"Jesse F. Hughes" <je…@phiwumbda.org> wrote ..

- — -

> Herc7 <ozd…@australia.edu> writes:

>> On Jun 7, 6:56 pm, "Kelpie" <kelp…@gmail.com> wrote:
>>> "A Little Bit" <ti…@beerlover.com.au> wrote in messagenews:0l9p0696j5dql6jcuh953h1c4qg8teo6n3@4ax.com…

>>> > Adriana Xenides Dead of Wheel of Fortune fame has died after an operation
>>> > in hospital. She
>>> > was only 54.

>>> I hope she’s turning letters in the sky.

>> Xen dies

> X denies.

>> it appears I *have* disproven higher infinity

> No, you haven’t.  The mysterious value X denies the validity of your
> "proof".

Fraid so, when I booted Ullrich out of my office due to Principal Transfer, the
reason I’m here, he forgot to take his little doggy with him, skit boy!

Funny nobody on sci.math will answer this question on higher infinity, on the same day
Xendies dies, a favorite Australian model who spent a career revealing what label was on the box!

>>   Given a set of labeled boxes containing numbers inside them,
>>   can you possibly find a box containing all the label numbers of boxes
>>   that don’t contain their own label number?

Have a go mate!

Then if you’re the first sci.mather to admit I disproved Cantor, Halt, Turing, and Godel
in front of his peers you can be my Assistant Principal.

Herc

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I’m addressing this post to people who like to solve basic problems
related to foundational questions.

Start with ZFC. Now let ZF0 consist of the restriction of ZFC to the
following axioms:

Extensionality
Pair-set
Unions
Subsets

Given this theory, we can define the concept of an ordinal, and prove
that the ordinals are linearly ordered, and, in fact, well ordered.
The definition I am thinking of is:

x is an ordinal iff x if epsilon-transitive, linearly ordered by
epsilon, and if for any subset y of x y contains an epsilon-least
element.

You can also define finite ordinals and prove induction (mathematical
or transfinite) and show how to do recursive definitions (on finite
ordinals or in the form of transfinite recursion). (Well, actually,
transfinite recursion probably needs the Axiom of Replacement, so just
say recursion on finite ordinals.)

Now, let ZF00 be ZF0 without the Axiom of Subsets. It is easy to show
that there are models for ZF00 in which the ordinals, as just defined,
are not linearly ordered.

Note however, that the null-set — 0 — is provable to be an element
of any ordinal.

Now, here’s the problem: within ZF00 can you add a clause to the
ordinal definition such that you restrict attention to the usual
finite ordinals 0,1,2… and maybe transfinite ordinals, such that all
these more restricted "ordinals" — call these "really really
ordinals" — are linearly ordered? