# [FOM] 724: Philosophical Geometry/9

Harvey Friedman hmflogic at gmail.com
Mon Oct 10 09:36:54 EDT 2016

```THIS POSTING IS SELF CONTAINED

We continue the development of Philosophical Geometry starting in one
dimension from http://www.cs.nyu.edu/pipermail/fom/2016-October/020109.html.

1. ONE DIMENSIONAL GEOMETRY
2. CONTINUUOUS BETWEEN FUNCTIONS
3. MIDPOINT FUNCTIONS
4. EQUIDISTANCE
5. EXTENDED CATEGORICITY
6. EXTENDED FIRST ORDER COMPLETENESS
7. BETWEENESS

1-6 has been covered in
http://www.cs.nyu.edu/pipermail/fom/2016-October/020109.html

We decided to change the notation a little bit. In sections 1-6 we
used E for the domain, which throughout is a closed line segment
(bottom edge of top sheet of pad of paper). We want this changed to D.
The reason is that in section 7 we want to use E in connection with
"endpoints". Also we are going to use "strict" to indicate
irreflexivity. Thus we work in sections 1-6 with strict dense linear
orderings with both endpoints, or sdlowbe.

NOTE: There was a typo in 4' of section 2. I wrote

B from  {(x,y): 0 < x < y < 1} into B

which should be

{(x,y): 0 < x < y < 1} into D.

where I have decided to change the domains in sections 1-6 from E to D.

7. BETWEENESS
still one dimension only
sections 1-6 were one dimension only

When contemplating the bottom edge of the top page of a paper pad, we
see left to right. However, if you are looking at this from somewhere
else in the room, my left to right might be viewed by you as your
right to left. So in sections 1-6 we were focused on what we see
together looking at the bottom edge of the top page of my pad of paper
from the same vantage point. But here we are focused on what EVERYBODY
sees.

Below we use the concept of a sdlowbe = strict dense linear ordering
with both endpoints in the following principles for betweenness.

1. Everybody sees points. We use D for the set of points.
2. Everybody sees the 3-ary relation BE(x,y,z) for "y is strictly
between x and z".
3. Everybody sees endpoints. We use E for the unary predicate "being
an endpoint". Everybody sees that E holds of exactly two points. Of
course, we have no notion of left and right endpoints, as not
everybody will agree on that.
4. Everybody sees that if v,w comprise the endpoints, then the
relation R given by BE(v,x,y) or BE(x,y,w) or (x = v and y = w) is an
sdlowbe with left endpoint v and right endpoint w, where BE(x,y,z) if
and only if (R(x,y) and R(y,z)) or (R(z,y) and R(y,x)).    .
5. For every nonempty set S of points, there is a minimal interval
(given by endpoints, possibly open, closed, or half open, possibly
with only zero or one element) which contains S.

I like the second order axiom 5 better than the second order axiom 5
used in sections 1-6. This same 5 here should replace the 5 used in
sections 1-6.

We could have avoided 3, and avoided any talking about endpoints
directly, and state 4 existentially - that there exist v,w such that
that relation R has the indicated properties. However, when discussing
what we or everybody sees, it is preferable to be more specific,
particularly when convenient.

Models of 1-4 are written as (D,BE,E).

THEOREM 7.1. Let (D,BE) be given. The following are equivalent.
i. There is a sdlowbe (D,<) such that BE(x,y,z) if and only if (x < y
< z or z < y < x).
ii. There is a (D,BE,E) satisfying 1-4.
If this holds then there are exactly two strict linear orderings (D,<)
in i, and they are reverses of each other. Furthermore There is a
unique (D,BE,E) satisfying 1-3, and for this unique E, the elements of
E are the endpoints of these two (D,<).

Proof: Assume ii, and let (D,BE,E) satisfy 1-4. By 4, i holds. Assume
i. Let (D,<) be a sdlowbe such that BE(x,y,z) if and only if (x < y <
z or z < y < x). Let E hold of exactly the endpoints of (D,<). We need
to verify 4. Let v,w comprise the endpoints of (D,<). Let R be given
by BE(v,x,y) or BE(x,y,w) or (x = v and y = w). We need to check that
R is an sdlowbe with left endpoint v and right endpoint w. We also
need to check that BE(x,y,z) if and only if (R(x,y) and R(y,z)) or
(R(z,y) and R(y,x)).

