
Mark's Elements, II
Posted:
Jul 24, 1996 2:00 PM


In this article, as indicated at the end of the previous one, I will illustrate the construction of closed metric spaces using Zeno sequences. In addition, alternative axiomitizations along the lines of that presented before will be given which will solve the logical equations:
Banach Space = Metric Space + (___) Banach Space = Metric Space + Vector Space + (___)
in the process also providing an originindependent definition of a vector space in terms of the operator r > [A, r, B] and just 3 axioms.
Contents: (7) Construction of Closed Metric Spaces (8) Construction of Hilbert Spaces (9) Characterization of Vector Spaces (10) Characterization of Banach Spaces (11) Summary & Conclusion. Incorporating Minkowski Spaces.
(7) Construction of Closed Metric Spaces First, this is a review of some of the definitions and results given before.
DEFINITION: A sequence (a_n) is said to be of bounded variation if the set of distances { a_m a_n: m, n >= 0 } is bounded above. THEOREM: If (a_n) and (b_n) are of bounded variation then the set { a_m b_n: m, n >= 0 } is bounded above. DEFINITION: Given sequences (a_n), (b_n) of bounded variation, define (a_n) (b_n) = inf (sup { a_n b_n: n >= m }: m >= 0) If (a_n) = (a, a, a, ...) then let a(b_n) = (a_n)(b_n), and if (b_n) = (b, b, b, ...) then let (a_n)b = (a_n)(b_n).
The axioms presented in the first section that are referred to later below are also rewritten here:
AXIOM 1: AA = 0 AXIOM 2: AB > 0 unless A = B AXIOM 3: AB = BA AXIOM 4: AB <= AC + CB AXIOM 8: Every Zeno sequence has a Zeno limit
Previous results will be used implicitly, so the theorems will be numbered anew.
THEOREM 1: Let (a_n), (b_n) be Zeno sequences. Then d = (a_n) (b_n) is the unique number for which:  a_n b_n  d  <= 2 (a_n a_{n+1} + b_n b_{n+1}) Proof: Let n >= 0 and e > 0 be given. Then we can find an N >= n such that
d < sum { a_p b_p: p >= N } < d + e
and an P >= N such that
d  e < a_P b_P < d + e Therefore, a_n b_n <= a_n a_P + a_P b_P + b_P b_n <= 2 a_n a_{n+1} + d + e + 2 b_n b_{n+1} a_n b_n  d <= 2 (a_n a_{n+1} + b_n b_{n+1}) + e and d <= a_P b_P + e <= a_n a_P + a_n b_n + b_n b_P + e <= 2 a_n a_{n+1} + a_n b_n + 2 b_n b_{n+1} + e d  a_n b_n <= 2 (a_n a_{n+1} + b_n b_{n+1}) + e
Therefore a_n b_n  d <= 2 (a_n a_{n+1} + b_n b_{n+1}) + e
Since e > 0 was chosen arbitrarily, it follows that
a_n b_n  d <= 2 (a_n a_{n+1} + b_n b_{n+1})
Suppose d' is another number satisfying the same condition. Then let e > 0 be given and choose M >= 0 so that
a_0 a_1 <= 2^M e/2, b_0 b_1 <= 2^M e/2 Then d  d' <= d  a_N b_N + a_N b_N  d' < e/2 + e/2 = e
Since the difference between d and d' can be made less than any e > 0, it follows that d  d' = 0, or d = d'.
DEFINITION: If (a_n), (b_n) are Zeno sequences, then write (a_n) == (b_n) if a_n b_n <= 2 (a_n a_{n+1} + b_n b_{n+1}), for all n >= 0
THEOREM 2: For Zeno sequences (a_n), (b_n), AXIOMS 1 and 2 hold in the following modified form: (1') If (a_n) == (b_n) then (a_n) (b_n) = 0 (2') (a_n) (b_n) > 0 unless (a_n) == (b_n) Proof: The equivalence (a_n) (b_n) = 0 <==> (a_n) == (b_n) is direct from the definition and THEOREM 1. Also, since
sup { a_n b_n: n >= m } >= 0 we have (a_n) (b_n) = inf (sup { a_n b_n: n >= m }: m >= 0) >= 0
Result (2') follows from this.
