# Convergence in a metric space: limit and uniqueness for an ultimately constant sequence on a finite metric table

[← Guide index](https://ocme.evemisslab.com/en/derivations/) · [Original MKO](https://ocme.evemisslab.com/en/mko/mko-metric-convergence/)

This is an owner-adopted natural-language teaching guide. Finite/arithmetic checks are not universal proofs; no Lean was run, and review scope is recorded per revision. Original source status labels are retained as historical data. Formulas retain exact TeX; no new semantic AST is invented.

derivation-mko-metric-convergence-01 · production-20261019-adopted-96



## Goal and scope

On the three-point set X={a,b,c} a complete metric table is given: d(a,b)=1, d(a,c)=2, d(b,c)=3/2. The sequence has x_0=a, x_1=c, x_2=a, and it is declared that x_n=b for every n≥3. (1) Check that d is a metric; (2) prove x_n→b from the ε-N definition and find the least valid N for each ε; (3) prove that limits in a metric space are unique, and give direct witnesses that x_n converges neither to a nor to c; (4) using a sequence that agrees with x_n on a finite tail but has a different limit, show that checking a finite tail cannot decide whether the limit is the specified b.

Treats only the finite metric space of G1 and the ultimately constant sequence model of G2 (MKO assumptions a1, a2). The uniqueness-of-limits lemma is proved in full for every metric space, and so is the ultimately-constant lemma. Cauchy sequences, completeness and continuous maps are not treated, and a check of a finite sampled tail is never taken as a proof of convergence. The canonical MKO and its English edition are not changed.





## Givens and symbols


- metric_def
- order_basics
- quantifier_negation
- forall_intro
- exists_intro


- G1 — X={a,b,c} with three distinct labels a, b, c. Metric table: d(p,p)=0 for every p∈X; d(a,b)=d(b,a)=1; d(a,c)=d(c,a)=2; d(b,c)=d(c,b)=3/2 (MKO assumption a1: a complete finite metric table).
- G2 — The sequence (x_n)_{n∈ℕ0}: x_0=a, x_1=c, x_2=a, and it is declared that x_n=b for every n≥3 (MKO assumption a2: ultimately constant model with prefix length 3 and tail value b).
- G3 — Comparison sequence (y_n)_{n∈ℕ0}: y_n=x_n for 0≤n≤10, and it is declared that y_n=c for every n≥11.


- X=\{a,b,c\} — the three-point carrier
- d — the complete metric table of G1
- (x_n)_{n\in\mathbb{N}_0} — the ultimately constant sequence: prefix a, c, a and x_n=b for n≥3
- (y_n)_{n\in\mathbb{N}_0} — the comparison sequence equal to x_n for n≤10 and to c for n≥11
- \varepsilon,\ N — tolerance (ε>0) and cutoff index
- (s_n),\ p,\ q,\ M — the generic sequence, candidate limits and prefix length in the lemmas





## 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).
- 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).
- 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).
- 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.
- numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
- 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.
- def_intro · prerequisite — Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable.
- forall_elim · prerequisite — ∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds.
- eq_replace · prerequisite — Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true.
- modus_ponens · prerequisite — Modus ponens: from P and P⇒Q infer Q.
- exists_elim · prerequisite — ∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a.
- contradiction · prerequisite — Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds.
- contrapositive · prerequisite — Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q.
- and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
- mko_metric_conv · cited — A sequence x_n in a metric space converges to x if, for every ε>0, there exists N such that all n≥N satisfy d(x_n,x)<ε. Passing a sampled finite tail is insufficient for general convergence; an explicitly ultimately constant infinite sequence can instead provide a complete witness. · mko-metric-convergence
- L_metric_table · lemma — The table d of G1 is a metric on X: d≥0 and d(p,q)=0 ⇔ p=q; d is symmetric; the triangle inequality holds for all 27 ordered triples.
- L_tail_const · lemma — In any metric space, if a sequence (s_n) has s_n=p for every n≥M, then s_n→p, and N=M is a valid cutoff for every ε>0.
- L_limit_unique · lemma — In any metric space, if s_n→p and s_n→q, then p=q.