We rewrite the relation R(x,y) by

BE(v,x,y) or BE(x,y,w) or (x = v and y = w) iff
v < x < y or y < x < v or x < y < w or w < y < x or (x = v and y = w) iff
v < x < y or x < y < w or (x = v and y = w).

Irreflexity is clear of this relation of x,y. For transitivity,
suppose R(x,y) and R(y,z). Therefore

v < x < y or x < y < w or (x = v and y = w)
v < y < z or y < z < w or (y = v and z = w)

Suppose v < x < y. Then y < z, and so v < x < z. Hence R(x,z).

Suppose x < y < w. Then y < z and so x < z < w. Hence R(x,z).

Suppose x = v and y = w. This is impossible.

For connectivity, suppose x not= y. First assume x < y. If x = v then
clearly R(x,y). If v < x then clearly R(x,y). Now assume y < x. Then
R(y,x).

The left endpoint of R is v since R(x,v) is impossible. The right
endpoint of R is w since R(w,y) is impossible.

It remains to check that BE(x,y,z) iff (R(x,y) and R(y,z)) or (R(z,y)
and R(y,x)). I.e., that

(x < y < z or z < y < x) iff
(R(x,y) and R(y,z)) or (R(z,y) and R(y,x).

Suppose x < y < z. If x,z are not endpoints then R(x,y) and R(y,z). If
x = v and y not= w, then R(x,y) and R(y,z). If x not= v and y = w,
then R(x,y) and R(y,z). If x not= v and y not= w then R(x,y) and
R(y,z).

Suppose z < y < x. By the previous paragraph, R(z,y) and R(y,x). This
completes the proof of the equivalence of i,ii.

Now assume (D,BE,E) satisfies 1-4. Let (D,<) be a strict linear
ordering such that BE(x,y,z) if and only if (x < y < z or z < y < x).
We determine (D,<) uniquely according to whether v < w or w < v, where
E holds of exactly v,w.

case 1. v < w. Let x be not v,w. Then B(v,x,w), and so v < x < w or w
< x < v, and so v < x < w. Hence v is the left endpoint of < and w is
the right endpoint of <. Now let v < x,y < w. Then BE(v,x,y) if and
only if x < y.

case 2. w < v. We get the same < in case 1 except reversed.

Now suppose (D,BE,E) and (D,BE,E') satisfy 1-4, and we determine E as
follows. According to 4, if v,w comprise E and v',w' comprise E', then
there are sdlowbe, one with endpoints v < w, one with endpoints w < v,
one with endpoints v' < w', one with endpoints w' < v', satisfying i.
But we have shown that there are at most two such, with interchanged
endpoints. Hence {v,w} = {v',w'}, and E = E'. The last claim is by 4.
QED

THEOREM 7.2. Any two countable models of 1-4 are isomorphic. 1-4 is
consistent and complete.

Proof: Let countable (D,BE,E) and (D',BE',E') satisfy 1-4. By Theorem
7.1, let (D,<) and (D',<') be sdlowbe such that BE(x,y,z) iff (x < y <
z or z < y < x) and BE'(x,y,z) iff (x <' y < < z' or z <' y <' x) with
endpoints given by E and E' respectively. Let h be an isomorphism from
(D,<) onto (D',<'). Then h is an isomorphism from (D,BE) onto (D',BE')
because of these explicit definitions. Also h maps the endpoints of
(D,<) onto the endpoints of (D',<'), and hence E onto E'. QED

Now before moving forward with the discussion of system 1-6 and system
1-7 and various extension and variants, like we did in sections 1-6,
let's step back and talk about a distinguishing feature of the actual
betweenness relation on any sdlowbe (D,<,0,1), where it is understood
that 0,1 are the left and right endpoints.

We are looking for a relation R on the sdlowbe (D,<), of any number of
variables, that

*) has a definition over (D,<) without parameters from < which remains
the same if the definition is read with < replaced by >. (Can be seen
by all observers).

Betweenness is BE(x,y,z) if and only if x < y < z or x > y > z.

THEOREM 7.3. (characterizing betweenness). Let a sdlowbe (D,<,0,1) be
given. BE is the strongest relation satisfying *) in the following
sense. Every relation satisfying *) is definable in R,= without
parameters. BE is not definable from any binary relation, even with
parameters. BE is also powerful in the sense that  < is definable in
BE,= with any endpoint as the parameter, and any two distinct non
endpoints as parameters.