THEOREM 3: AXIOMS 3 and 4 hold for Zeno sequences. Proof: The symmetry property (AXIOM 3) is obvious, since
a_n b_n  d = b_n a_n  d Therefore (a_n) (b_n) = (b_n) (a_n).
To prove the Triangle Inequality (AXIOM 4), let (a_n), (b_n) and (c_n) be Zeno sequences with (a_n) (b_n) = d, (b_n) (c_n) = d'. Let e > 0 be chosen. Then we can find an N >= 0 such that for all n >= N,
a_n b_n <= d + e/2, b_n c_n <= d' + e/2 so that a_n c_n <= a_n b_n + b_n c_n <= d + d' + e Therefore sup { a_n c_n: n >= N } <= d + d' + e and (a_n) (c_n) = inf (sup { a_n c_n: n >= N }: m >= 0) <= d + d' + e.
Since e was chosen arbitratily, it follows that
(a_n) (c_n) <= d + d' = (a_n) (b_n) + (b_n) (c_n).
That the definition of Zeno equivalence (==) also provides a direct generalization of Zeno convergence is seen by the following:
COROLLARY 4: The relation == is an equivalence. Proof: Reflexivity: AA = 0, therefore A == A Symmetry: A == B > AB = 0 > BA = 0 (by THEOREM 2) > B == A Transitivity: A == B, B == C > AB = BC = 0 > AC <= AB + BC = 0 + 0 = 0 (by THEOREM 3) > 0 <= AC <= 0 (by THEOREM 2) > A == C
THEOREM 5: If (b_n) is the constance sequence (a, a, a, ...), then (a) (b_n) is a Zeno sequence (b) For all Zeno sequences (a_n): (a_n) == (b_n) <==> a_n >> a Proof: Condition (a) is obvious since b_{n+1} b_{n+2} = 0 <= 1/2 b_n b_{n+1} and is only stated for completeness. For part (b), note that
(a_n) == (b_n) <==> a_n b_n <= 2 (a_n a_{n+1} + b_n b_{n+1}) <==> a_n a <= 2 a_n a_{n+1} <==> a_n >> a
THEOREM 6: Let (a_n), (b_n) be Zeno sequences, with a_n >> a. Then (a) If b_n >> b, then (a_n) (b_n) = ab (b) If b_n >> a, then (a_n) == (b_n) (c) If (a_n) == (b_n) then b_n >> a Proof: All these results can be derived directly from THEOREM 3. Part (a): If b_n >> b, then (a_n) (b_n) <= (a_n) a + ab + b (b_n) = 0 + ab + 0 = ab ab <= a (a_n) + (a_n)(b_n) + (b_n) b = 0 + (a_n)(b_n) + 0 = (a_n)(b_n)
Part (b): If b_n >> a, then by the previous result (a_n)(b_n) = aa = 0, therefore (a_n) == (b_n).
Part (c): If (a_n) == (b_n) then (a_n)(b_n) = 0. Therefore a(b_n) <= a(a_n) + (a_n)(b_n) = 0 + 0 = 0, a(b_n) = 0 (since it can't be negative by THEOREM 2), b_n >> a
However, it is more illustrative to prove these results directly. Part (a): Suppose a_n >> a, b_n >> b. Then ab  a a_n  b_n b <= a_n b_n <= ab + a a_n + b_n b a_n b_n  ab <= a_n a + b_n b <= 2 (a_n a_{n+1} + b_n b_{n+1}) Therefore, by THEOREM 1, (a_n)(b_n) = ab.
