε for all m,n≥N. (2) Prove that the same sequence is still Cauchy in X=(0,1] with the restricted metric, but has no limit in X. (3) Prove that y_n=(−1)^n is not Cauchy. (4) Prove that the ultimately constant sequence z (z₁=0, z₂=7, z_n=5 for n≥3) is Cauchy. Also explain why a finite tail cannot prove the Cauchy property.

Stays inside the MKO's slice: complete certificates only for an explicit ultimately constant model, reciprocal-sequence bounds and one exact counterexample family (the alternating sequence). The metric is |s−t| on ℝ and its restriction to (0,1]. The general theory of complete metric spaces and Banach spaces is not treated; finitely listed tails serve only as bounded audits; the canonical MKO and its English edition are not changed.

Givens and symbols

  • metric_def
  • seq_limit_def
  • archimedean
  • abs_rules
  • order_basics
  • quantifier_negation
  • G1 — (ℝ,d) with d(s,t)=|s−t|; X=(0,1]={t ∣ 0<t≤1}, with the same formula d(s,t)=|s−t| restricted to X.
  • G2 — x_n=1/n for n≥1.
  • G3 — y_n=(−1)^n for n≥1.
  • G4 — z₁=0, z₂=7, and z_n=5 for every n≥3.
  • d(s,t)=|s-t| — the standard metric on ℝ
  • X=(0,1] — the subspace with the restricted metric
  • x_n=\tfrac1n — the reciprocal sequence
  • y_n=(-1)^n — the alternating sequence
  • z_n — the ultimately constant sequence
  • \varepsilon,\ N — the tolerance and the start of the tail

Rules and sources

  • metric_def · prerequisite — Metric: d:X×X→ℝ with d(x,y)≥0 and d(x,y)=0 ⇔ x=y; d(x,y)=d(y,x); triangle inequality d(x,z)≤d(x,y)+d(y,z).
  • seq_limit_def · prerequisite — Limit of a sequence: a_n→L means that for every ε>0 there is N such that |a_n−L|<ε for all n≥N.
  • archimedean · prerequisite — Archimedean property: for every real x there is a natural number n>x; hence for every ε>0 there is n with 1/n<ε.
  • abs_rules · prerequisite — Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r.
  • order_basics · prerequisite — Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw.
  • quantifier_negation · prerequisite — Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x).
  • def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true.
  • case_analysis · prerequisite — Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility.
  • mul_inverse_nonzero · prerequisite — Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined.
  • frac_rules · prerequisite — Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0.
  • forall_intro · prerequisite — ∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain.
  • exists_intro · prerequisite — ∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x).
  • contradiction · prerequisite — Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds.
  • forall_elim · prerequisite — ∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds.
  • and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
  • power_laws · prerequisite — Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1.
  • neg_mul · prerequisite — Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv.
  • numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
  • mko_cauchy · cited — A sequence (x_n) is Cauchy if for every ε>0 there exists N such that all m,n≥N satisfy d(x_m,x_n)<ε. The definition does not first assume that a limit exists or belongs to the space. A finitely listed tail supplies only a bounded audit, not a proof of the general Cauchy property. · mko-cauchy-sequence
  • max_two · prerequisite — Maximum of two numbers: max(s,t) is the larger of s and t; max(s,t)≥s and max(s,t)≥t; if s≤K and t≤K, then max(s,t)≤K.
  • subspace_metric · prerequisite — Subspace metric: for X⊆ℝ, d(s,t)=|s−t| restricted to X is still a metric. A sequence in X converges in X to L when L∈X and for every ε>0 there is N with |x_n−L|<ε for n≥N.
  • L_absmax · lemma — If s≥0 and t≥0, then |s−t|≤max(s,t).
  • L_recip · lemma — For reals m≥N>0: 0<1/m≤1/N; and if m>N, then 1/m<1/N.

Main derivation

S01 · equivalent

Before

(x_n)\ \text{Cauchy in}\ (\mathbb{R},d)

After

\forall\varepsilon>0\ \exists N\ \forall m,n\ge N:\ \left|\tfrac1m-\tfrac1n\right|<\varepsilon

Conditions: none

Rule: mko_cauchy · Uses: G1, G2, mko_cauchy

Leaf basis: mko_cauchy + def_unfold

Unfold the MKO's definition: what is compared is the distance between any two terms of the tail, and no limit appears in the definition. N may depend on ε but not on m or n.

