Re: Well Ordering the Reals

Tony Orlow wrote:
Matt Gutting said:
Tony Orlow wrote:
David R Tribble said:
David R Tribble said: 
But this is still much too short to be an uncountably infinite list,
and therefore it cannot possibly denumerate all the reals.
Tony Orlow wrote: 
Actually untrue. Consider this. We have a countably infinite number of
iterations starting with 0. We generate 2^n elements at iteration n. Therefore,
with aleph_0 such iterations, we generate 2^aleph_0-1 elements, an uncountably
infinite set. Is there a rule that says we must generate only one element at a
time? My set produces 2, so it grows like the power set.
David R Tribble said: 
It really is annoying that you don't have a clue about infinite series
or the actual meaning of Aleph_0.  You are confusing arithmetic
exponentiation and the cardinality of powersets - they are not the
same thing, you know.
Tony Orlow wrote: 
It's annoying that you can't just address the issue without an insult. It's not
my fault that your system doesn't work right. Exponentiation is exponentiation.

Heh.  No, ordinal exponentiation is not the same as cardinal
exponentiation.

It's provably true that Aleph_0^2 = Aleph_0, for instance.
That's patent nonsense. If it's provable in your system, the system sucks.
In your opinion. That's fine. But make it clear that your system is entirely different from the standard mathematical one. This might be made easier by introducing new terminology; or at least by making it clear that old terminology was not intended to

True. it is different fromt he standard system. I am using some different terminology, but the term "infinite" was not coined by Cantor. I think I may end up simply calling them "spaces" and elements "points", do emphasize the strong geometrical basis of my perspective. Right now, when I have moments of peace, I am boiling things down to primitives from more complex statements, as MoeBlee suggested last week, which seems liek a good complementary approach to trying to define the atoms of the theory first. We'll see how it goes.

If set theorists didn't want the power set confused with exponentiation, they
should have made upa new symbols, but I guess they had too many already.

Point taken.  The fact that cardinal exponentiation is close to ordinal
exponentiation is the probably the reason that mathematicians use
the same notation for both.  But the fact that cardinal exponentiation
is not identical to ordinal exponentiation can be a big source of
confusion, as you illustrate.  So yes, it might have been better if
they had chosen a different notation.
Actually, the best approach would be to have exponentiation, and if you have some other unrelated function, call it something else.
That's precisely what's being requested of you, as I understand it.
Yes, and that's why I coined Bigulosity, and the T-riffic numbers (as infinite naturals and reals). But, "infinite" is a common word, and not owned by set theory.
It's kind of like the
claim that cardinality=size, but then again, not really the same. One really needs to try to keep their terms to a minimum possible size for the problem at hand, not randomly generate lots of extra terms and special rules and exceptions. The power set is 2^n for set of size n, and that means actual, real, normal exponentiation. There is no need for special eponents for cardinals, and any apprent need for it is the result of a kludged system. 
The fact is, the power set, like the binary strings, DO comprise a set witha
relationship of 2^n elements for every n elements in the root set or n digits
in the binary string. For you, 2^aleph_0 doesn't really mean 2^aleph_0. For me,
it does. There is no reason to make the distinction.

Except that the math does not work right if you don't make that
distinction.  For us, 2^Aleph_0 is just a bigger cardinal than Aleph_0.
Then call it aleph_1, if you don't actually mean 2 to the aleph_0 power. What is that exponentiation doing there anyway, if you don't think exponentiation has anythign to do, really, with the relationship? The fact is, it does, and it should be there, and it should be interpreted as exponentiation in the usual sense. That works just fine, as long as you're not shying away from actual infinities.
They do "actually mean 2 to the aleph_0 power".
I don't remember who, but a couple people seemed to be saying that it doesn't really mean exponentiation in the normal sense. I am glad to see you don't have that opinion, but I am not sure about consensus on that matter. 
But it's provably true that 3^Aleph_0 = 2^Aleph_0, for instance, so
again it's obvious that cardinal exponentiation is not the same as
ordinal exponentiation.
That should not be true, although I think it is in standard theory. 3^n>2^n for n>0, and aleph_0 is certainly bigger than 0, is it not? These are precisely the types of expressions I want to see distinguished, not all lumped together as if arithmetic becomes meaningless for infinite values. It doesn't.
No, it doesn't. But it does become different. I don't see anything wrong with that. 
Why should 3^aleph_0 be equal to 2^aleph_0, when 3^n>2^n for all n>0? Is aleph_
0=0? If not, then those two are not equal.


3^n is not necessarily greater than 2^n for every number n, only for every *finite* number n. Since aleph_0 doesn't share this property of being finite, it's not necessarily true that "those two are not equal"

The infinite sum
  s = 1 + 2 + 4 + 8 + ...
is a sum composed of a countably infinite number of finite terms,
thus it can be said to be a countable (denumerable) infinite number,
of the same order as Aleph_0 (or omega).  Since each term
represents the number of points your mapping defines at each step,
we conclude that your mapping defines only a countably infinite
number of points in total.

