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Topic: 2.65 - Does bijection between infinite sets mean they have the
same number of elements?

Replies: 5   Last Post: Aug 21, 2014 3:51 AM

 Messages: [ Previous | Next ]
 Graham Cooper Posts: 4,495 Registered: 5/20/10
Re: 2.65 - Does bijection between infinite sets mean they have the
same number of elements?

Posted: Aug 20, 2014 3:27 PM

On Thursday, August 21, 2014 1:05:59 AM UTC+10, John Gabriel wrote:
> Of course not. Cantor was a deluded mythmatician.
>
>
>
> The applet at http://tube.geogebra.org/student/m145489 explains it all.
>
> A bijection between two infinite sets does not mean they have the same cardinality.

In order for these bad ideas to be true, it must be possible to have an infinite number of functions, but as the applet will show, in order to do this, ALL points in the given real interval must be well defined, that is, distinct and describable by a number.

****** LONG REPLY ON DEFINABLE SET THEORY OF REALS FOLLOWS ******

If you meant describable by a FINITE FORMULA then your assertion would be clearer.

Let's list all reals by listing all computer programs anyway!

UniversalTuringMachine( # , DigitPos ) = 0..9

Now we have a TERMINATING process to inspect any digit in the list.

"GIVEN ANY INFINITE LIST -THEN- CONSTRUCT A NEW REAL"
is a NON-TERMINATING PROCESS
which is why Cantor's Theory is part of ABSTRACT MATHEMATICS

GIVEN AN INFINITE LIST he has an algorithm
that runs forever and makes a new missing real.

Fair enough if you define a real as an infinite string of digits
where every digit position in N has a digit value in {0..9}.

Then any infinite string defined by computer must
also run forever.

Let UTM( n , pos ) = TM_n(pos)

Assume a Halt Function exists, then

UHM( h , pos ) = UTM( n , pos ) where n is the hth TM
where all inputs halt.

DUHM( h , pos ) = UHM( h , pos ) MOD 10

gives the computable reals list (with no missing values).

Each (halting) TM corresponds to 1 Real so

r1 = SUM ( DUHM( 1 , p ) * 1/(10^p) )

A Computable Real is defined as:
rn = SUM ( DUHM( n , p ) * 1/(10^p) )

--------------------------------------------

So given any rn, Virgil is right he can
calculate any digit of an infinite string
in such a way all digits mismatch all(n) rn.

LONG VERSION

Let's forget about Russell's Set for a moment
and allow the Naive Set theory definition

ALL(p):WFF

EXIST(S) A(X) XeS <-> p(X)

--------

Let the above formula = s.

Since s<->s
~(s<->~s)

its already possible to Logically Deduce that
~E(russell) A(X) X e Russell <-> X~eX

-----------------------

So although N.S.T. is inconsistent.

ALL(p):WFF

EXIST(S) A(X) XeS <-> p(X)

-----------------

the specification formula can work if p is stratified not to cause a

BETTER:

EXIST(p):WFF

EXIST(S) A(X) XeS <-> p(X)

----------------

an increasing set of WFF'
can be Logically constructed.

In essence

phi e WFF' <-> phi XOR ~phi

There is no deterministic transitive closure however since

F -> ~G
G -> ~F

either may be theorems, but not both.

(Well Spotted Jesse F Hughes)

You might have a disjoint set theory with several possible 'Universes'.

So given:

ALL(p):WFF' <-- notice the dash (wff stratification)
EXIST(S) A(X) XeS <-> p(X)

and some rudimentary Starting Axioms a large transitive closure
over inference can be calculated w.r.t. the base of starting axioms.

-----------------------

Now we have AVOIDED the pitfalls of ZFC Specification Axiom.

ZFC avoids Russell's set in a different way
by avoiding bi-implication in set specification.

ALL(p):WFF
EXIST(S) A(X) XeS <- p(X)

this is equivalent to ZFC Specification

XeS <-> p(X) & SCZ
XeS <-> p(X) & XeZ (usual form)

-------------

By Avoiding The Bi-Conditional a Class System is introduced.

EXIST(S) A(X) XeS <- p(X)

which is where never-ending bigger infinities seem to derive!

--------------

This is in direct opposition to Naive Set Theory where

EVERY MEMBER OF EVERY SET is DEFINABLE BY A PREDICATE

So: Axiom Of Choice was added :

A CHARACTERISTIC ELEMENT DEFINES A SET.

e.g.

midpoint( L : LINE )

if a midpoint exists, then so does a line!

-----------------------------

Or in Cantor's Case:

if an Anti-Diagonal Exists, then so does a SET OF THOSE REALS!

---------------------------

AOC: E(X) XeS -> E(S)

(Well executed Charlie Boo!)

********************************************

But A.O.C. only complicates matters further.

It seems to define uncountable many elements
with a specification of 1 element <=> 1 set.

THE GODEL NUMBER OF A CHOICE FUNCTION

2130415
a1(0,1)
MIDPOINT(0,1)

But now you have uncountable many choice functions!
One for each SET!

| N | = | GODEL NUMBERS |
| GODEL NUMBERS | = | FUNCTIONS |
| FUNCTIONS | = | CHOICE FUNCTIONS |
| CHOICE FUNCTIONS | = | ELEMENTS |
| ELEMENTS | = | SETS |
| SETS | > | N |

Once of these must be incorrect!

The only further resolution for Z.F.C. is
Uncountable Many Functions

| GODEL NUMBERS | < | FUNCTIONS |

Which means the WFF include Infinite Length Formula!

Such As:

1 e CS <-> 1 ~e SET1
AND
2 e CS <-> 2 ~e SET2
AND
3 e CS <-> 3 ~e SET3
AND
...

CANTORS MISSING SET FORMULA in its expanded term format.

You probably recognise the above as:

CANTORS SET = { n | n ~e f(n) }

---------------------------------

These are INFINITE SEQUENCES of CONDITIONAL STATEMENTS
not INFINITE SUMS of ARITHMETIC!

That is why the definition of a Real
that each individual digit over N digit positions is specified
is insufficient to define the LIMIT EXISTS

BETTER:

A LIMIT for A Computable Real
rn = SUM ( DUHM( n , p ) * 1/(10^p) )

-----------------------------------

Every rn in R is definable by a finite formula

R = ( n | SUM ( DUHM( n , p ) * 1/(10^p) ) }

EVERY ELEMENT is defined.

------------------------------------

SUMMARY:

for any constructed predicate Pn

NAIVE
Pn(X) <-> X e SETn

ZFC
Pn(X) <- X e SETn

PROLOG
Pn(X) -> X e SETn

In PROLOG Set Membership is a predicate goal to prove

e( 2 , evens ) ?

Set Specification defines which predicate determines set membership

e( X , evens ) <- even(X).

A predicate formalises each numerical function.

even( 0 ).
even( s(s(X)) ) <- even(X).

--
Have You Seen www.MUD.com/news Today?

Date Subject Author
8/20/14 Scott Berg
8/20/14 Tucsondrew@me.com
8/20/14 Graham Cooper
8/20/14 Dan Christensen
8/21/14 Graham Cooper