S02 · implies

Before

s,t\ge 0

After

|s-t|\le\max(s,t)

Conditions: s\ge 0; t\ge 0

Rule: L_absmax · Uses: L_absmax

Leaf basis: L_absmax

Lemma L_absmax: the absolute difference of two nonnegative numbers is at most the larger of them. The proof in S02.1 and S02.2 splits into the cases s≥t and s<t. It will be used with s=1/m and t=1/n.

S03 · implies

Before

m,n\ge N\ge 1

After

0<\tfrac1m\le\tfrac1N,\ 0<\tfrac1n\le\tfrac1N

Conditions: N\ge 1

Rule: L_recip · Uses: L_recip

Leaf basis: L_recip

By Lemma L_recip (proved in S03.1), m≥N>0 gives 0<1/m≤1/N, and likewise for n. Every term of the tail lies in (0,1/N].

S04 · implies

Before

m,n\ge N\ge 1

After

\left|\tfrac1m-\tfrac1n\right|\le\max\left(\tfrac1m,\tfrac1n\right)\le\tfrac1N

Conditions: N\ge 1

Rule: max_two · Uses: S02, S03, max_two

Leaf basis: L_absmax + max_two + order_basics

L_absmax (s=1/m≥0, t=1/n≥0) gives the first inequality; by S03 both are ≤1/N, so their maximum is ≤1/N. This bound is uniform over all m, n in the tail, whatever their relative size.

S05 · implies

Before

\varepsilon>0\ \text{arbitrary}

After

\exists N\in\mathbb{N},\ N\ge 1:\ \tfrac1N<\varepsilon

Conditions: \varepsilon>0

Rule: archimedean · Uses: archimedean

Leaf basis: archimedean

Take an arbitrary ε>0, with no other assumption. By the Archimedean property there is a natural number N with 1/N<ε; N≥1 can be taken. For example, with ε=1/10, N=11 works.

S06 · implies

Before

m,n\ge N,\ \tfrac1N<\varepsilon

After

\left|\tfrac1m-\tfrac1n\right|\le\tfrac1N<\varepsilon

Conditions: none

Rule: order_basics · Uses: S04, S05, order_basics

Leaf basis: order_basics + forall_intro

For the N chosen in S05 and any m,n≥N, S04 gives |1/m−1/n|≤1/N, and 1/N<ε with transitivity of order gives |1/m−1/n|<ε.

S07 · implies

Before

\text{for arbitrary } \varepsilon>0:\ N\ \text{found with S06}

After

(x_n)=(\tfrac1n)\ \text{is Cauchy in}\ (\mathbb{R},d)

Conditions: none

Rule: forall_intro · Uses: S01, S06, forall_intro, exists_intro

Leaf basis: forall_intro + exists_intro

For each ε an N was found (∃-introduction), and ε was arbitrary (∀-introduction); by S01, (1/n) is Cauchy in (ℝ,d). This is a complete certificate covering every ε (the reciprocal-sequence bound allowed by MKO assumption a2).

S08 · implies

Before

x_n\in X?\ \text{and}\ d_X(x_m,x_n)

After

x_n\in(0,1],\ d_X(x_m,x_n)=\left|\tfrac1m-\tfrac1n\right|;\ (x_n)\ \text{Cauchy in}\ X

Conditions: none

Rule: subspace_metric · Uses: G1, S03, S07, subspace_metric

Leaf basis: subspace_metric + L_recip

By L_recip with N=1, 0<1/n≤1, so every term lies in X. The distance on X is the same formula, so every inequality of S04 to S07 is unchanged and (x_n) is Cauchy in X as well. The Cauchy property depends only on distances between terms.

S09 · implies

Before

\text{suppose } L\in X\ \text{and}\ x_n\to L\ \text{in}\ X

After

\text{contradiction; no } L\in X\ \text{is a limit of}\ (x_n)

Conditions: none

Rule: contradiction · Uses: S08, subspace_metric, contradiction

Leaf basis: contradiction + seq_limit_def

By contradiction: suppose some L∈X (so L>0) is a limit. S09.1 takes N₁ for ε=L/2; S09.2 uses the Archimedean property to find n≥N₁ with 1/n<L/2; S09.3 shows that then |1/n−L|>L/2, contradicting S09.1. So (x_n) has no limit in X.

