Main derivation
S01 · definition
Before
(\mathbb{R}^2,\|\cdot\|_1)\ \text{Banach}After
\|\cdot\|_1\ \text{norm}\ \land\ d_1=\|x-y\|_1\ \text{metric}\ \land\ (\mathbb{R}^2,d_1)\ \text{complete}Conditions: none
Rule: mko_banach_def · Uses: G1, G2, mko_banach_def
Leaf basis: mko_banach_def + def_unfold
By the MKO, a Banach structure binds the one carrier ℝ², its vector operations, the norm ‖·‖₁, the metric d₁ induced by that norm and completeness. So three things must be proved: ‖·‖₁ is a norm (S02 to S05), d₁ is a metric (S06), and (ℝ²,d₁) is complete (S08 to S12).
S02 · equivalent
Before
\|x\|_1=|x_1|+|x_2|=0
After
x=(0,0);\ \ \text{and}\ \|x\|_1\ge 0\ \text{for all}\ xConditions: none
Rule: abs_rules · Uses: G2, abs_rules
Leaf basis: abs_rules + order_basics
|x₁|≥0 and |x₂|≥0, so the sum is ≥0. A sum of two nonnegative numbers is 0 exactly when both are 0 (if one were positive the sum would be positive), and |t|=0 exactly when t=0, so ‖x‖₁=0 ⇔ x₁=x₂=0 ⇔ x=0.
S03 · equivalent
Before
\|\lambda x\|_1=|\lambda x_1|+|\lambda x_2|
After
\|\lambda x\|_1=|\lambda|(|x_1|+|x_2|)=|\lambda|\,\|x\|_1
Conditions: none
Rule: abs_rules · Uses: G1, G2, abs_rules
Leaf basis: abs_rules + distributive
The coordinates of λx are λx₁ and λx₂ (G1). By |uv|=|u||v|, |λx_k|=|λ||x_k|, and distributivity pulls out |λ|. This holds for every real λ and every x.
S04 · implies
Before
\|x+y\|_1=|x_1+y_1|+|x_2+y_2|
After
\|x+y\|_1\le(|x_1|+|y_1|)+(|x_2|+|y_2|)=\|x\|_1+\|y\|_1
Conditions: none
Rule: abs_rules · Uses: G1, G2, abs_rules
Leaf basis: abs_rules + order_basics + add_assoc + add_comm
Apply the real triangle inequality |x_k+y_k|≤|x_k|+|y_k| in each coordinate, add the two inequalities (the order axioms allow adding inequalities in the same direction), and regroup to get ‖x‖₁+‖y‖₁.
S05 · implies
Before
\text{S02},\ \text{S03},\ \text{S04}After
(\mathbb{R}^2,\|\cdot\|_1)\ \text{is a real normed vector space}Conditions: none
Rule: mko_norm_def · Uses: S02, S03, S04, mko_norm_def
Leaf basis: mko_norm_def + and_intro_elim
Nonnegativity and definiteness (S02), homogeneity (S03) and the triangle inequality (S04) are exactly the three clauses of the MKO's norm definition, so ‖·‖₁ is a norm on ℝ².
S06 · theorem_application
Before
d_1(x,y)=\|x-y\|_1
After
d_1\ \text{is a metric on}\ \mathbb{R}^2Conditions: none
Rule: L_induced_metric · Uses: S05, G2
Leaf basis: metric_def + mko_norm_def
By Lemma L_induced_metric (for any norm, d(x,y)=‖x−y‖ is a metric), d₁ is a metric. The lemma is proved in sub-steps S06.1 to S06.3 using only the three norm axioms of S02 to S04.
S07 · implies
Before
x=(3,-1),\ y=(1,2)
After
d_1(x,y)=|2|+|-3|=5;\ \ \max(|2|,|-3|)=3\neq 5
Conditions: none
Rule: numeric_evaluation · Uses: G2, G3
Leaf basis: numeric_evaluation + abs_rules
x−y=(2,−3), so d₁(x,y)=2+3=5. For contrast, the ℓ^∞ norm gives max(2,3)=3. The two numbers differ, which shows the ℓ^∞ metric is not the metric induced by ‖·‖₁; the MKO forbids taking norm and metric from different formulas.
S08 · implies
Before
x,y\in\mathbb{R}^2,\ k\in\{1,2\}After
|x_k-y_k|\le d_1(x,y)
Conditions: none
Rule: L_coord_le · Uses: G2
Leaf basis: abs_rules + order_basics
The key lemma for completeness: the distance in each coordinate is at most the d₁ distance (Lemma L_coord_le). The proof is in sub-step S08.1.