## Main derivation

S01 · definitionBefore

x_n\to p\ \text{in}\ (X,d)

After

\forall\varepsilon>0\ \exists N\in\mathbb{N}_0\ \forall n\ge N:\ d(x_n,p)<\varepsilon

Conditions: none

Rule: mko_metric_conv · Uses: G1, mko_metric_conv

Leaf basis: mko_metric_conv + def_unfold

Unfold 'x_n converges to p' by the MKO's definition: an arbitrary ε>0 is given first, then N is sought, and every n≥N must satisfy d(x_n,p)<ε. The quantifier order is ∀ε ∃N ∀n and the inequality is strict (an open ball). Every later claim of convergence or non-convergence returns to this formula.



S02 · impliesBefore

d:\ \text{the table of G1}

After

(X,d)\ \text{is a metric space}

Conditions: none

Rule: L_metric_table · Uses: G1, L_metric_table

Leaf basis: L_metric_table + metric_def

MKO assumption a1 uses complete finite metric tables, but whether a table really is a metric must be checked first, or the later uniqueness proof (which uses the triangle inequality and separation) has no basis. By Lemma L_metric_table (proved in S02.1 to S02.3), the table of G1 satisfies every metric axiom.



S03 · definitionBefore

x_0=a,\ x_1=c,\ x_2=a

After

x_n=b\ \text{for every}\ n\ge 3\ \text{(declared tail)}

Conditions: none

Rule: def_intro · Uses: G2

Leaf basis: def_intro

Per MKO assumption a2, the ultimately constant model declares that every term after the prefix (length 3: a, c, a) equals the tail value b. This is a rule covering infinitely many n, not a finite list; precisely because it speaks of every n≥3, a complete witness can be given later.



S04 · theorem_applicationBefore

x_n=b\ \ (n\ge 3)

After

x_n\to b

Conditions: M=3; p=b

Rule: L_tail_const · Uses: S01, S03, L_tail_const

Leaf basis: L_tail_const + forall_elim

Lemma L_tail_const: if a sequence equals p for every n≥M, it converges to p. With M=3 and p=b, S03 gives x_n→b. The lemma is proved in S04.1 to S04.3 for any metric space and any M.



S05 · case_splitBefore

\varepsilon>0

After

N_{\min}(\varepsilon)=0\ (\varepsilon>\tfrac32),\quad 2\ (1<\varepsilon\le\tfrac32),\quad 3\ (0<\varepsilon\le 1)

Conditions: \varepsilon>0

Rule: case_analysis · Uses: S01, S03, G1

Leaf basis: case_analysis + order_basics

Although N=3 works for every ε, the least valid N still changes with ε. The distances to b are d(x_0,b)=d(a,b)=1, d(x_1,b)=d(c,b)=3/2, d(x_2,b)=1, and 0 from then on. The three cases are mutually exclusive and cover every ε>0; they are treated in S05.1 to S05.3. This shows that the least valid N may depend on ε, consistently with the single valid cutoff N=3 working for every ε.



S06 · theorem_applicationBefore

x_n\to b\ \land\ x_n\to q

After

q=b

Conditions: (X,d)\ \text{metric space}

Rule: L_limit_unique · Uses: S02, S04, L_limit_unique

Leaf basis: L_limit_unique + modus_ponens

Lemma L_limit_unique: in a metric space, if one sequence converges to p and to q, then p=q. S02 confirms that (X,d) is a metric space and S04 gives x_n→b, so every limit q of x_n equals b. The lemma is proved in S06.1 to S06.4 using the triangle inequality and separation (MKO paragraph 2: metric separation gives uniqueness of the limit).



S07 · impliesBefore

a\neq b,\ c\neq b

After

x_n\not\to a,\ \ x_n\not\to c

Conditions: none

Rule: contrapositive · Uses: S06, G1

Leaf basis: contrapositive + L_limit_unique

S06 says: if x_n→q then q=b. Its contrapositive: if q≠b then x_n does not converge to q. Both a and c differ from b (G1), so x_n converges neither to a nor to c.



