Re: Well Ordering the Reals

Tony Orlow wrote:
>> 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.
>

David R Tribble said:
>> 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.
>

Tony Orlow wrote:
> 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?

Because the Alephs are not defined using exponentiation or powersets.
So while it's true that Aleph_0 < Aleph_1, it's indeterminate whether
Aleph_1 = 2^Aleph_0.

The Beth cardinals, on the other hand, are defined that way.
By definition, Beth_1 = 2^Beth_0. I don't know why they are not
more popular.


David R Tribble said:
>> 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.
>

Tony Orlow wrote:
> 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.

Arithmetic for infinite ordinals and infinite cardinals is still
meaningful, it just doesn't conform to your intuitive expectations
about how it "should" be. Get used to disappointment.


Tony Orlow wrote:
>> 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.
>

David R Tribble said:
>> 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.
>

Tony Orlow wrote:
> 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.

These "foolish" rules are not "made up", but derived from the axioms
and rules of the system. You obviously don't comprehend how this
works.

The rules for cardinality can, and are, applied to solve abstract
problems of set theory, topology, etc. Abstract math was not invented
to solve real-world problems, and so what if it doesn't?


Tony Orlow wrote:
> 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.

Actually, it's provable that they are exactly equal. Which only shows
that intuition breaks down for infinite arithmetic.

I know you'd like to treat infinite numbers just like really really
big finite numbers, but that approach just doesn't work.


David R Tribble said:
>> It is provably true that Aleph_0 - 1 = Aleph_0, for instance.
>

Tony Orlow wrote:
> 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? 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.

Your first problem is that you can't accept that an infinite set is the
same size as any of its infinite proper subsets. Galileo recognized
this.

Your second problem is that you don't accept mathematical proofs.
If the result of a proof disagrees with your intuition, you say the
proof as faulty. You'd make a good creationist, but a terrible
logician.


David R Tribble said:
>> 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.
>

Tony Orlow wrote:
> 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.

Keep saying that, it might become true. Where is 3 in your list?


David R Tribble said:
>> 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.
>

Tony Orlow wrote:
> 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. 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.

You're saying that a bijection between P(N) and R does not,
in fact, exist. Obviously, these "other approaches" of yours must be
more correct. Of course, you're still of the opinion that (-oo,+oo)
has more points than [0,1], so maybe not. But feel free to post your
proof in more formal terms for all the world to see this amazing
discovery.

You might start by proving that x+1 > x if x is infinite.


Tony Orlow wrote:
>> Isn't c related to an ordinal?
>

David R Tribble said:
>> Since c is a cardinal, there must be a corrresponding ordinal for it.
>> However, there can be an infinitude of ordinals between two
>> cardinals.
>

Tony Orlow wrote:
> And we don't know what ordinal it corresponds to, how many are above or beneath
> it, whether there is anything between that cardinality and that of the
> naturals, or whether it is aleph_1. It's really hard for me to even fathom how
> anyone can be satisfied with this system, and why no one has suplanted it with
> something better yet.

We know that c = 2^Aleph_0 = 2^Beth_0 = Beth_1, if that's any help.

.

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