Part (b): Same proof as above
Part (c): This is basically a reproof of a theorem from the previous artlcle. Let n >= 0 be given and choose and e > 0. Then we can find a N >= n such that a_0 a_1 + b_0 b_1 < e 2^N/3 from which it follows that: a_N a <= 2 a_N a_{N+1} <= 2 2^{N} a_0 a_1 < e/3 Therefore a b_n <= a a_N + a_N b_N + b_N b_n < e/3 + 2 (a_N a_{N+1} + b_N b_{N+1}) + 2 b_n b_{n+1} < e/3 + 2 2^{N} (a_0 a_1 + b_0 b_1) + 2 b_n b_{n+1} < e/3 + 2 e/3 + 2 b_n b_{n+1} Noting the arbitrariness of e > 0, it follows that a b_n <= 2 b_n b_{n+1} Therefore, b_n >> a.
THEOREM 7: AXIOM 8 holds for Zeno sequences. Proof: Let (A_m) be a Zeno sequence with A_m = (a_mn) a Zeno sequence for each m >= 0. The diagonal of the matrix (a_mn): A = (a_nn: n >= 0) will be a Zeno limit of (A_m). Define d_m = A_m A_{m+1}. Then by definition, d_{m+1} <= 1/2 d_m for all m >= 0. Let e > 0 and n > m >= 0 be given, and define N_k so that
a_kn a_{k+1}n <= d_k + e/(nm) whenever m <= k < n
Define N = max { N_m, N_{m+1}, ..., N_{n1} }. Then for all n >= N, we have: a_mn a_nn <= a_mn a_{m+1}n + ... + a_{n1}n a_nn <= (d_m + e/(nm)) + ... + (d_{n1} + e/(nm)) = (d_m + ... + d_{n1}) + e <= d_m (1 + 1/2 + ... + 1/2^{n1})) + e < 2 d_m + e Therefore a_mn a_nn < 2 d_m sup { (A_m)_n A_n: n >= m } < 2 d_m A_m A = inf ( sup { (A_m)_n A_n: n >= m }: m >= 0 ) < 2 d_m or A_m A <= 2 A_m A_{m+1}
which proves that A_m >> A.
This leads directly to our final result.
THEOREM 8: Any space satisfying AXIOMS 14 (i.e., Metric space) can be extended to one satisfying AXIOM 8. Proof: Let V be a Metric space, as indicated in the condition of the theorem. By THEOREM 4, we can define the following space:
V* = { [A]: A is a Zeno sequence in V } where [A] = { B: B is a Zeno sequence in V, B == A } is the equivalence class of A.
A distance measure can be unambiguously defined on equivalence classes by [A][B] = AB, since
A == A', B == B' > AB <= AA' + A'B' + B'B = 0 + A'B' + 0 = A'B' A'B' <= A'A + AB + BB' 0 0 + AB + 0 = AB > AB = A'B'.
The map a > [a] will embed V into V*: by THEOREM 6b, 6c: [a] = { A: A >> a }, for any point a in V. by THEOREM 6a, [a][b] = (a)(b) = ab
AXIOMS 14 and 8 can be verified directly:
AXIOM 1: [A][A] = AA = 0 AXIOM 2: [A][B] = AB >= 0 [A][B] = AB = 0 <==> A == B <==> [A] = [B]. AXIOM 3: [A][B] = AB = BA = [B][A] AXIOM 4: [A][B] = AB <= AC + CB = [A][C] + [C][B] AXIOM 8: If { [A_n]: n >= 0 } is a Zeno sequence, then for each n >= 0: A_{n+1} A_{n+2} = [A_{n+1}] [A_{n+2}] <= 1/2 [A_n] [A_{n+1}] = 1/2 A_n A_{n+1} Therefore, { A_n: n >= 0 } is also a Zeno sequence. By THEOREM 7, the diagonal sequence A will be a Zeno limit. Therefore, [A_n] [A] = A_n A <= 2 A_n A_{n+1} = 2 [A_n] [A_{n+1}] and [A_n] >> [A].