S08 · equivalentBefore

\neg\,\forall\varepsilon>0\ \exists N\ \forall n\ge N:\ d(x_n,q)<\varepsilon

After

\exists\varepsilon>0\ \forall N\ \exists n\ge N:\ d(x_n,q)\ge\varepsilon

Conditions: none

Rule: quantifier_negation · Uses: S01, quantifier_negation

Leaf basis: quantifier_negation + order_basics

Rewrite 'x_n does not converge to q' in positive form: flip each quantifier (∀ to ∃, ∃ to ∀) and negate d<ε to d≥ε. So to prove non-convergence it suffices to find one ε>0 such that after every N some term is at distance at least ε from q.



S09 · impliesBefore

q=a,\ \varepsilon=1,\ N\ \text{arbitrary}

After

n=\max(N,3)\ge N:\ d(x_n,a)=d(b,a)=1\ge 1

Conditions: \varepsilon=1>0

Rule: exists_intro · Uses: S08, S03, G1

Leaf basis: exists_intro + forall_intro + order_basics

Take ε=1. For any N, let n=max(N,3); then n≥N and n≥3, so x_n=b and d(x_n,a)=d(b,a)=1≥1 (for example N=7 gives n=7, see N7). By S08, x_n does not converge to a. Note that the strict inequality is used exactly here: d=1 is not less than ε=1; any 0<ε≤1 would also serve as a witness.



S10 · impliesBefore

q=c,\ \varepsilon=\tfrac32,\ N\ \text{arbitrary}

After

n=\max(N,3)\ge N:\ d(x_n,c)=d(b,c)=\tfrac32\ge\tfrac32

Conditions: \varepsilon=\tfrac32>0

Rule: exists_intro · Uses: S08, S03, G1

Leaf basis: exists_intro + forall_intro + order_basics

The same construction: take ε=3/2, and for any N let n=max(N,3); then x_n=b and d(b,c)=3/2≥3/2 (N9). By S08, x_n does not converge to c. This agrees with the conclusion S07 drew from uniqueness.



S11 · impliesBefore

x_n\to b,\ x_n\not\to a,\ x_n\not\to c

After

\{q\in X\mid x_n\to q\}=\{b\}

Conditions: none

Rule: and_intro_elim · Uses: S04, S09, S10

Leaf basis: and_intro_elim + case_analysis

X has only the three points a, b, c: S04 shows b is a limit, and S09 and S10 exclude a and c. Together, the set of limits of x_n is exactly {b}. S06 reaches the same conclusion from general uniqueness.



S12 · theorem_applicationBefore

y_n=c\ \ (n\ge 11)

After

y_n\to c\ \land\ y_n\not\to b

Conditions: M=11; p=c; b\neq c

Rule: L_tail_const · Uses: G3, L_tail_const, L_limit_unique, G1

Leaf basis: L_tail_const + L_limit_unique + contrapositive

G3 declares y_n=c for n≥11, and Lemma L_tail_const (M=11, p=c) gives y_n→c. By L_limit_unique the limit of y_n is unique, and b≠c, so y_n does not converge to b.



S13 · impliesBefore

x_n=y_n=b\ \ (3\le n\le 10)

After

\lim x_n=b\neq c=\lim y_n

Conditions: none

Rule: L_tail_const · Uses: S03, S04, S12, G3

Leaf basis: L_tail_const + L_limit_unique + forall_intro

x and y agree for n=0 to 10, and in particular the finite tail x_3 to x_10 is all b. Checking only this finite tail, the two look identical, yet x_n→b while y_n→c. So no data listing finitely many terms can decide the limit. In general, for any K the sequence equal to x up to K and constantly c afterwards converges to c by L_tail_const (M=K+1); this is the MKO's point that passing a sampled finite tail is insufficient for general convergence.



S14 · impliesBefore

\text{S02},\ \text{S04},\ \text{S05},\ \text{S11},\ \text{S13}

After

x_n\to b\ \text{with witness}\ N=3\ \text{for every}\ \varepsilon;\ \text{limit unique}

Conditions: none

Rule: and_intro_elim · Uses: S02, S04, S05, S11, S13

Leaf basis: and_intro_elim

Summary: d is a metric (S02); x_n→b, with N=3 valid for every ε and the least valid N equal to 0, 2 or 3 depending on ε (S04, S05); the limit is exactly b (S11); this complete witness rests on the declaration of G2 for all n≥3, and a finite tail by itself is not enough (S13).