S09 · implies
Before
(x^{(m)})\ \text{Cauchy in}\ (\mathbb{R}^2,d_1)After
(x^{(m)}_k)_{m\ge1}\ \text{Cauchy in}\ \mathbb{R}\ \ (k=1,2)Conditions: none
Rule: cauchy_def · Uses: S08, cauchy_def
Leaf basis: cauchy_def + L_coord_le + order_basics
Take an arbitrary d₁-Cauchy sequence (no specific values). Given ε>0, the Cauchy definition gives N with d₁(x^{(m)},x^{(p)})<ε for m,p≥N; by L_coord_le, |x^{(m)}_k−x^{(p)}_k|≤d₁(x^{(m)},x^{(p)})<ε. The same N works for both coordinates, so each coordinate sequence is a Cauchy sequence of reals.
S10 · theorem_application
Before
(x^{(m)}_k)_{m\ge1}\ \text{Cauchy in}\ \mathbb{R}After
x^{(m)}_k\to a_k\in\mathbb{R}\ (k=1,2),\ \ a=(a_1,a_2)\in\mathbb{R}^2Conditions: none
Rule: real_cauchy_complete · Uses: S09, real_cauchy_complete
Leaf basis: real_cauchy_complete
By the Cauchy completeness of the reals, the two coordinate sequences converge to some reals a₁ and a₂. Put a=(a₁,a₂); a is a point of ℝ², so the candidate limit stays in the space. This is the only analytic fact the guide imports.
S11 · implies
Before
x^{(m)}_1\to a_1,\ x^{(m)}_2\to a_2After
d_1(x^{(m)},a)\to 0Conditions: none
Rule: cauchy_def · Uses: S10, cauchy_def, G2
Leaf basis: cauchy_def + seq_limit_def + abs_rules
We must show x^{(m)}→a in d₁. Since d₁(x^{(m)},a) is the sum of the two coordinate errors, it suffices to make each error less than ε/2. Sub-step S11.1 picks a common threshold and S11.2 completes the estimate.
S12 · implies
Before
\text{arbitrary Cauchy}\ (x^{(m)})\ \text{converges in}\ (\mathbb{R}^2,d_1)After
(\mathbb{R}^2,d_1)\ \text{complete};\ (\mathbb{R}^2,\|\cdot\|_1)\ \text{real Banach space}Conditions: none
Rule: forall_intro · Uses: S05, S06, S09, S10, S11, mko_complete_def, mko_banach_def
Leaf basis: forall_intro + mko_complete_def + mko_banach_def
S09 to S11 hold for an arbitrary d₁-Cauchy sequence, so by universal generalization every Cauchy sequence converges to a point of ℝ²: (ℝ²,d₁) is complete. Together with the norm of S05 and the induced metric of S06, all three parts bind one structure, so by the MKO (ℝ²,‖·‖₁) is a real Banach space.
S13 · theorem_application
Before
\dim\mathbb{R}^n=n<\infty,\ \ \|\cdot\|\in\{\|\cdot\|_1,\|\cdot\|_\infty\}After
(\mathbb{R}^n,\|\cdot\|)\ \text{Banach (named-theorem certificate)}Conditions: n\ \text{finite}
Rule: thm_findim_complete · Uses: S12, thm_findim_complete
Leaf basis: thm_findim_complete
The MKO's certificate is the named theorem: every finite-dimensional real normed space is complete in its induced metric. Its hypotheses are finite dimension and a metric induced by the same norm. S08 to S12 are a complete proof of the case n=2, ℓ¹, relying only on completeness of the reals; the ℓ^∞ and general-n cases are only cited here, not proved (the same coordinate argument would work but is not written out).
S14 · implies
Before
u=(1,0),\ v=(0,1):\ \|u+v\|_1^2+\|u-v\|_1^2=2^2+2^2=8
After
8\neq 4=2\|u\|_1^2+2\|v\|_1^2\ \Rightarrow\ \|\cdot\|_1\ \text{induced by no inner product}Conditions: none
Rule: L_parallelogram · Uses: G3, S12
Leaf basis: contrapositive + numeric_evaluation
u+v=(1,1) and u−v=(1,−1) both have ℓ¹ norm 2, so the left side is 4+4=8, while ‖u‖₁=‖v‖₁=1 makes the right side 2+2=4. By the contrapositive of Lemma L_parallelogram, ‖·‖₁ is induced by no inner product. The MKO notes that a Banach space's specified norm need not be induced by an inner product or define a compatible Hilbert structure; its underlying vector space may still carry another inner product. The lemma is proved in sub-steps S14.1 and S14.2.
S15 · implies
Before
\text{S06},\ \text{S07},\ \text{S12},\ \text{S14}After
(\mathbb{R}^2,\|\cdot\|_1)\ \text{Banach};\ d_1((3,-1),(1,2))=5;\ \|\cdot\|_1\ \text{not from an inner product}Conditions: none
Rule: and_intro_elim · Uses: S06, S07, S12, S13, S14
Leaf basis: and_intro_elim
Combine: ‖·‖₁ is a norm, d₁ is its induced metric and d₁((3,−1),(1,2))=5; (ℝ²,d₁) is complete, so the space is a real Banach space; the ℓ¹ norm violates the parallelogram law and does not come from an inner product.
