Re: Galileo's Paradox and the Project of the Reals

Tony Orlow wrote:
cbrown@xxxxxxxxxxxxxxxxx wrote:
Tony Orlow wrote:
cbrown@xxxxxxxxxxxxxxxxx wrote:
Tony Orlow wrote:

Okay, define me an infinite set which doesn't use successor or order in
the definition, or in the definition of something used in the definition.

The set of all lines in the Euclidean plane.
Define "line" without '<'.


That is an odd but easily satisfied request. "Line" is a primitive in
Euclidean geometry; as such it has no "definition" at all, with or
without '<'.

See:

http://en.wikipedia.org/wiki/Hilbert%27s_axioms#The_Axioms


To say there is no definition is hardly to satisfy a request for a
definition. Hilbert's axioms treat points, lines and planes as
"primitive", defining the relationships between them. But, if you want
to talk about the set of all lines in the Euclidean plane, then you need
to do a little better than that. Which of Hilbert's axioms do you intend
to use to calculate the size of that set?


You requested a set which was infinite, that didn't use the successor
function or '<' in its definition.

To see that the set of lines is infinite, we can proceed as follows:

By I.8, there exists at least 2 points.

Then by I.1 and I.2, there exists at least one line; and by I.3, there
exists at least one point not on that line.

The set of all lines must of course contain the above line, as well as
every line which passes through the above point.

If P is a point on the above line, then there is a line which passes
through the above point and P (again by I.1 and I.2).

Thus, if the set of points in the above line is not finite, then the
set of lines which pass through the above point is not finite; and as
this is obviously a subset of the set of /all/ lines, it follows that
then the set of /all/ lines is not finite.

By II.2 for any two points A and C on a line, there exists at least one
point B on that line which is distinct from A and C and between A and
C. From this and II.3, it follows that, given /any/ finite set of
points on a line, there is always a point on that line which is not in
that set of points.

Therefore, the set of /all/ points on any line is not a finite set.

Thus, the set of points on any line is infinite (i.e., is not finite);
and therefore the set of all lines is infinite (i.e., not finite).

Or:

The set of all triangles in the Euclidean plane.
Define "triangle" without "line".


Obviously, that would be silly. Presumably, you didn't realize that I
don't "need" '<' to "define" "line" in Euclidean geometry.


Euclidean geometry does not address the question we're considering.


When you asked for an infinite set, I assumed you meant an infinite set
of mathematical objects. Do you agree that the objects described by
Euclidean geometry are mathematical objects? Or did you have something
else in mind?

Neither of these sets has a "standard" ordering which allows us to say,
for any two elements a, b (lines or triangles) that exactly one of a <
b, a > b or a = b holds true.

Cheers - Chas

No, in the 2D plane, one needs to use something like a lexicographic
ordering by ordering the dimensions of the space, and then using the
order within each dimension.

Sure, one /can/ do something of the sort. But why does one "need" to
order these things at all?

In order to be able to express the set as a function of an independent
variable, "iterations".


Well, of course; but why does one "need" to the express the set as a
function of an independent variable, "iterations"?

How does such a thing in any way change the mathematical facts that can
be proven in Euclidean geometry, e.g., that the angle subtended by any
point on a circle and the points at each end of a diameter is congruent
to a right angle? Why do we "need" an ordering on the set of all lines
in order to ask and answer such questions, which have obvious
mathematical interest?


Anyway:

(i) You are talking about Analytic Geometry, in the Cartesian Plane.
The sets I defined are sets of elements from Euclidean Geometry, in the
Euclidean Plane. These are /different things/.

Points in the Euclidean plane are /not/ defined by some pair of real
numbers (x,y). They are primitives, just as lines are. Again, refer to:

http://en.wikipedia.org/wiki/Hilbert%27s_axioms#The_Axioms

where you will fail to find any reference to real numbers, or even
rational numbers.

Okay, then you simply say there are an uncountable set of points, and an
uncountable set of pairs of points determining lines, so you have an
uncountable set of lines. That's all very general.


Actually, I wouldn't go so far as to say the set was uncountable
(although I suspect it is by axioms V.1 and V.2) because I don't have a
proof in mind for that assertion.

On the other hand, I gave a /proof/ above that the set is certainly not
finite; and certainly not empty.

You can make a distinction between Euclidean and Cartesian definitions
of a line, but a line's a line.

No, actually a line is a thing which is specific to a particular
mathematical context. A line in Euclid's geometry is not a line in
Hyperbolic geometry: they have different properties.