And yet, if there are aleph_0 terms, this sum is one less than 2^aleph_0, and
this being the count of elements in the set, it would appear to have a
bijection with P(N) and to be an uncountable set.

Your cardinal math is incorrect.  The sum is Aleph_0 (omega), even
though there are Aleph_0 terms.  Infinite numbers don't behave like
finite numbers.
Yeah, so I've heard, over and over and over, here. It's nonsense to justify some rule with "why not?" "Infinite things need not behave like finite things" is no excuse for making up whatever foolish rules you want, just because no one will ever actually apply them to anything. If you have two infinite series, and one has tems that are always larger than the corresponding terms in the other, then the sum of the first is larger than the sum of the second. If you have sum
(n=1->oo: 1), it is smaller than sum(n=1->oo: n). The first is oo. The second is 2^oo-1. This is PROVABLE inductively.
*What* is provable inductively? According to my calculus book, the statement you offer about two infinite series apply only to series both of which converge. A series which does not converge, regardless of the value of its partial sums, is simply said to diverge. It's unfortunately the case that our notation for this uses the symbol for infinity; but infinity doesn't really enter into it.

But, indeed it does. The infinite series are actually inifite, in that the terms do not end. You are right that in standard math a series which does not converge to a specific finite value is simply said to diverge, but there are two forms of divergence: an infinite sum, or an oscillating sum. the sum of (-
1)^n oscillates. The sum of 1's is precisely N, or aleph_0. The harmonic series is obviously a smaller divergent series, since every term (except the initial 1) is smaller than 1, so the sum must be smaller. Now, you say this rule only applies to convergent series, such that if one series is known to converge, and another is less in every term, then it also converges. But, there is no reason not to be able to apply this logic to divergent series. After all, by the same token, if one series is known to diverge, and another is greater in every term, then it also is known to diverge. It is only because there is no established system of infinite numbers that these divergent sums cannot be compared, but I am not the only one who sees sum(n) as being larger than sum(1), which is larger than sum(1/n). 

I'm not familiar offhand with anyone else who sees things that way.

> sum(x=1->n:1) is obviously n. sum(x=1->n:x) is 2^n-1.
That's what's provable inductively. I would like to figure out what the sum of the harmonic series is. This may be one of the smallest infinities available. It's definitely smaller than aleph_0, in any case.

If it's provable inductively, then it's true for every finite number. It may or may not be true for infinite numbers such as aleph_0.

> Any system which denies the inequality
there is bonkers, which is the case with transfinite set theory.
Again, you appear to fail to understand that inductive proofs apply only to finite sets/numbers.

First of all, standard induction is defined to apply to the entire set of natural numbers, which is not considered a finite set, though you may say it applies only to finite values, since that's what the set contains.


No. Standard induction is defined to apply to *every element of* the entire set of natural numbers, not to the set itself.

Second, I do not fail to understand this concept. I disagree with it. Where you have an inductive proof of some kind of inequality based on a difference which has a limit of zero at n=oo, you cannot use such a proof to assert the inequality for the infinite case. This is true. However, I see no reason why, if proving a relationship of equality between two expressions, which holds for each successive iteration, why this should not hold for the infinite case. 

Because you haven't said anything about infinity at any point.

> Even
in cases where the proof is of an inequality, if it can be shown that the inequality does not disappear in the infinite case, the proof should be considered valid. The only justification for limiting such proofs to the finite case is based on the natural numbers which are considered finite, but then of course the inductive proof that the naturals are all finite is based on an assumption of the conclusion, and is therefore invalid.

So, if for every n, a string of n 1's in binary represents a value of 2^n-1, then I see no reason for this not to hold in the infinite case. It really doesn't matter what n is. If you think it does, can you give a real reason why this is wrong?

I'd certainly agree that this holds for every natural number n. But since there is no natural number q such that q is infinite, there isn't any "infinite case" at all for this expression.

So, in your mind there is a difference in cardinality between the power set of set S with size N, and the set of binary strings of length N? Funny how a difference of 1 can change an infinite cardinality, eh? You know, funny in a ha ha sense. More of a HAHAHA AAAHAHAHA HA HA HA HA WAHAHAHAAAAA Oh my god!!! kind of sense.

It is provably true that Aleph_0 - 1 = Aleph_0, for instance.
You can use that word, "provably", until it's the only word on the page. I don't care. I have made myself clear, have I not, that a system that "proves" such things is useless in my estimation?
Again, that's fine. You can hold anything you want in whatever esteem you like. That doesn't prevent others from being able to use it.
Sure, anyone can use it if they wish. They just can't claim that it is the be-
all and end-all of the topic of infinity, expecially when it is agreed that this pure mathematics is not beholden to any empirical evidence, or even agreement with any other treatments of infinity, and only needs to be self-
consistent. Sure, it can play with itself, but it can't throw all the other infinity-kids out of the playground.

It doesn't claim to. The claim is that this particular concept is useful for the kinds of things mathematicians tend to be interested in, and that your concept is not only not useful in such a sense, but self-contradictory to boot.

