Re: Subtleties with Cantor's diagonal.

[Graham Cooper]

On Nov 4, 8:48 am, George Greene <gree...@email.unc.edu> wrote:
> If you decide to look at it in the more limited way and ask simply
> about
> a set of n subsets of a set of cardinality n, the the theorem just
> says that
> FROM this set of n susbets, WE CAN CONSTRUCT AN n+1st SET,
> and that's ALL it says!
> THAT version DOESN'T say anything about the whole ...
>


We'll give you that much!

|PS( |n| )| > |n|



===========

INDUCTION:

p(1) & p(n)->p(n+1)
--> ALL(n) p(n)

SO: p(n) <-> |PS(|n|)|>|n|

===========

SUBSTITUTE |PS(|n|)|>|n| for p(n)

|PS(|1|)|>|1| & |PS(|n|)|>|n| -> |PS(|n+1|)|>|n+1|
--> ALL(n) |PS(|n|)|>|n|

===========

Once again Cantor's proof is not supported by proof by induction.


> we jump from this newly-constructed thing to
> the conclusion that "hyper-infinity exists". They are not willing to
> buy THAT because you could just tack this
> 1 new thing onto the head of your existing infinite list of subsets
> and get a list of subsets OF UNCHANGED,
> NOT HIGHER, cardinality.


Considering COMPLETENESS OF SETS of subsets
and COMPLETENESS OF SETS of reals
is the underlying issue you cannot prove incompleteness
for the base case with a process harness methodology.

MISSING SUBSET ELEMENT 1 = { 1, .. }
MISSING REAL DIGIT 1 = 0.4 ..

This information based on one case in meaningless.

When you SAY:

ROWn_n =/= ANTIDIAG_n
ROWn+1_n+1 =/= ANTIDIAG_n+1

AND SO ON....

It doesn't hold for 1 single case so you have proven p(n) -> p(n+1)
without p(1).

Herc

[Gary Forbis]

Would you just think rather than keep opening new threads on this matter
without thinking.

While Peano's original axiom was 1 was a natural number most use
the axiom that 0 is a natural number. That doesn't matter. Would
you just read the section at
http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
You go on and on about some stupid thing you know nothing about.

A countably infinite set has a one to one correspondence with
the set of natural numbers. Any sets of sets where the number
of elements in no set is more than countably infinite can be
mapped onto the set of natural numbers and has no more than
countably infinite elements.

I don't know why you keep introducing new symbols without
explaining how you are using them. I believe you are using
the pipe characters to represent the cardinality of the sets
What in the heck do you mean by:

SUBSTITUTE |PS(|n|)|>|n| for p(n)

But n isn't a set. n is a natural number so there's nothing
to stubstitute since there isn't type-type correspondence.

The set theoretic formation of PA defines 0 as corresponding
with the cardinality of the empty set. 0 isn't the empty set.
The set theoretic formation of PA defines the successor set
to a set corresponding to a natural number as the injection
of the set containing the set into the set then defines the
successor to the natural number as the cardinality of the
successor set.

Remember the axioms

0 is a natural number
every natural number has a successor that is a nautral number.
There is no natural number whose successor is 0.

Here's the set theoretic version.

{} has cardinality 0
the injection of the set {} into {} is {{}} and this set has cardinality 1.
the injection of the set {{}} into {{}} is {{},{{}}} and has cardinality 2.

Forget about powersets for a moment. Think about what is being done here.
This isn't about defining all sets that don't cotain themselves, its about
defining the natural numbers by way of the cardinality of sets.

0 corresponds with the set that has no elements and there is no set inside
the set that has no elements so this meets the axioms that 0 is a
natural number and that 0 has no successor. It doesn't really matter
that {{},{{}}} and {0,1} have the same cardinality. The important
part for the set theoretic version of PA is that there are regular
procedure that map onto the axioms. Now that you've introduce the
symbolic representation of the cardinality of a set n as |n| we can
write:

|{{},{{}}}| = |{0,1}|

Don't get sidetracked into the weeds. Don't get bogged down with
terminology. What makes 1 a natural number and 0.5 not a natural
number? It doesn't matter. PA doesn't talk about 0.5 but only
about the natural numbers. Think about cardinality of sets for
a moment. Can a set have a cardinality of 0.5? What would it mean
for a set to have cardinality of 0.5?