Substeps and lemmas
S06.1 · equivalent
Before
d(x,y)=\|x-y\|
After
d(x,y)\ge 0,\ \ d(x,y)=0\iff x-y=0\iff x=y
Conditions: none
Rule: mko_norm_def · Uses: S02
Leaf basis: mko_norm_def + vector_space_axioms
A norm is nonnegative, so d≥0; a norm is 0 exactly for the zero vector, and x−y=0 exactly when x=y (add y to both sides).
S06.2 · equivalent
Before
d(y,x)=\|y-x\|=\|(-1)(x-y)\|
After
d(y,x)=|-1|\,\|x-y\|=d(x,y)
Conditions: none
Rule: mko_norm_def · Uses: S03
Leaf basis: mko_norm_def + vector_space_axioms + abs_rules
y−x=(−1)(x−y) (vector space axioms), and by homogeneity ‖(−1)z‖=|−1|‖z‖=‖z‖, so d is symmetric.
S06.3 · implies
Before
x-z=(x-y)+(y-z)
After
d(x,z)\le d(x,y)+d(y,z)
Conditions: none
Rule: mko_norm_def · Uses: S04
Leaf basis: mko_norm_def + vector_space_axioms + metric_def
Split x−z as (x−y)+(y−z) and apply the triangle inequality of the norm. S06.1 to S06.3 together prove Lemma L_induced_metric: nonnegativity and separation, symmetry and the triangle inequality all hold.
S08.1 · implies
Before
d_1(x,y)=|x_1-y_1|+|x_2-y_2|
After
d_1(x,y)\ge|x_1-y_1|\ \text{and}\ d_1(x,y)\ge|x_2-y_2|Conditions: none
Rule: order_basics · Uses: G2, abs_rules
Leaf basis: abs_rules + order_basics + add_zero
Since |x₂−y₂|≥0, d₁(x,y)=|x₁−y₁|+|x₂−y₂|≥|x₁−y₁|+0; replacing the first term by 0 likewise gives the second inequality. This proves Lemma L_coord_le.
S11.1 · implies
Before
\varepsilon>0;\ \exists N_k:\ m\ge N_k\Rightarrow|x^{(m)}_k-a_k|<\tfrac{\varepsilon}{2}After
N=\max(N_1,N_2):\ m\ge N\Rightarrow|x^{(m)}_1-a_1|<\tfrac{\varepsilon}{2}\ \land\ |x^{(m)}_2-a_2|<\tfrac{\varepsilon}{2}Conditions: \varepsilon>0
Rule: seq_limit_def · Uses: S10, seq_limit_def
Leaf basis: seq_limit_def + order_basics
Given ε>0, also ε/2>0; by the definition of the limit of a real sequence, for each k there is N_k such that the coordinate error is below ε/2 for m≥N_k. Take N=max(N₁,N₂); then m≥N implies both m≥N₁ and m≥N₂.
S11.2 · implies
Before
m\ge N
After
d_1(x^{(m)},a)=|x^{(m)}_1-a_1|+|x^{(m)}_2-a_2|<\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilonConditions: none
Rule: order_basics · Uses: S11.1, G2
Leaf basis: order_basics + frac_rules + cauchy_def
Adding the two strict inequalities gives a sum below ε/2+ε/2=ε (N7 checks this algebraic identity). Since ε is an arbitrary positive number, x^{(m)}→a by the definition of convergence.
S14.1 · equivalent
Before
\langle u\pm v,u\pm v\rangle
After
\langle u,u\rangle\pm 2\langle u,v\rangle+\langle v,v\rangle
Conditions: none
Rule: inner_product_def · Uses: inner_product_def
Leaf basis: inner_product_def + distributive
Expand by linearity in both slots: ⟨u±v,u±v⟩=⟨u,u⟩±⟨u,v⟩±⟨v,u⟩+⟨v,v⟩; then symmetry ⟨v,u⟩=⟨u,v⟩ merges the two middle terms.
S14.2 · equivalent
Before
\langle u+v,u+v\rangle+\langle u-v,u-v\rangle
After
2\langle u,u\rangle+2\langle v,v\rangle,\ \text{i.e.}\ \|u+v\|^2+\|u-v\|^2=2\|u\|^2+2\|v\|^2Conditions: none
Rule: eq_replace · Uses: S14.1
Leaf basis: eq_replace + add_inverse + inner_product_def
Add the two expansions of S14.1: the terms ±2⟨u,v⟩ cancel; then rewrite with ‖z‖²=⟨z,z⟩. S14.1 and S14.2 together prove Lemma L_parallelogram.