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Topic: Mark's Elements, II
Replies: 0

 Mark Hopkins Posts: 51 Registered: 12/12/04
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 origin-independent 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 re-proof 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/(n-m) whenever m <= k < n

Define N = max { N_m, N_{m+1}, ..., N_{n-1} }. Then for all n >= N, we
have:
a_mn a_nn <= a_mn a_{m+1}n + ... + a_{n-1}n a_nn
<= (d_m + e/(n-m)) + ... + (d_{n-1} + e/(n-m))
= (d_m + ... + d_{n-1}) + e
<= d_m (1 + 1/2 + ... + 1/2^{n-1})) + 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 1-4 (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 1-4 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 5-7
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 1-4, 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 1-4, 7-8 and THEOREM 10,
or equivalently
AXIOMS 1-8.

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 (1-r) 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(1-t), [B, s, C]] = [[A, rt(1-s), B], t, [A, rs(1-t), C]]

These three properties are true in any vector space, where [A, r, B] is
defined as (1-r)A + rB. The third axiom reduces to the identity:

(1-rt(1-t)) A + rt(1-t)(1-s) B + rt(1-t)s C
= ((1 - rt(1-s))(1-t) + (1 - rs(1-t))t) A + rt(1-s)(1-t) B + rs(1-t)t C

which is true since

(1-t)(1 - rt(1-s)) + t(1 - rs(1-t))
= (1-t) + t - rt(1-s)(1-t) - rs(1-t)t
= 1 - rt(1-t)

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/(1-m), s = 0, t = 1-m 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/(1-m), s = 1 and t = m. Then

[A, m, [A, n, B]] = [[A, rt(1-s), A], t, [A, rs(1-t), B]]
= [A, rt(1-t), [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(1-n)). Then

[A, m, [B, n, C]] = [A, rt(1-t), [B, s, C]]
= [[A, rt(1-s), B], t, [A, rs(1-t), 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(1-t), [B, s, B]]
= [[A, rt(1-s), B], t, [A, rs(1-t), 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, 2-m, [A, 1/(2-m), 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(1-t)).
Then
[A, m, [B, 1/m, C]] = [A, rt(1-t), [B, s, C]]
= [[A, rt(1-s), B], t, [A, rs(1-t), C]]
= [[A, 1, B], 2-m, [A, 1/(2-m), C]]
= [B, 2-m, [A, 1/(2-m), C]]

THEOREM 7: If m is not 2, then
[A, 1/(2-m), [B, m, C]] = [B, 1/(2-m), [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 = (2-m)/(1-m), s = m and
t = 1/(2-m). Then

[A, 1/(2-m), [B, m, C]] = [A, rt(1-t), [B, s, C]]
= [[A, rt(1-s), B], t, [A, rs(1-t), C]]
= [[A, 1, B], 1/(2-m), [A, m, C]]
= [B, 1/(2-m), [A, m, C]]

THEOREM 8: [A, 1/2, [B, m, C]] = [[A, 1-m, B], 1/2, [A, m, C]]
Proof:
Take r = 2, s = m, t = 1/2. Then

[A, 1/2, [B, m, C]] = [A, rt(1-t), [B, s, C]]
= [[A, rt(1-s), B], t, [A, rs(1-t), C]]
= [[A, 1-m, 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 1-3 define a Vector Space with
[A, r, B] = (1-r)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, 1-1/2, 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, 1-1/2, 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, 1-r, A], 1/2, [O, r, B]] by THEOERM 8
= 2 [(1-r)A, 1/2, rB]
= (1-r)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 1-4: The Metric Space Axioms:
1) AA = 0
2) AB > 0 unless A = B
3) AB = BA
4) AB <= AC + CB
AXIOMS 5-7: The Vector Space Axioms:
5) [A, 0, B] = A
6) [A, 1, B] = B
7) [A, rt(1-t), [B, s, C]] = [[A, rt(1-s), B], t, [A, rs(1-t), 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 1-8, then for any point O in
V, V_O is a Banach Space satisfying the properties:
[A, r, B] = (1-r)A + rB, AB = |A - B|
Conversely, if B is a Banach space, then B satisfies AXIOMS 1-8
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| = |(1-r)A + rB - (1-r)A - rC|
= |r(B - C)|
= |r| |B - C| = |r| BC

Given a space V, which satisfies AXIOMS 1-8, 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 = (1-r) AD^2 + r BD^2 - r(1-r) AB^2

THEOREM 11: AXIOMS 1-7, 8' are equivalent to (AXIOMS 1-7 + 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 1-7 and
THEOREM 10 of sections 1-2, then AXIOMS 1-7 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 1-7, 8' here. Then
it's clear that AXIOMS 1-6 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, 1-r, B] [B, 1-r, A] = |1-r| BA = |1-r| 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 = (1-r) AC = |1-r| 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 + (1-r) AC AD^2
= AC (r CD^2 - BD^2 + (1-r) AD^2)
= AC (r(1-r) AC^2)
= AC r AC (1-r) 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 = (1-r) AD^2 + r CD^2 - r(1-r) AC^2
AD^2 = (1-r) AA^2 + r AB^2 - r(1-r) AB^2 = r^2 AB^2
CD^2 = (1-r) AC^2 + r BC^2 - r(1-r) AB^2
Thus
DE^2 = (1-r) r^2 AB^2 + r(1-r) AC^2 + r^2 BC^2 - r^2(1-r) AB^2
- r(1-r) 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 Counter-Axiom -- 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(1-t), [B, s, C]] = [[A, rt(1-s), B], t, [A, rs(1-t), 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 = (1-r) AD^2 + r BD^2 - r(1-r) 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 "pseudo-Euclidean 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.