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<\varepsilonConditions: \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}\ XConditions: 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}\ XAfter
\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}\ XAfter
\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_nAfter
|y_n-y_{n+1}|=|2y_n|=2|y_n|=2\ \text{for every } n\ge 1Conditions: 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+1After
|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=3After
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>0Conditions: 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 L2Conditions: 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.