COROLLARY 9: Any metric space can be extended to a closed metric space (i.e., one closed under limits of Cauchy sequences).
(8) Construction of Hilbert Spaces The same construction above will also yield a Hilbert space (i.e. an inner product space closed under limits of Cauchy sequences), if the original space V is an inner product space. For completeness, AXIOMS 57 of section (1) and THEOREM 10 of section (2) are also presented below:
THEOREM 10: Corresponding to any two points A and B, and real number r, there is a unique point C = [A, r, B] such that AC = r AB, BC = 1  r AB AXIOM 5: [A, 1/N, B] exists for all N = 2, 3, 4, ... AXIOM 6: [A, M, B] exists, for all M = 2, 3, 4, ... AXIOM 7: If AB + BC = AC, then AB CD^2  AC BD^2 + BC AD^2 = AB AC BC
In the previous sections, we found the following characterizations:
THEOREM: An Inner Product Space is one which satisfies AXIOMS 14, 7 and THEOREM 10. [The statement I gave previously included AXIOM 8, which is only true, in general, for Hilbert spaces].
THEOREM: A Hilbert Space is one which satisfies AXIOMS 14, 78 and THEOREM 10, or equivalently AXIOMS 18.
The direct proof that V* satisfies AXIOM 7 and THEOREM 10 if V does would be to define [(a_n)] + [(b_n)] = [(a_n + b_n)] r [(a_n)] = [(r a_n)] and show that <a_n, b_n> > <(a_n), (b_n)> = (0(a_n)^2 + 0(b_n)^2  (a_n)(b_n)^2)/2 as n > infinity
However, (a_n + b_n) need not be a Zeno sequence. To fix this problem, we will need the following technical results:
THEOREM: If B is a Zeno subsequence of a Zeno sequence A, then B == A. Proof: Let A = (a_n) and B = (b_n) = (a_fn), where f: n > fn is an increasing function of n. Then
b_n a_n = a_fn a_n <= 2 a_n a_{n+1} <= 2 (a_n a_{n+1} + b_n b_{n+1})
Therefore, BA = 0, and B == A.
THEOREM: If C is a finite family of Cauchy sequences, then there is an increasing function f: n > fn, such that
For all A in C: (A_fn: n >= 0) is a Zeno subsequence of A.
In THEOREM 9, of section (6) we had proven the result that every Cauchy sequence has a Zeno subsequence. This proof works here just as well, with the following minor modification:
f0 is defined to be 0 as before but f(k+1) is defined to be the least N >= 0 such that A_m A_n <= 1/3 A_fk (A_n), for all m, n >= N, for all A in C
The resulting function will provide common Zeno subsequences to all the Cauchy sequences in the set C. As a result, we have the following:
THEOREM: For every [A], [B] in V*, there are Zeno sequences A' in [A], B' in [B], such that (A'_n + B'_n) and ([A'_n, r, B'_n]) are both Zeno sequences.
We can therefore define [A] + [B] = [A + B], [ [A], r, [B] ] = [[A, r, B]], taking appropriate representatives for the equivalence classes [A] and [B]. A direct, and constructive, proof of these results can also be carried out, much like part (c) of THEOREM 6, but this isn't really central to our focus here. Likewise, the construction of Hilbert spaces isn't really central to our focus here either, therefore these remarks will be left in outline form only.
(9) Characterization of Vector Spaces A similar approach can be adopted to find a characterization for Vector Spaces, and with a few modifications to the original system of axioms, incorporating the results of this section, we will have an axiomatic definition of Banach Spaces (i.e. normed Vector Spaces).
For vector spaces, the only basic concept is the function [A, r, B]. The following then provides an equivalent axiomitization of Vector Spaces. If you attempt to diagram the third axiom, interpreting [A, r, B] as (1r) A + r B, you may notice that it looks similar to something you may see in Projective Geometry. In each of the applications of the axiom below, there will invariably be exceptional cases that have to be handled separately or omitted. These usually have to do with the occurrence of parallel lines where one requires lines to intersect.