It requires a long proof to show that the theorems of Euclidean
Geometry can be mapped one to one with theorems of Analytic Geometry.
Are you claiming that you accept this proof, despite the fact that it
relies on many, many assertions which you have also claimed are
"nonsense", "bogus", and (my favorite) "to poodles with it", amongst
other uncomplimentary things?

Unless Euclid offers some way to
calculate the number of points in the plane, that distinction doesn't
help much, does it?


Read the proof. There is no "calculation" involved; except in the sense
that every proof is a "calculation" in logic.


(ii) It is not in dispute that in general a set /can/ be ordered. The
Axiom of Choice for example implies that every set can be well-ordered
(which is a stronger statement than simply claiming a total order).

Right. All I am saying is, to generate an infinite set, to define it at
all, involves the use of SOME order.

Yes, that is what you have been saying; and you are WRONG when you say
that. That's why I provided a counterexample.

If you try to define what it means
for a line to be part of your set, you're going to have to order the
points to express how the unique lines are determined by them.


No, that is /not/ required in Euclidean geometry. Given two points
tweedle-dee and tweedle-dum, I don't "need" to know which of those two
points "comes first" in order to have defined a particular line in the
set of lines: its existence and uniqueness follow immediately from I.1
and I.2.


The point in dispute is your claim that there /must be/ some specific
"natural" order for /every/ infinite set, which is "inherent" in its
definition.

No, I am saying two different things,

First, in order to explicitly
define any infinite set one must employ some notion of order, whether in
terms of intermediate values or successors, which associates the
generation of a particular element with a particular iteration of
element generation.

And that is false (at least in mathematical terms).

We need only show that, given /any/ finite set of objects, we have the
/existence/ of /at least/ one object which is not in that finite set.
From this it follows that the totality of all such objects cannot be
found in /any/ finite set. Therefore the totality of such objects
cannot be a finite set.

(Of course, whether that totality is a set or not relies on other
assumptions of what we mean by "a set"; but generally, we allow for the
existence of infinite sets if the above is true).

Regardless of the fact that these objects /can/, in general, be ordered
in any number of ways, we simply don't "need" to have any ordering
whatsoever on the objects themselves to show that there cannot be
"only" a finite set of them..

Second, when dealing with subsets of the reals there
is a common quantitative order, which can be employed to compare the
sets over any given value range, including the entire real line.

But the set of Euclidean lines, or Euclidean triangles, is not a
"subset of the reals"; so your statement is not very general; and at
least for the mathematics I am interested in, not very useful.

Believe it or not, /most/ mathematical objects are /not/ subsets of the
real numbers; and that's why in general density measurements are
important but of necessarily limited use.

Those
are two different statements. I am not saying that every set has one
correct order, but that an ordering on a set can be used on its subsets
to provide a common yard stick between them.


There's nothing particularly controversial in that observation. But it
has also been observed that each such set can have many different
orderings; and under different orderings, your "common yardstick" and
my "common yardstick" will give different "measurements" for the same
subset of the same set.


This is false; as my examples show: given two lines A and B in the
Euclidean plane, there is /no inherent order/ in the primitives,
definitions and axioms which /forces/ us to say A < B or A > B. Any
such ordering is a completely arbitrary structure "tacked on" /after/
the definition of L = "the set of all lines in the Euclidean plane".

How do you distinguish one line from another?

By the points which are contained by those lines (see axioms I.1 and
I.2, which guarantee existence and uniqueness).

If they are elements of a
set, they must be distinguished from each other, but in the Euclidean
plane, they have no particular identity.


Erm??? Given non-colinear points A, B, and C, there is no problem
showing that the lines which pass through AB, BC, and AC exist and are
distinct.


(iii) This is a serious problem for your notions of "set size". If we
do not agree on the "natural" ordering of a countable infinite set,
then we will also in general /disagree/ on the "density measure" of
some proper subset of that set, "lim d(n)/n as n-> oo".

Yes, which is why your definition of the set doesn't lend itself to any
better estimation than "uncountable".

Well, so far I'm only claiming that I can show that it is /at least/
countable. It would indeed be a "better estimation" (in the sense that
it would provide more information) if I /could/ prove that it was
uncountable; but I can't at the moment.

If there is a generating process
with an independent variable counting iterations, it's a different
story, right?


Assuming we can agree on this "generating process", and guarantee that
this "generating process" actually "generates" the set in question,
maybe.