## Substeps and lemmas

S02.1 · impliesBefore

d(p,p)=0,\ d(a,b)=1,\ d(a,c)=2,\ d(b,c)=\tfrac32\ \text{(with mirror entries)}

After

d\ge 0,\ \ (d(p,q)=0\Leftrightarrow p=q),\ \ d(p,q)=d(q,p)

Conditions: none

Rule: metric_def · Uses: G1, metric_def

Leaf basis: metric_def + numeric_evaluation + order_basics

Read the table cell by cell: the three diagonal cells are 0; the six off-diagonal cells are 1, 2, 3/2 and their mirror entries, all greater than 0. So d≥0, and d(p,q)=0 happens exactly when p=q. The three mirror pairs d(a,b)=d(b,a), d(a,c)=d(c,a), d(b,c)=d(c,b) agree, so symmetry holds. This is a complete enumeration of all 9 cells.



S02.2 · impliesBefore

x,y,z\ \text{pairwise distinct}

After

2\le 1+\tfrac32,\quad 1\le 2+\tfrac32,\quad \tfrac32\le 1+2

Conditions: x\neq y,\ y\neq z,\ x\neq z

Rule: metric_def · Uses: S02.1

Leaf basis: metric_def + numeric_evaluation + order_basics

For pairwise distinct points, the triangle inequality d(x,z)≤d(x,y)+d(y,z) is determined by the middle point y: y=b gives d(a,c)=2≤d(a,b)+d(b,c)=5/2; y=c gives d(a,b)=1≤d(a,c)+d(c,b)=7/2; y=a gives d(b,c)=3/2≤d(b,a)+d(a,c)=3. Swapping x and z only replaces each entry by its mirror entry (equal by S02.1), so the 6 ordered triples of distinct points reduce to these 3 inequalities, all of which hold.



S02.3 · case_splitBefore

x=z\ \lor\ y=x\ \lor\ y=z

After

d(x,z)\le d(x,y)+d(y,z)

Conditions: none

Rule: case_analysis · Uses: S02.1, case_analysis

Leaf basis: case_analysis + metric_def + order_basics

The remaining 27−6=21 ordered triples have at least two equal points, in three cases: if x=z the left side is d(x,x)=0 and the right side is a sum of two nonnegative numbers; if y=x the right side is 0+d(x,z); if y=z it is d(x,z)+0. All three cases hold (they may overlap, but together they cover every triple with a repeated point). With S02.2, the triangle inequality holds for all 27 triples, and S02.1 to S02.3 complete Lemma L_metric_table.



S04.1 · impliesBefore

s_n=p\ \text{for all}\ n\ge M

After

d(s_n,p)=d(p,p)=0\ \text{for all}\ n\ge M

Conditions: n\ge M

Rule: eq_replace · Uses: metric_def

Leaf basis: eq_replace + metric_def

Let (s_n) be a sequence in any metric space with s_n=p for every n≥M. For such n, replacing s_n by p (substitution of equals) gives d(s_n,p)=d(p,p), and the metric axiom d(x,y)=0 ⇔ x=y gives d(p,p)=0.



S04.2 · impliesBefore

\varepsilon>0\ \text{arbitrary};\ d(s_n,p)=0\ (n\ge M)

After

\exists N\ (N=M)\ \forall n\ge N:\ d(s_n,p)<\varepsilon

Conditions: \varepsilon>0

Rule: exists_intro · Uses: S04.1, exists_intro

Leaf basis: exists_intro + order_basics

Take any ε>0. Since 0<ε, S04.1 gives d(s_n,p)=0<ε for every n≥M. So N=M is a valid witness, and ∃-introduction gives ∃N. Here N does not depend on ε, which is special to ultimately constant sequences; a general convergent sequence may require N to be adjusted with ε, but the definition requires neither a change every time nor strict monotonicity of arbitrarily chosen witnesses.



S04.3 · impliesBefore

\varepsilon\ \text{arbitrary}:\ \exists N\ \forall n\ge N:\ d(s_n,p)<\varepsilon

After

\forall\varepsilon>0\ \exists N\ \forall n\ge N:\ d(s_n,p)<\varepsilon,\ \text{i.e.}\ s_n\to p

Conditions: none

Rule: forall_intro · Uses: S04.2, forall_intro, S01

Leaf basis: forall_intro + def_unfold

S04.2 made no assumption on ε beyond ε>0, so ∀-introduction gives the statement for every ε>0; folding back the definition of S01 gives s_n→p. S04.1 to S04.3 together prove Lemma L_tail_const.