AXIOM 1: [A, 0, B] = A AXIOM 2: [A, 1, B] = B AXIOM 3: [A, rt(1t), [B, s, C]] = [[A, rt(1s), B], t, [A, rs(1t), C]]
These three properties are true in any vector space, where [A, r, B] is defined as (1r)A + rB. The third axiom reduces to the identity:
(1rt(1t)) A + rt(1t)(1s) B + rt(1t)s C = ((1  rt(1s))(1t) + (1  rs(1t))t) A + rt(1s)(1t) B + rs(1t)t C
which is true since
(1t)(1  rt(1s)) + t(1  rs(1t)) = (1t) + t  rt(1s)(1t)  rs(1t)t = 1  rt(1t)
For completeness, we state the definition of a Vector Space
DEFINITION: A Vector Space V is a set with the following: (a, b) > a + b in V (r, v) > rv in V 0 is in V for a, b in V, and real number r, such that: (a+b)+c = a+(b+c), a+b = b+a, a+0 = a, a + a = 0 1a = a, r(a+b) = ra+rb, (r+s)a = ra+sa, (rs)a = r(sa) where a is defined as (1)a.
The following results lead us to our final conclusion (again, theorems will be numbered anew).
THEOREM 1: [A, t, A] = A Proof: Simply take r = 0 in AXIOM 3, and apply AXIOM 1.
THEOREM 2: [A, m, B] = [B, 1  m, A] Proof: If m is not 1, we can let r = 1/(1m), s = 0, t = 1m in AXIOM 3 and write [A, m, B] = [A, m, [B, 0, C]] = [[A, 1, B], 1  m, [A, 0, C]] = [B, 1  m, A] The case for m = 1 follows since [B, 0, A] = B = [A, 1, B].
THEOREM 3: [A, m, [A, n, B]] = [A, mn, B] Proof: If m = 1, then [A, 1, [A, n, B]] = [A, n, B] = [A, 1n, B]. Otherwise, let r = n/(1m), s = 1 and t = m. Then
[A, m, [A, n, B]] = [[A, rt(1s), A], t, [A, rs(1t), B]] = [A, rt(1t), [A, s, B]] = [A, mn, B]
THEOREM 4: [A, m, [B, n, C]] = [[A, m, B], n, [A, m, C]] Proof: If n = 0, then the result follows trivially since both sides would be equal to [A, m, B], and if n = 1, they both become [A, m, C]. So assume n is neither 0 nor 1, and let s = t = n and r = m/(n(1n)). Then
[A, m, [B, n, C]] = [A, rt(1t), [B, s, C]] = [[A, rt(1s), B], t, [A, rs(1t), C]] = [[A, m, B], n, [A, m, C]]
THEOREM 5: [[A, m, B], 1/2, [A, n, B]] = [A, (m+n)/2, B] Proof: If m+n = 0, we can write [A, n, B] = [A, m, B] = [A, 1, [A, m, B]] = [[A, m, B], 2, A]. Then [[A, m, B], 1/2, [A, n, B]] = [[A, m, B], 1/2, [[A, m, B], 2, A]] = [[A, m, B], 1, A] = A = [A, (m+n)/2, B] Otherwise, let r = 2(m+n), s = n/(m+n), t = 1/2. Then [A, (m+n)/2, B] = [A, (m+n)/2, [B, n/(m+n), B]] = [A, rt(1t), [B, s, B]] = [[A, rt(1s), B], t, [A, rs(1t), C]] = [[A, m, B], 1/2, [A, n, B]]