S10 · implies

Before

(x_n)\ \text{Cauchy in}\ X,\ \text{no limit in}\ X

After

\text{Cauchy does not presuppose a limit in the space}

Conditions: none

Rule: and_intro_elim · Uses: S08, S09, mko_cauchy

Leaf basis: and_intro_elim + mko_cauchy

In X=(0,1], (1/n) is Cauchy (S08) yet has no limit (S09). This is exactly the MKO's point: the definition does not first assume that a limit exists or belongs to the space. X is not a complete metric space, but this guide does not develop the theory of completeness.

S11 · implies

Before

y_{n+1}=(-1)^{n+1}=-y_n

After

|y_n-y_{n+1}|=|2y_n|=2|y_n|=2\ \text{for every } n\ge 1

Conditions: none

Rule: abs_rules · Uses: G3, abs_rules, power_laws

Leaf basis: power_laws + abs_rules + neg_mul

(−1)^(n+1)=(−1)^n·(−1)=−y_n, so y_n−y_(n+1)=2y_n, and |y_n|=|±1|=1, giving |y_n−y_(n+1)|=2. This holds for every n, not just for some listed terms.

S12 · equivalent

Before

\neg\,(y_n)\ \text{Cauchy}

After

\exists\varepsilon>0\ \forall N\ \exists m,n\ge N:\ |y_m-y_n|\ge\varepsilon

Conditions: none

Rule: quantifier_negation · Uses: mko_cauchy, quantifier_negation

Leaf basis: quantifier_negation + order_basics

Negating the statement ∀ε∃N∀m,n flips each quantifier in turn and replaces <ε by ≥ε. So to refute the Cauchy property one ε is enough, such that beyond every N there are two terms at distance at least ε.

S13 · implies

Before

\varepsilon=1,\ N\ \text{arbitrary},\ m=N,\ n=N+1

After

|y_N-y_{N+1}|=2\ge 1;\ (y_n)\ \text{is not Cauchy}

Conditions: none

Rule: exists_intro · Uses: S11, S12, exists_intro

Leaf basis: exists_intro + forall_intro + order_basics

Take ε=1. For any N take the adjacent m=N and n=N+1 (both ≥N); by S11 their distance is 2≥1. By the equivalence of S12, (y_n) is not Cauchy. This is the MKO's alternating counterexample family: for every N there is an adjacent counterexample.

S14 · implies

Before

\varepsilon>0\ \text{arbitrary},\ N=3

After

m,n\ge 3\Rightarrow|z_m-z_n|=|5-5|=0<\varepsilon;\ (z_n)\ \text{Cauchy}

Conditions: \varepsilon>0

Rule: forall_intro · Uses: G4, mko_cauchy, forall_intro

Leaf basis: forall_intro + exists_intro + numeric_evaluation

Ultimately constant model: for any ε>0 take N=3, independent of ε. For m,n≥3, z_m=z_n=5 and the distance is 0<ε. Since ε is arbitrary, (z_n) is Cauchy. The first two terms 0 and 7 do not affect the conclusion, because the definition looks only at the tail (MKO para1).

Substeps and lemmas

S02.1 · implies

Before

s\ge t\ge 0

After

|s-t|=s-t\le s=\max(s,t)

Conditions: s\ge t; t\ge 0

Rule: abs_rules · Uses: abs_rules, max_two, order_basics

Leaf basis: abs_rules + max_two + order_basics

When s≥t, s−t≥0, so |s−t|=s−t; since t≥0, s−t≤s; and when s≥t, max(s,t)=s.

S02.2 · implies

Before

t>s\ge 0

After

|s-t|=t-s\le t=\max(s,t)

Conditions: t>s; s\ge 0

Rule: case_analysis · Uses: S02.1, abs_rules, max_two, case_analysis

Leaf basis: abs_rules + max_two + case_analysis

When t>s, s−t<0 and |s−t|=t−s; since s≥0, t−s≤t=max(s,t). The cases s≥t and t>s cover everything, so S02.1 and S02.2 together prove L_absmax.