S05.1 · impliesBefore

\varepsilon>\tfrac32

After

N=0\ \text{valid, hence}\ N_{\min}=0

Conditions: \varepsilon>\tfrac32

Rule: order_basics · Uses: S05

Leaf basis: order_basics + numeric_evaluation

Every d(x_n,b) lies in {0,1,3/2}, whose maximum is 3/2 (N5), and 3/2<ε, so with N=0 every term satisfies d(x_n,b)<ε. N cannot be below 0, so the least valid N is 0.



S05.2 · impliesBefore

1<\varepsilon\le\tfrac32

After

N\in\{0,1\}\ \text{invalid},\ N=2\ \text{valid, hence}\ N_{\min}=2

Conditions: 1<\varepsilon\le\tfrac32

Rule: order_basics · Uses: S05

Leaf basis: order_basics + numeric_evaluation

With N=0 or N=1 the index n=1 is in the tail, and d(x_1,b)=3/2≥ε, so the strict inequality fails and these N are invalid. With N=2 the tail is x_2=a (distance 1<ε, N10) followed by b (distance 0), all below ε. So the least valid N is 2.



S05.3 · impliesBefore

0<\varepsilon\le 1

After

N\le 2\ \text{invalid},\ N=3\ \text{valid, hence}\ N_{\min}=3

Conditions: 0<\varepsilon\le 1

Rule: order_basics · Uses: S05

Leaf basis: order_basics + numeric_evaluation

Every tail with N≤2 contains n=2, and d(x_2,b)=d(a,b)=1≥ε, so such N are invalid. With N=3 the whole tail is b, at distance 0<ε. So the least valid N is 3; for example with ε=1/1000, N must be at least 3.



S06.1 · impliesBefore

s_n\to p,\ s_n\to q,\ \varepsilon>0

After

\exists N:\ d(s_N,p)<\tfrac{\varepsilon}{2}\ \land\ d(s_N,q)<\tfrac{\varepsilon}{2}

Conditions: \tfrac{\varepsilon}{2}>0

Rule: forall_elim · Uses: S01, forall_elim, exists_elim

Leaf basis: forall_elim + exists_elim + order_basics

Take any ε>0; then ε/2>0. Applying the definition of s_n→p at ε/2 gives N_1 with d(s_n,p)<ε/2 for n≥N_1; applying s_n→q at ε/2 gives N_2. Let N=max(N_1,N_2); then N≥N_1 and N≥N_2, so the term s_N satisfies both inequalities.



S06.2 · impliesBefore

d(s_N,p)<\tfrac{\varepsilon}{2},\ d(s_N,q)<\tfrac{\varepsilon}{2}

After

d(p,q)\le d(p,s_N)+d(s_N,q)<\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon

Conditions: none

Rule: metric_def · Uses: S06.1, metric_def

Leaf basis: metric_def + order_basics + forall_intro

The triangle inequality gives d(p,q)≤d(p,s_N)+d(s_N,q); symmetry replaces d(p,s_N) by d(s_N,p). Adding the two strict inequalities gives less than ε/2+ε/2=ε (N6 checks that ε/2+ε/2=ε is an identity). Since ε was arbitrary in S06.1, d(p,q)<ε holds for every ε>0.



S06.3 · impliesBefore

\forall\varepsilon>0:\ d(p,q)<\varepsilon

After

d(p,q)=0

Conditions: d(p,q)\ge 0

Rule: contradiction · Uses: S06.2, contradiction

Leaf basis: contradiction + order_basics

Since d≥0, it suffices to rule out d(p,q)>0. Suppose d(p,q)>0; then ε=d(p,q) is an admissible positive number and S06.2 gives d(p,q)<d(p,q), a contradiction. Hence d(p,q)=0.



S06.4 · impliesBefore

d(p,q)=0

After

p=q

Conditions: none

Rule: metric_def · Uses: S06.3, metric_def

Leaf basis: metric_def

Separation, d(p,q)=0 ⇔ p=q, gives p=q. S06.1 to S06.4 together prove Lemma L_limit_unique; the proof works in any metric space and does not use that X is finite.







