Re: Logarithm of transfinite numbers

Jonathan Hoyle said:
You are failing to see your own proof here. You have indeed
proven that sets of the form {1,2,...,n} are finite, for each and
every natural number n. That is what your induction proof
demonstrated. It does not however say anything about the
set {1,2,...}.

It says soemthign about each and every n in that set...

Actually, it does not. It speaks only of each of the subsets
{1,2,...n}. That's not the same thing.

It speaks of a property of the set defined by n. P(S(n)) may be considered a
compound property of n, as n is the only parameter to the property statement.


...which are all the points in the space of that set, that that
space is not infinite at any one of those points.

I think you are being a bit sloppy in your terminology, but I agree
with your principle that each and every subset {1,2,...,n} for all
possible n, is finite.

Sorry, I am admittedly a little sloppy.


True, but again, this applies only to sets of the form
{1,2,...,n}. Your induction proof says nothing about
unbounded sets.

It doesn't have to. It says that as the set increases
without finite bound the element values increase without
bound...

Agreed.

...but that as long as the element values are finite, the
set size is equally finite.

This is the part that is false. No such conclusion can be drawn from
the proof you outline.

Isn't it inductively provable that if we increment the max value of a finite
set and simultaneously increment the set size, if they are equal, they will
remain equal through any number of simulatneous increments? If this is the
case, can one achieve infinity while the other remains finite?


Acutally, there is a universal rule about this. If a set has
a minumum finite difference between any two elements,
or more generally, has a finite average difference between
successive elements, then it can only have a finite
number of elements in any finite range. That is, if no two
elements are infinitely different, then no two elements can
have an infinite number of elements between them...

Everything up to this part is correct.

Whew! :)


...and no subset of the set is infinite in that sense.

This is not true. The set { n, n+1, ... } is a subset, but is
infinite. Perhaps you meant to say "no subset of the form {1,2,...,n}
is infinite"? If so, then we are still in agreement.

Maybe why I say this becomes more apparent.....


If two elements are infinitely different, on the other hand,
one or another of them must have an infinite value, since
the difference between any two finite values is a finite
value.

This is correct reasoning. It is essentially a proof as to why any two
finite natural numbers is only a finite distance apart.

Right, and if there are only finite sequences within the set from any element
to any other, then where do infinite sequences in the set come from? To me, the
question is whether there are actually an infinite NUMBER of steps, and if
every element is a finite number of steps from every other, there can't be. If
we put this in spatial terms, can a 3D (for instance) space be infinite in
volume, if every point in that space is within a finite distance of every
other? I don't think so.


Now, a set may have an infinite number of elements in a
finite range, provided that is has at least one point of
condensation...

Correct. What you are loosely speaking of is more rigorously described
in the Bolzano-Weierstrass Theorem: Every bounded infinite set in R^n
has an accumulation point. However, this does not apply to N, the set
of natural numbers, since you cannot have a bounded infinite set in N.

It especially doesn't apply to the naturals because there is no such
accumulation point. The set can't achieve infinity that way, so it must be
infinite in extent, and at least from an internal standpoint, there is no
infinite difference or distance between elements within the set. I understand
that the standard theory has it that the size of this set must be larger than
any finite value, but this is based on the von Neumann ordinals, and I believe
it is a mistake to take that as a good model of the naturals. To consider an
element to be the successor of the max element of a set representing it leads
to the conclusion that the number representing all the finite naturals is
something bigger than any finite number. However, starting at 1 instead of 0,
we alternatively have a set where n is in the nth position, and unless n is
infinite, there is no infinite position in the set.


To say that each natural is the set of all its predecessors
says that this natural is one greater than the max value in the
equivalent set.

This is not true at all. It is true only of successors. It is not
true of limit values. The only limit natural number is 0, and as you
can see, 0 is not one greater than its set of predecessors. 0 is the
empty set, which has no predecessor.

Sure, that's the root case, the empty set. But at the other end of the
spectrum, we have the set of all finite naturals being equated with some number
which must be larger than all finite naturals, based on the finite case where
the "value" of the set is always greater than all values within the set. The
problem with this is that, in the finite case, this "value" is only one greater
than the max value of the set, and conceiving of omega this way produces all
the contradictions that arise from the assumption of any largest finite. This
is why I have to reject the von Neumann ordinals as a model for the naturals.


Is omega one greater than the maximum finite? Uh, no,
that would lead to a contradiction. There is a logical lapse there.

The logical lapse here is in the assumption that omega one is "one
greater than the max value in the equivalent set". And this is a lapse
you are making; it is not one that we are making.