THEOREM 6: If m is not 0 or 2, then [A, m, [B, 1/m, C]] = [B, 2m, [A, 1/(2m), C]] Proof: The case where m = 1 is easily handled since both sides would then reduce to C. Assuming m is not 1, let s = 1/m, t = 2  m and r = m/(t(1t)). Then [A, m, [B, 1/m, C]] = [A, rt(1t), [B, s, C]] = [[A, rt(1s), B], t, [A, rs(1t), C]] = [[A, 1, B], 2m, [A, 1/(2m), C]] = [B, 2m, [A, 1/(2m), C]]
THEOREM 7: If m is not 2, then [A, 1/(2m), [B, m, C]] = [B, 1/(2m), [A, m, C]] Proof: Again, the case m = 1 is easily handled since then both sides will reduce to C. Otherwise if m is not 1, let r = (2m)/(1m), s = m and t = 1/(2m). Then
[A, 1/(2m), [B, m, C]] = [A, rt(1t), [B, s, C]] = [[A, rt(1s), B], t, [A, rs(1t), C]] = [[A, 1, B], 1/(2m), [A, m, C]] = [B, 1/(2m), [A, m, C]]
THEOREM 8: [A, 1/2, [B, m, C]] = [[A, 1m, B], 1/2, [A, m, C]] Proof: Take r = 2, s = m, t = 1/2. Then
[A, 1/2, [B, m, C]] = [A, rt(1t), [B, s, C]] = [[A, rt(1s), B], t, [A, rs(1t), C]] = [[A, 1m, B], 1/2, [A, m, C]]
DEFINITION: Let O be a point in V. Define V_O as the Vector Space with 0 = O rA = [O, r, A] A + B = 2 [A, 1/2, B]
THEOREM 9: AXIOMS 13 define a Vector Space with [A, r, B] = (1r)A + rB, O = 0 in V_O. Proof: Let a point O be given. We will show that V_O is then a vector space satisfying the conditions above.
A + B = 2 [A, 1/2, B] = 2 [B, 11/2, A] = 2 [B, 1/2, A] = B + A
(rs)A = [O, rs, A] = [O, r, [O, s, A]] = [O, r, sA] = r(sA)
(A+B)+C = C+(A+B) = 2 [C, 1/2, [O, 2, [A, 1/2, B]]] = 2 [O, 3/2, [C, 2/3, [A, 1/2, B]]] by THEOREM 6 = 2 [O, 3/2, [A, 2/3, [C, 1/2, B]]] by THEOREM 7 = 2 [A, 1/2, [O, 2, [C, 1/2, B]]] by THEOERM 6 = A+(C+B) = A+(B+C)
1 A = [O, 1, A] = A
A + O = 2 [A, 1/2, O] = 2 [O, 11/2, A] = 2 (1/2 A) = (2/2) A = 1 A = A
A + A = 2 [A, 1/2, [O, 1, A]] = 2 [A, 1/2, [A, 1(1), O]] = 2 [A, 1/2, [A, 2, O]] = 2 [A, 1, O] = 2 O = [O, 2, O] = O
(r+s)A = 2 (r+s)/2 A = 2 [O, (r+s)/2, A] = 2 [[O, r, A], 1/2, [O, s, A]] by THEOREM 5 = 2 [rA, 1/2, sA] = rA + sA
r(A+B) = r (2 [A, 1/2, B]) = (2r) [A, 1/2, B] = 2 (r [A, 1/2, B]) = 2 [O, r, [A, 1/2, B]] = 2 [[O, r, A], 1/2, [O, r, B]] by THEOREM 4 = 2 [rA, 1/2, rB] = rA + rB
[A, r, B] = 2 (1/2 [A, r, B]) = 2 [O, 1/2, [A, r, B]] = 2 [[O, 1r, A], 1/2, [O, r, B]] by THEOERM 8 = 2 [(1r)A, 1/2, rB] = (1r)A + rB
(10) Characterization of Banach Spaces. Using the results above, we can then provide an axiomitization for Banach spaces, which extends the axiom set for Metric Spaces.