## Reverse audits


- S14 · premise_audit · S02, S04, S05, S11, S13 — This only conjoins conclusions already proved; there is no new inference.
- S13 · premise_audit · S04, S12, L_tail_const — The conclusion is only that finite data do not decide the limit; it does not deny x_n→b, because the convergence of x comes from the declaration of S03 for all n≥3, not from the finite tail.
- S12 · premise_audit · G3, L_tail_const, L_limit_unique — Both lemmas were proved for any metric space; this step only instantiates them.
- S11 · premise_audit · S04, S09, S10, G1 — Saying 'exactly' needs the complete list of X (G1); excluding only some points would not suffice.
- S10 · premise_audit · S08, S03, exists_intro — The same premises as S09; only the target point and ε change.
- S09 · premise_audit · S08, S03, exists_intro, forall_intro — One ε is enough; an n must be found for every N, and the construction max(N,3) works for arbitrary N. The prefix terms x_0=a and x_2=a do not help the opposite conclusion at all.
- S08 · inverse_equivalence · quantifier_negation, order_basics — Quantifier negation holds in both directions; trichotomy of the reals makes ¬(d<ε) equivalent to d≥ε.
- S07 · premise_audit · S06, contrapositive — The contrapositive is equivalent to the original statement, so this is a valid one-way inference; S08 to S10 give direct witnesses as a cross-check.
- S06.4 · premise_audit · metric_def — This uses the ⇒ direction of separation; for a pseudometric (allowing d(p,q)=0 with p≠q) limits need not be unique.
- S06.3 · premise_audit · S06.2, contradiction, order_basics — The key point is that S06.2 holds for every ε>0, so ε=d(p,q) may be used; if it held only for some ε, this choice would not be allowed.
- S06.2 · premise_audit · metric_def, order_basics, forall_intro — Uses the triangle inequality, symmetry and the addition of inequalities; only an upper bound is obtained, no equality is claimed.
- S06.1 · premise_audit · forall_elim, exists_elim, order_basics — Instantiation needs ε/2>0; taking the max puts one index in both tails at once. N_1 and N_2 are fresh names from ∃-elimination, and the final conclusion p=q does not mention them.
- S06 · premise_audit · S02, S04, L_limit_unique — Requires d to be a genuine metric (S02); if the table violated the triangle inequality, S06.2 could not be taken.
- S05.3 · premise_audit · S05, S03, order_basics — Rejecting N≤2 uses the term n=2; accepting N=3 uses the declaration of S03 for all n≥3.
- S05.2 · premise_audit · S05, order_basics — Rejecting an N needs one bad index (n=1); accepting N=2 needs every tail term, which the prefix values and the declaration of S03 supply.
- S05.1 · premise_audit · S05, order_basics — Uses the complete list of distance values (three prefix terms plus the constant tail of S03) and transitivity d≤3/2<ε.
- S05 · premise_audit · S03, case_analysis — The intervals (3/2,∞), (1,3/2] and (0,1] are pairwise disjoint with union (0,∞), so the cases are exhaustive.
- S04.3 · premise_audit · S04.2, forall_intro, S01 — The generalization needs ε to be arbitrary; no particular ε was chosen in the proof.
- S04.2 · premise_audit · S04.1, order_basics — ε>0 is needed for 0<ε; if ε=0 were allowed, the strict inequality 0<0 would fail, and the definition does not ask for it.
- S04.1 · premise_audit · eq_replace, metric_def — This is claimed only for n≥M; the terms with n<M are unrestricted and do not affect the conclusion.
- S04 · premise_audit · S03, L_tail_const — The only premises are 'x_n=b for every n≥3' and d(b,b)=0. The converse (convergence implies ultimately constant) is not needed here; it does hold in finite metric spaces, but fails in general metric spaces such as ℝ (the separately positive-indexed sequence 1/n, n≥1, tends to 0 without being ultimately constant).
- S03 · premise_audit · G2, def_intro — This step only records the declaration of G2; without a declaration for all n≥3, no finite list could determine the behaviour of the tail.
- S02.3 · premise_audit · S02.1, case_analysis — The cases must cover every triple with a repeated point: if the three points are not all distinct, then x=y, y=z or x=z, which are exactly these cases.
- S02.2 · premise_audit · S02.1, numeric_evaluation — Each inequality is recomputed independently in N1, N2 and N3; the reduction to 3 inequalities uses the symmetry from S02.1.
- S02.1 · premise_audit · G1, metric_def, numeric_evaluation — Only the enumeration of all 9 cells supports these three axioms; leaving out a cell would make it a partial check.
- S02 · premise_audit · L_metric_table, metric_def — Only this table is claimed to be a metric, not any other table. Here the triangle inequality amounts to finitely many inequalities, and the conclusion follows only after all of them are enumerated.
- S01 · premise_audit · mko_metric_conv — This step only unfolds the definition and produces no new fact; the definition presupposes that (X,d) is a metric space, which S02 checks.