Proof: Let (D,<,0,1) be the sdlowbe, where of course we use < and
endpoints 0,1 for mathematical purposes, still being sensitive to the
philosophical points - all the while reasoning mathematically.

Clearly x < y iff (x not= 1 and y = 1) or BE(x,y,1). Also x < y iff (x
= 0 and y not= 0) or BE(0,x,y). Now fix non endpoints a < b. x < y if
and only if for all u, if u is an endpoint such that B(a,b,u), then
(B(x,y,u) or (y = w and x not= y)). We can do this without
quantifiers:
i. BE(x,y,a); or BE(x,y,b); or BE(x,a,y) or BE(x,b,y) or BE(a,x,y) or
BE(b,x,y); or
ii. x = a and y = b; or
iii. x = a and (B(a,y,b) or BE(a,b,y)); or
iv. x = b and BE(a,b,y).

Now let R be any relation satisfying *). Write R(x,y) iff phi(x,y).
Changing < to > in phi makes no change. From the above, there is a
formula psi in BE,=,x,y, allowing for two parameters a,b, such that if
a < b then the formula defines x < y, and if a > b then the formula
defines  x > y. So we can then define phi(x,y) in BE,= by asserting
that for any a not= b, either phi(x,y) holds with < defined according
to psi with parameters a,b or phi(x,y) holds with < defined according
to psi with parameters b,a.

Let R be a binary relation obeying *). Clearly R is symmetric. By
quantifier elimination, R is definable in 0,1,=. If BE is definable in
R,= without parameters, then BE is definable in 0,1,=. This is
impossible by quantifier elimination. QED

TO BE CONTINUED.

***********************************************
My website is at https://u.osu.edu/friedman.8/ and my youtube site is at
This is the 724th in a series of self contained numbered
postings to FOM covering a wide range of topics in f.o.m. The list of
previous numbered postings #1-699 can be found at

700: Large Cardinals and Continuations/14  8/1/16  11:01AM
701: Extending Functions/1  8/10/16  10:02AM
702: Large Cardinals and Continuations/15  8/22/16  9:22PM
703: Large Cardinals and Continuations/16  8/26/16  12:03AM
704: Large Cardinals and Continuations/17  8/31/16  12:55AM
705: Large Cardinals and Continuations/18  8/31/16  11:47PM
706: Second Incompleteness/1  7/5/16  2:03AM
707: Second Incompleteness/2  9/8/16  3:37PM
708: Second Incompleteness/3  9/11/16  10:33PM
709: Large Cardinals and Continuations/19  9/13/16 4:17AM
710: Large Cardinals and Continuations/20  9/14/16  1:27AM
700: Large Cardinals and Continuations/14  8/1/16  11:01AM
701: Extending Functions/1  8/10/16  10:02AM
702: Large Cardinals and Continuations/15  8/22/16  9:22PM
703: Large Cardinals and Continuations/16  8/26/16  12:03AM
704: Large Cardinals and Continuations/17  8/31/16  12:55AM
705: Large Cardinals and Continuations/18  8/31/16  11:47PM
706: Second Incompleteness/1  7/5/16  2:03AM
707: Second Incompleteness/2  9/8/16  3:37PM
708: Second Incompleteness/3  9/11/16  10:33PM
709: Large Cardinals and Continuations/19  9/13/16 4:17AM
710: Large Cardinals and Continuations/20  9/14/16  1:27AM
711: Large Cardinals and Continuations/21  9/18/16 10:42AM
712: PA Incompleteness/1  9/2316  1:20AM
713: Foundations of Geometry/1  9/24/16  2:09PM
714: Foundations of Geometry/2  9/25/16  10:26PM
715: Foundations of Geometry/3  9/27/16  1:08AM
716: Foundations of Geometry/4  9/27/16  10:25PM
717: Foundations of Geometry/5  9/30/16  12:16AM
718: Foundations of Geometry/6  101/16  12:19PM
719: Large Cardinals and Emulations/22
720: Foundations of Geometry/7  10/2/16  1:59PM
721: Large Cardinals and Emulations//23  10/4/16  2:35AM
722: Large Cardinals and Emulations/24  10/616  1:59AM
723: Philosophical Geometry/8  Oct 8 13:47:36 EDT 2016

Harvey Friedman
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