So here we are. The cardinality of {0.5} and {1} are the same.
We can biject these sets many ways. The easiest way to biject
two ordered sets is to index elements of the sets based upon the
natural numbers. I'll name the first set S and the second S'
I can identify the elements of S by their index and S' by their
index:
S[1] = 0.5
S'[1] = 1

The set of natural numbers {0,1,2,3,...}, where ... just mentions
the subsequent successors but doesn't name them, has 0 as the first
element. Most normal people want to declare the index as the
cardinal position of the element but computer programmers often
choose the positional map against the set of natural numbers. They
do this because they are concerned about the addressability of fixed
sized elements and by multiplying the index by the size of the elements
they get the offset from the base. I know you are fond of computers
but for our purpose I'd like to use the cardinal position of the element
in the set as the index. This means for the set of the natural numbers,
N, N[1] = 0. N is the name I've given to the set of natural numbers
and have identified the index as the cardinality of the set of natural
numbers less than the cardinality of the set.
N[2] = N[|{0,1}|] = 1
The ordered set of all natural numbers has a cardinality but since
it's an infinite set we can't identify it. The best we can do is
name it. We name it aleph 0. Aleph 0 is the cardinality of the
set of all natural numbers.

The easiest way to biject a set against the set of natural numbers
is to define a function or procedure that maps between the elements.
For instance when we want to count marbles we might have different
spacial regions for the marbles counted so far and the marbles that
haven't been counted yet. The marbles needn't be unique or ordered
for the procedure to work they merely need to be in the appropriate
region of counted and uncounted marbles. At the start there are
no marbles in the counted region, this is mapped to the first natural
number, 0. The set of marbles in the uncounted region starts with
no members. I call this set Cm. When the set of marbles Cm is
empty the cardinality is 0. The function mapping Cm to elements
in the set of natural number is n = N[|Cm|] and the set Cm at
evrey step has the property |Cm| = f(n). In order to talk about
the bijection of two sets logically I need to define the existential
quantifier function EXISTS, the universal quantifier function FORALL,
and the membership function MEMBER. There are common symbols for
these functions but they don't translate to ascii. While google
keeps the text in a character set that has the normal symbols usenet
is suppose to be transmitted in ascii. The arity of MEMBER is 2
where the first argument is the name of a set and the second the
name of an element. MEMBER(s,m) returns TRUE when m is a member of s.
I guess I need to define the function |- which means maps to

The bijection would go something like this like this

(NOT (EXISTS(x) (MEMBER(Cm,x))) |- 0
0 |- (NOT (EXISTS(x) (MEMBER(Cm,x)))
FORALL(x) EXISTS(y) (((MEMBER(N,x) AND (x > 0)) -> ((y |- x) AND MEMBER(Cm,y))
FORALL(x) EXISTS(y) ((MEMBER(Cm,x) -> ((y |- x) AND MEMBER(N,y))
FORALL(X) EXISTS(y) FORALL(z) (((x |- y) AND NOT (y = z)) -> NOT (x |- z))

During the procedure the counting goes:

x = N[|Cm|]

What if we had an infinite set of marbles to be counted?
The procedure would be the same. As long as the procedure
works we can count the marbles and the cardinality of the
infinite but countable set is aleph 0, just as we named it
before.

I want to name a set of sets that can be bijected with the
set of natural numbers. I name that set S. I'd also like
to name the set of cardinality of infinite sets Aleph.

Aleph[0]=|N|

This means
FORALL(x)(MEMBER(S,x) <-> (|x| = |N|))

So, here's the problem. We get into some goofy set theoretic
issues if we let S be a member of S. This alone should be
sufficient to say
NOT (|S| = |N|)
and that means Aleph has at least 2 members |N| and |S|, that
is unless S has a finite number of elements.

Since I can swap N[1], 0, with any other element of N it sure
seems like the cardinality of S is infinite but it isn't the
same as |N| so |S| > |N|.

Do you accept that there can be a set S whose members are
sets whose cardinality is the same as the set of natural numbers?
do you accept that there are infinite sets that aren't a member of S?