## Error controls and negative controls

A finite tail is not a proof of convergenceThe misconception named by the MKO is to assert convergence because a long but finite tail looks close. The sequence y_n of S13 equals x_n for n=3 to 10 (all equal to b), yet it converges to c. x_n→b holds because G2 declares x_n=b for every n≥3; that is a rule covering infinitely many terms, not the result of a finite check.

Strict inequality and quantifier orderThe definition uses d(x_n,x)<ε, a strict inequality. In S09, d(x_n,a)=1 does not satisfy < for ε=1, so ε=1 already excludes a; with ≤ by mistake this particular witness would fail (although ε=1/2 would still exclude a). The quantifier order allows N to depend on ε but does not require dependence: N=3 works for every ε>0 here, while S05 shows that the least valid N changes from 0 to 2 to 3 as ε shrinks.

Check the metric table firstThe uniqueness proof S06.2 uses the triangle inequality. If the table were miswritten with d(a,c)=3 (everything else unchanged), then d(a,c)=3>d(a,b)+d(b,c)=5/2, the triangle inequality fails, the table is not a metric, and the argument of S06 cannot be used. So S02 must be completed before any convergence reasoning.


- x_3 to x_10 all equal b, so every sequence with these terms converges to b → reject — y_n has exactly these terms but converges to c (S12, S13).
- Since x_0=a and x_2=a, x_n→a → reject — Convergence depends only on the tail: with ε=1, for every N the index n=max(N,3) gives d(x_n,a)=1≥1 (S09).
- For n≥3, d(x_n,a)=1≤1, so taking ε=1 shows that x_n converges to a → reject — The definition requires every ε>0 and a strict <; at ε=1 already 1<1 fails, and at ε=1/2 it fails as well.
- In some metric space a sequence can converge to both b and c → reject — L_limit_unique holds in every metric space: the distance between two limits is below every ε>0, hence 0, and separation gives b=c; here d(b,c)=3/2>0.





## Result and limitations

The table of G1 is a metric; the sequence x_n (prefix a, c, a, then constantly b for n≥3) converges to b, with N=3 valid for every ε>0 and least valid N equal to 0, 2, 3 for ε>3/2, 1<ε≤3/2, 0<ε≤1 respectively; limits in a metric space are unique, and x_n converges neither to a (witness ε=1) nor to c (witness ε=3/2). The sequence y_n, which agrees with x_n on a finite tail, converges to c instead, both sequences converge but have different limits, so a finite tail check cannot decide whether the limit is the specified b. This pair by itself does not distinguish existence from nonexistence of a limit.


- This guide does not treat Cauchy sequences, completeness or continuous maps (the MKO's formal next).
- The remark in the audit of S04 that convergence implies ultimately constant in a finite metric space (provable by taking ε to be the least positive distance) is only a note; the argument of this guide does not depend on it.
- The least valid N of S05 holds only for this sequence and this table.





## Verification and source binding

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


- source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
- adopted source SHA-256: 9f78ba61eeae5e730d7f5c30614aceca18c228eb2343bf50bee946868c588393
- original source SHA-256: c050f16c9639e75c444249ed4ba0351bd36dc14296bbf9c4278170c501246d93
- Lean: NOT_RUN · independent AI: NOT_MEASURED

[Same-source full JSON (formulas, ASTs, controls and revision log)](https://ocme.evemisslab.com/data/derivations/production-20261019/guides/derivation-mko-metric-convergence-01.json)