S03.1 · implies

Before

m\ge N>0

After

\tfrac1N=m\cdot\tfrac{1}{mN}\ge N\cdot\tfrac{1}{mN}=\tfrac1m>0

Conditions: N>0

Rule: order_basics · Uses: order_basics, mul_inverse_nonzero, frac_rules

Leaf basis: order_basics + mul_inverse_nonzero + frac_rules

m≥N>0 gives m>0 and mN>0, so 1/(mN)>0. Multiplying m≥N by the positive number 1/(mN) keeps the direction: 1/N≥1/m; if m>N it is strict: 1/N>1/m. Also 1/m>0, since the reciprocal of a positive number is positive. This proves L_recip.

S09.1 · implies

Before

x_n\to L\in X,\ \varepsilon=\tfrac L2>0

After

\exists N_1\ \forall n\ge N_1:\ \left|\tfrac1n-L\right|<\tfrac L2

Conditions: L>0

Rule: seq_limit_def · Uses: subspace_metric, seq_limit_def

Leaf basis: seq_limit_def + forall_elim

L∈X means L>0, so ε=L/2>0 is a legitimate tolerance; the definition of limit gives N₁.

S09.2 · implies

Before

n_0>\tfrac2L\ \text{(Archimedean)},\ n=\max(n_0,N_1)

After

n\ge N_1,\ \tfrac1n\le\tfrac1{n_0}<\tfrac L2

Conditions: L>0

Rule: archimedean · Uses: S09.1, archimedean, L_recip, max_two

Leaf basis: archimedean + L_recip + max_two

By the Archimedean property take a natural number n₀>2/L>0. Let n=max(n₀,N₁); then n≥N₁ and n≥n₀. By L_recip, 1/n≤1/n₀, and since n₀>2/L, 1/n₀<1/(2/L)=L/2.

S09.3 · implies

Before

\tfrac1n<\tfrac L2,\ n\ge N_1

After

\left|\tfrac1n-L\right|\ge L-\tfrac1n>L-\tfrac L2=\tfrac L2

Conditions: none

Rule: abs_rules · Uses: S09.1, S09.2, abs_rules

Leaf basis: abs_rules + order_basics + contradiction

|w|≥−w holds for every real w; with w=1/n−L this gives |1/n−L|≥L−1/n. From 1/n<L/2, L−1/n>L/2. So |1/n−L|>L/2, yet n≥N₁ and S09.1 requires |1/n−L|<L/2, a contradiction.

Reverse audits

  • S14 · premise_audit · G4, mko_cauchy, forall_intro — It needs the constant value 5 for all n≥3 (stated in G4), not only for some listed terms.
  • S13 · premise_audit · S11, S12, exists_intro, forall_intro — It needs a witness for every N; S11 holds for all n, so a witness exists for every N.
  • S12 · inverse_equivalence · quantifier_negation, order_basics — Quantifier negation and ¬(d<ε) ⇔ d≥ε are both two-way.
  • S11 · premise_audit · abs_rules, power_laws — It uses the exponent laws and |uv|=|u||v|; it holds for all n.
  • S10 · premise_audit · S08, S09, and_intro_elim — It only conjoins two proved conclusions; the sentence on completeness is a remark.
  • S09.3 · premise_audit · S09.1, S09.2, abs_rules, order_basics — The same n satisfies two contradictory inequalities; the contradiction comes from the assumption that some L∈X is a limit.
  • S09.2 · premise_audit · archimedean, L_recip, max_two — The same n satisfies both n≥N₁ and 1/n<L/2, which is the purpose of taking the maximum.
  • S09.1 · premise_audit · seq_limit_def, subspace_metric — It needs L>0; this is exactly where 0∉X matters.
  • S09 · premise_audit · subspace_metric, contradiction — L is an arbitrary point of X, so every possible limit is ruled out; the claim is only that there is no limit in X (in ℝ the limit 0 exists, but 0∉X).
  • S08 · premise_audit · S03, S07, subspace_metric — It needs all terms in X and the same metric; no limit is involved.
  • S07 · premise_audit · S01, S06, forall_intro, exists_intro — It needs the arbitrariness of ε and the existence of N; the argument uses no limit at all.
  • S06 · premise_audit · S04, S05, order_basics — Chaining ≤ and < gives <; no converse is claimed.
  • S05 · premise_audit · archimedean — N depends only on ε, which is exactly what the position of ∃N in the Cauchy definition allows.
  • S04 · premise_audit · S02, S03, max_two — It needs both terms nonnegative and both ≤1/N; equality is not claimed.
  • S03.1 · premise_audit · order_basics, mul_inverse_nonzero, frac_rules — It needs N>0 so that the reciprocals exist and the multiplier is positive; L_recip itself is not used.
  • S03 · premise_audit · L_recip — It needs N>0; no converse is claimed.
  • S02.2 · premise_audit · abs_rules, max_two, case_analysis — It uses s≥0; the two cases cover all pairs of reals; L_absmax itself is not used.
  • S02.1 · premise_audit · abs_rules, max_two, order_basics — It uses t≥0; L_absmax itself is not used.
  • S02 · premise_audit · L_absmax — It needs s and t nonnegative; it fails for negatives, for example |(−1)−1|=2>1=max(−1,1).
  • S01 · inverse_equivalence · mko_cauchy, def_unfold — Unfolding and folding the definition undo each other, so the step holds both ways.