It is based on the finite case where it is always true, and applied to the
infinite case, dropping the "1", so that omega is not "1 greater than the
maximum value in the set", but simply, "greater than all values in the set".
You get this result by starting at 0. If you start at 1, then you get the
opposite result. Does adding this extra element, 0, make the finite set
infinite? It can't.


You are in effect creating a straw man argument here: you construct an
argument that neither of us agrees with, tear it down, and then somehow
suggest that this undermines our original argument. If it helps any, I
will agree with your point that ordinals are not (in general) "one
greater than the max value in the equivalent set". I think though we
both can agree that it remains true of successor ordinals.

Well, concept of limit ordinals is based on this notion, that the value of the
set is greater than all values within the set, isn't it? It seems liek a
misapplication of the construction in the finite case to the infinite case, and
a near miss with direct self-contradiction.


If there is not an infinite number of elements between ANY
pair of elements, then how do you get an infinite number of
elements in the set?

Simple: by merely appending an infinite number of finite successors.
As agreed to earlier, each finite successor is only a finite distance
away from the origin. Doing this an infinite number of times, we get
an infinite number of elements in the set, each of which remains
finite.

If each successive element is one greater than the last, and you have an
infinite number of such successions, isn't that equivalent to performing an
infinite number of increments on a finite value? Doesn't that mean it is now an
infinite number of units away from that finite value, and now infinite in
value, being a finite plus an infinite?


By adding those two elements and making a finite set
infinite by the addition of two? No, that doesn't work.
Between any two distinct reals, there are an infinite
number of other reals. That's why any finite interval
denoted by such a pair of distinct reals contains an
actually infinite number of real numbers. If you think my
conclusion doesn't follow, please explain why, and leave
out aleph_0 if you can. It's not a number, or even a valid
concept.

I'm not sure what the relevance to real numbers are to this discussion.

I am just pointing out that, in any set I consider infinite, there are elements
which are infinitely far apart, in terms of the number of intermediate
elements.

I think the essential problem you are struggling with is the existence
of an infinite number of finite numbers.

There are an infinite number of finite numbers in [0,1], but not naturals. I
have never been able to accept that you can add 1 an infinite number of times
to a finite and not get an infinite value, and I don't ever expect to.


I think once you've
understood this, all your other issues go away. I know that you find
the concept to be counter-intuitive, but mathematics is not obligated
to follow your intuition. It is counter-intuitive to me that if you
folded a piece of paper in half, and did it again, and continued for 25
times (assuming the paper is long enough), your wad of paper would be a
mile thick at the end of the process. Some people find it
counter-intuitive that if you give a man a penny at the beginning of
the month, double it each day, he would be a millionaire at the end of
the month.

You said that months ago. That's about not being able to make a good guess
because you don't have an intuition developed for that kind of exponential
calculation. But, that's not the same thing here. there are basic mathemtical
ideas that contradict this notion. With the unit increments, one unit of value
for each element, there is an identity relationship between element count and
value which makes it impossible for the set to be infinite and the elements not
to be infinite, if infinite is a word used at all consistently. If the element
values can't become infinite by increments, then neither can the set size. They
are the same number as the set increases without bound.


If we let go of intuition and look merely at pure logic, the answer is
easy. It is a straight forward proof by contradiction. Here is the
proof: The set of natural numbers is either finite or it is not.
Suppose it is finite. Then it is of size n for some natural number n.
Yet, the set {1,2,...,n+1} is a subset of N and is larger than n. Thus
the assumption that N is finite leads us to a contradiction.
Therefore, N cannot be finite. N is thus infinite.

Right, that proof rests on the Dedekind definition of a finite set and one with
no injection into any of its proper subsets, and therefore means a set with a
distinct finite natural as its size. I would call that a finite and bounded
set. I don't see the set of finite naturals as finite, despite the fact that
it's unbounded, becuse there is not an infinite sequence within it. it's liek
the 3D space with all points finitely distant from each other. It's finite.


If we leave our emotion at the door, this proof is obvious, simple
enough for a middle school student to understand. It is only when we
have a "religious" axe to grind can something as trivially obvious as
this be disputed.

This isn't a matter of "religion". Sometimes it seems like that with those
defending the status quo in this area, and to be honest, the methods in
transfinite set theory seem rather like hocus pocus and not mathemtics to me.
This is a matter of finding consistency between different mathemtical
treatments of infinity, and finding that this theory doesn't play well with
others. Mathematicians seem to be willing to leave all intuition at the door,
but intuition is what guides creativity. Where it's strong, it shouldn't be
ignored. The nth natural is n, if you start at 1, so if there are aleph_0
naturals, the aleph_0th is aleph_0. Is that infinite?


Regards,

Jonathan Hoyle



--
Smiles,

Tony
.

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