But the set of /all/ lines in the Euclidean plane are /not/ "generated"
in this way, and yet I would consider Euclidean geometry to be a rich
and interesting mathematical pursuit.


(iv) You seem to feel that the "natural" order of points in the
Cartesian plane is to order them lexicographically by their x and y
coordinates in R^2.

I prefer to order them lexicographically by their /polar/ coordinates
(theta, r); so that points near the origin come "before" points far
from the origin. I find this ordering more in keeping with the notion
of "distance".

Hmmm, well those aren't Cartesian coordinates, but that's okay. If you
want points close to the origin to come first, each point should be the
pair {r,theta}.

(theta, r) or (r, theta); it's somewhat irrelevant. Just as ordering
(x,y) lexicographically can be done in two ways (again with different
implications).



So I have a "smallest" element (the origin); whereas you do not. And
every interval in my ordering is bounded; whereas you have many
unbounded intervals (e.g., the interval ((0,0), (1,1)) is unbounded in
either x or y, depending on your lexicographic ordering).

Why do I need a smallest element? I just need to be able to compare any two.


I didn't claim you "needed" it; in fact I don't claim that we "need"
any such ordering at all. I am simply demonstrating that the /choice/
of "natural" ordering has implications for any attempt to compare
subset sizes via some notion of "as n-> oo".

As a result, the notion that "there are twice as many foos as there are
bars" is not a meaningful statement unless we /explicitly/ state
/which/ ordering we are using to determine this.

Equally, if I say "there are more foos than bars", it reasonable for
/you/ to ask "what on Earth do you mean by that"? If I respond "I mean
the cardinality of foos is greater than that of bars", then I have
identified the ordering I meant.

We were talking about lines, and now you're talking about intervals. I
don't mind unbounded intervals.


If there is a total ordering '<' , then for any points p, q in the
plane with p < q, we can talk about the interval (p, q); by which I
mean the set of all points {r: p < r < q). That is "an interval".

When we talk about the relative density "lim d(n)/s(n) as n->oo", we
mean to evaluate d(n) and s(n) as "the total number of "d"'s and "s"'s,
resp., in the interval (0,n)" (or some other suitable notion).


Suppose we want to compare the number of lines x = n for n an integer,
to the number of circles with integer radius centered at the origin.

Since each vertical line is tangent to exactly one circle and each
circle to exactly two verticals, I'd say there are twice as many
verticals as circles. That doesn't require any ordering, really, since
it's a linear relationship.

That's certainly one way of approaching the problem.

However, by that logic, you must also agree that there are twice as
many negative integers of the form 1 - 6*n or -1 - 6*n, as there are
(strictly) positive integers ; because for each positive integer n,
there are exactly two negative integers 1 - 6*n and -1 - 6*n, and for
each negative integer of the form 1 - 6*n or -1 - 6*n, there is exactly
one positive integer n.

Since the latter numbers are always odd, they form a subset of all odd
negative integers; so you are also claiming that there are /at least/
twice as many negative odd integers as there are positive integers.

I expect you will disagree with this result; but it uses the same
logical argument you described when claiming that there are twice as
many lines as circles.

Why should the argument work in one context, but not the other; unless
you're merely claiming that it "feels" right in one context and "feels"
wrong in the other?



If we use my ordering, the number of such lines contained in the
interval (0,0), (theta, r) is always 0; and the number of circles is
floor(r). So I find that that there are more circles than lines, in the
limit.

You mean completely contained within r of the origin? Of course not, but
the portion of the line including the point of tangency is contained,
and the correspondence apparent.

If we are to include /partial/ containment, then any interval in your
ordering certainly intersects an /infinite/ number of circles. So the
number of circles must be always infinitely greater than the number of
lines, and (according to you) this must hold "as n->oo".


If we use your ordering, the number of such lines contained in the
interval (-x, -y), (x, y) is always 2*floor(x) + 1; and the number of
circles is floor(x). So you find that that there are more lines than
circles, in the limit.

I guess the circles would be floor(x)+1, including the circle of radius
0 at the origin. Forgot about that one. Yes, there are more lines than
circles. In any case, none of the vertical lines is completely contained
in that interval either, so I don't see the distinction here.


In your ordering, I assume that (x,y) > (u,v) iff x > u or x=u and y >
v.

What is the set of points, in this ordering, which are greater than
(0,0) and less than (2,0)? Does it not at least include all points of
the form (1, y) for all real y? That /is/ the vertical line x = 1.

Cheers - Chas

.

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