> It has no applications, can not be
tested, and fails to jibe with any other approach to these questions. I can "prove" that 1-1=1, too. But, as you are well aware, that proof involves a hidden division by zero. Your system includes such hidden problems that give rise to such faulty results.
To date, you haven't (as far as I've been aware) been able to point out anything that would be commonly considered a problem. That you consider it a problem is irrelevant.

Well, if I came up with a perfectly internally consistent system where 2=3 and 3<2 and also 2-3=3, and claimed you were wrong when you said 2<>3 and 2<3 and 2-3=-1, you would tell me I was wrong, at least in your system. So, I see problems in the system, not as far as internal consistency (although in some cases it's right on the line, in that respect), but as far as jibing with the rest of the world or any other understandings. What other theory concurs that you can cut a solid ball into a set of pieces and reassemble them into two solid balls the same size as the original? What other eveidence do you have that half of something is as large as the whole, if it has any size at all? How does one put 10 balls in a vase, take one out, repeat these two steps forever, and end up with an empty vase? You may not think such conclusions are a problem. In fact, it seems many mathematicians revel in the magically counterintuitive results. For my part, I say they are wrong. Don't get me wrong. Magic's a wonderful thing, but this is hocus pocus. Of course, that's just my opinion. And others'.

Which others? And as far as "jibing with the rest of the world", no statement about infinity can (as far was we know) jibe with the rest of the world, since we don't seem to be able to detect anything infinite. So the theory of infinities you're attempting to develop doesn't "jibe with the rest of the world" any better, or worse, than the usual one.


Your mapping function does not "generate" anything more than
a countably infinite number of points, which is far too small to
cover the reals.

Okay, then 2^aleph_0 is the size of a countable set, and Ross is correct in
saying that all infinite sets are equivalent and countable. If a countably
infinite set of countably infinite sets contains a countably infinite number of
elements in those sets, then the only thing uncountable is the continuum, not
the power sets of countaboly infinite sets.

You're confusing things.  Your mapping defines only Aleph_0 points,
no matter how wide the bifurcations are at each step.  Every point
you define can be denumerated by a natural number, so there are
at most Aleph_0 points defined.
Every level is denumerated by a natural number, but there are 2^n elements at level n. At level aleph_0 there are 2^aleph_0 elements. 
The continuum, c, is uncountable, and it is provably true that card(c)
and card(P(N)) are identical.  It is also provably true that P(N) is a
larger set than N.
Actually, it is not true that the power set of the naturals describes the continuum. I know in standard theory it is understood that way, but other approaches show that this is simply not true.
Not quite. Other approaches *may* show (I'm not familiar with any others) that the statement is not true *given that approach*. They don't show that the statement is not true.

Well, it all depends on the robustness of the approach, when evaluating contradictory conclusions. 

I'm not sure what you mean here.


> For any value range from 1 to x,
the number of elements of the form log2(n) for n in N is 2^(x-1), or 2^x/2. Over the entire value range of the real line, therefore, there are 2^aleph_0/2 log2's. Now picture the logs base 2 on the real line. Does that look like it includes every real number? Half of them? Not by a long shot. If N is the number of naturals, 2^N is not the number of reals in any real sense, at least outside of transfinite set theory.
Did you look at this part? Did it make sense? probably not, but this is the other approach I am talking about that makes 2^aleph_0 an unworthy choice as a representation of the continuum. If you want to map the naturals to the reals formulaically one thing we can do is use logN(x), but that leads to problems where we have N elements in [0,1], but instead of N^2 overall, we get N^N reals over the number line, which is inconsistent. The other mapping, which is essentially what Ross is advocating, is x/N, so that we have N infinitesimals within each unit. The inverse to f(x)=x/N is g(x)=xN, which over the entire real line will give N^2 reals, which works perfectly for N reals in each of N unit intervals. Ross's infinitesimals DO represent the proper mapping fucntion between the naturals and reals.

You see why we keep saying that cardinal exponentiation is not
the same as ordinal exponentiation?
I see that it's the case, but I don't see any justification for it. It's wrong, as far as I can tell.
What, precisely, makes something wrong?

That it makes no sense, adversely affects understanding, leads to false conclusions, is useless....you know, things like that. I mean, it's not "evil". It's just misguided and steers us into corners that don't serve us well. Exponentiation is exponentiation. Didn't you just say they actually DO mean "2 to the aleph_0 power"? The special treatment of ordinals and cardinals is, in my opinion, unnecessary. So, I guess I just mean it's not right.

If it's internally consistent, and doesn't contradict any observations from other areas of the same field, then I would certainly say that it makes sense. I believe this is true of the notion of infinity that's been expressed by the majority of posters here. Similarly, I don't see that it leads to false conclusions. Counter-intuitive ones, certainly; but a lot of things are true but counter-intuitive. As far as being useless - it certainly seems to have provided a lot of useful material for further mathematics.

As far as saying "What, precisely, makes something wrong? -- ...I guess I just mean it's not right": That's a tautology, and says nothing.

Matt
.

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