Error controls and negative controls

A close finite tail does not prove Cauchy

The mistake named by the MKO is to call the whole infinite sequence Cauchy because a long finite tail is close together. For the harmonic partial sums H_n=Σ_{k=1}^{n} 1/k the adjacent differences 1/(n+1) keep shrinking, and any listed finite tail may look close; but for every N, H_{2N}−H_N=Σ_{k=N+1}^{2N} 1/k≥N·1/(2N)=1/2 (each term 1/k≥1/(2N)), so the Cauchy condition fails for ε=1/2.

Compare any two terms of the tail, not only adjacent ones

Cauchy requires all m,n≥N, not only n and n+1. The harmonic partial sums have adjacent differences tending to 0 yet are not Cauchy. On the other hand, adjacent terms suffice to refute (−1)^n (S13), because a refutation needs only one pair of witnesses.

Do not presuppose a limit

S01 to S07 never use a limit. If 'Cauchy' is misread as 'close to some candidate limit', then in X=(0,1] there is no candidate (S09), and yet the sequence is still Cauchy (S08).

  • The adjacent differences 1/(n+1) of the harmonic partial sums H_n tend to 0, so (H_n) is Cauchy → reject — For every N, H_{2N}−H_N≥1/2.
  • (y_n)=((−1)^n) takes only the two values −1 and 1, so it is Cauchy → reject — With ε=1, |y_N−y_(N+1)|=2≥1 for every N.
  • (1/n) is Cauchy in X=(0,1], so it converges in X → reject — S09: no point of X is its limit.
  • Listing the first 1000 terms of 1/n and checking their mutual distances proves that (1/n) is Cauchy → reject — A finite tail is only a bounded audit (MKO assumption a1); the proof is the argument of S02 to S07 for every ε.

Result and limitations

(1/n) is Cauchy in (ℝ,|·|): for every ε>0 take N with 1/N<ε; then |1/m−1/n|≤1/N<ε for all m,n≥N. It is still Cauchy in X=(0,1], yet has no limit in X. (−1)^n is not Cauchy: with ε=1, for every N two adjacent terms are 2 apart. The ultimately constant sequence z is Cauchy, with N=3 working for every ε.

  • N3 and N5 are finite-range checks (m, n from 1 to 30; n from 1 to 40); they support only those points and do not prove L_absmax or the general claim of S11, which come from the arguments of S02.1, S02.2 and S11.
  • The harmonic partial sums in the error controls are not among the certificate types listed by the MKO and serve only as a counterexample illustration; their lower-bound argument is given in full in the error control, and N10, N11 check only N=2 and N=3.

Verification and source binding

12 declared checks for this guide; only the listed expressions are covered.

  • source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
  • adopted source SHA-256: 3cf4cca5b06ba03d3f526ddbf2d5e9556d77fa989091a6210c0ad46c7169d2a5
  • original source SHA-256: 3cf4cca5b06ba03d3f526ddbf2d5e9556d77fa989091a6210c0ad46c7169d2a5
  • Lean: NOT_RUN · independent AI: NOT_MEASURED

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