DEFINITION: A Banach Space is a vector space with an operation .: V > R, v > v satisfying the properties A >= 0, with equality if and only if A = 0 A + B <= A + B
We will show that the following system of axioms completely characterise Banach spaces. The basic concepts here are the POINT, the DISTANCE between points with the distance from point A to B denoted by AB, and the mapping r > [A, r, B], which is assumed to define a unique point for all numbers r.
AXIOMS 14: The Metric Space Axioms: 1) AA = 0 2) AB > 0 unless A = B 3) AB = BA 4) AB <= AC + CB AXIOMS 57: The Vector Space Axioms: 5) [A, 0, B] = A 6) [A, 1, B] = B 7) [A, rt(1t), [B, s, C]] = [[A, rt(1s), B], t, [A, rs(1t), C]] AXIOM 8: If D = [A, r, B] and E = [A, r, C] then DE = r BC.
We extend the definition of V_O now to include a definition for the norm:
A = AO
This leads to the final result.
THEOREM 10: If V is a space satisfying AXIOMS 18, then for any point O in V, V_O is a Banach Space satisfying the properties: [A, r, B] = (1r)A + rB, AB = A  B Conversely, if B is a Banach space, then B satisfies AXIOMS 18 with the same definitions provided above. Proof: The converse is easy to prove since we only need to verify AXIOM 8. Supping that D = [A, r, B] and E = [A, r, C], then
DE = D  E = (1r)A + rB  (1r)A  rC = r(B  C) = r B  C = r BC
Given a space V, which satisfies AXIOMS 18, we will only need to prove the properties of the norm. First, we have
A = AO >= 0, with equality if and only if A = O. and rA = O [O, r, A] = [O, r, O] [O, r, A] = r OA = r A
To show that A + B <= A + B, let C = 2A, D = A + B and E = 2B. Then
D = A + B = 2 [A, 1/2, B] = [O, 2, [A, 1/2, B]] = [[O, 2, A], 1/2, [O, 2, B]] = [2A, 1/2, 2B] = [C, 1/2, E] and [C, 1/2, O] = [O, 1/2, C] = [O, 1/2, [O, 2, A]] = [O, 1, A] = A
Therefore, AD = 1/2 OE = 1/2 2B = B. The triangle inequality gives us
A + B = OA + AD <= OD = D = A + B
To prove the property AB = A  B, let C = A  B. Then A = C + B, and by a similar proof to that above we can show that BA = C. Therefore
AB = BA = C = A  B
The definition of an inner product space can also be recovered from this axiom set if we replace AXIOM 8 by the following.
AXIOM 8': If C = [A, r, B] then CD^2 = (1r) AD^2 + r BD^2  r(1r) AB^2
THEOREM 11: AXIOMS 17, 8' are equivalent to (AXIOMS 17 + THEOREM 10) of sections 1 and 2, both providing a characterization of inner product spaces. Proof: Since we derived the properties of vector spaces from AXIOMS 17 and THEOREM 10 of sections 12, then AXIOMS 17 as just stated above will follow. Also, AXIOM 8' was proven as a theorem in section 2. Conversely, suppose V is a space given by AXIOMS 17, 8' here. Then it's clear that AXIOMS 16 of section 1 are satisfied. The definition given for [A, r, B] is also consistent with this operator since we have:
A [A, r, B] = [A, r, A] [A, r, B] = r AB B [A, r, B] = [B, 1r, B] [B, 1r, A] = 1r BA = 1r AB
To prove AXIOM 7 of section 1, let A, B and C be points for which AB + BC = AC. If AC = 0, then we have trivially AB = BC = AB = 0, and A = B = C, from which it follows that
AB CD^2  AC BD^2 + BC AD^2 = 0 = AB AC BC
Otherwise, if AC > 0, then let r = AB/AC. Then 0 <= r <= 1 and we have AB = r AC = r AC and BC = (1r) AC = 1r AC. Therefore, B = [A, r, C]. By AXIOM 8' above, we then have:
AB CD^2  AC BD^2 + BC AD^2 = r AC CD^2  AC BD^2 + (1r) AC AD^2 = AC (r CD^2  BD^2 + (1r) AD^2) = AC (r(1r) AC^2) = AC r AC (1r) AC = AC AB BC = AB AC BC
It would also be illustrative to derive AXIOM 8 above directly from AXIOM 8':
If D = [A, r, B] and E = [A, r, C], then
DE^2 = (1r) AD^2 + r CD^2  r(1r) AC^2 AD^2 = (1r) AA^2 + r AB^2  r(1r) AB^2 = r^2 AB^2 CD^2 = (1r) AC^2 + r BC^2  r(1r) AB^2 Thus DE^2 = (1r) r^2 AB^2 + r(1r) AC^2 + r^2 BC^2  r^2(1r) AB^2  r(1r) AC^2 = r^2 BC^2 and DE = r BC
(11) Summary & Conclusion. Incorporating Minkowski Spaces To put all this in perspective, we list all the pertinent axioms and theorems below and classify the different kinds of spaces according to which ones are satisfied.
Axiom Set 1 (for Euclidean Spaces and other extensions of Metric Spaces): A function (A, B) > AB is given for which the following hold 1.1) AA = 0 1.2) AB > 0 unless A = B 1.3) AB = BA 1.4) AB <= AC + CB 1.5) If A and B are points and N = 2, 3, 4, ... then there is a point C such that AC + CB = AB = N AC (i.e. [A, 1/N, B] is defined for N = 2, 3, 4, ...) 1.6) If A and B are points and M = 2, 3, 4, ... then there is a point C such that AB + BC = AC = M AB (i.e., C = [A, M, B]). 1.7) If AB + BC = AC, then AB CD^2  AC BD^2 + BC AD^2 = AB AC BC 1.8) The Zeno CounterAxiom  every Zeno sequence has a Zeno limit Theorem 10, section 2: There is a unique point C = [A, r, B] defined such that AC = r AB and BC = 1  r AB Axiom Set 2 (for Vector Spaces, Banach Spaces and Inner Product Spaces): A function r > [A, r, B] is given for which the following hold. 2.1) [A, 0, B] = A 2.2) [A, 1, B] = B 2.3) [A, rt(1t), [B, s, C]] = [[A, rt(1s), B], t, [A, rs(1t), C]] Axiom Set 3 (requires 1.1  1.4 and 2.1  2.3): 3.1) If D = [A, r, B] and E = [A, r, C] then DE = r BD 3.1') If C = [A, r, B] then CD^2 = (1r) AD^2 + r BD^2  r(1r) AB^2
The results are summarised below.
Metric Space  Axioms 1.1  1.4 Close Metric Space  Axioms 1.1  1.4, 1.8 Inner Product Space  Axioms 1.1  1.4 plus either Axiom 1.7 and Theorem 10 or Axioms 2.1  2.3, 3.1' Hilbert Space  Axioms 1.1  1.4, 1.8 plus either Axioms 1.5  1.7 or Axiom 1.7 and Theorem 10 or Axioms 2.1  2.3, 3.1' Vector Space  Axioms 2.1  2.3 Banach Space  Axioms 1.1  1.4, 2.1  2.3, 3.1
A feeling of incompleteness may be haunting you if you are conversant with the Minkowski Geometry, which is used in Relativistic Physics. Often this is termed a "pseudoEuclidean geometry", in virtue of the modified Pythagorean theorem which holds therein.
It is possible to provide a deeper foundation which will yield all the (n+1) dimensional Minkowski spaces, and with respect to which the basic concepts of POINT and DISTANCE can even be defined! The formulation is decidedly different at the outset, as the basic concept used is time, not space. As it turns out, all the spatial properties can be proven solely in terms of the logic of time relations in a Minkowski space, so that (in this sense) space is a redundant concept!
This formulation may be presented in a future article, time permitting.

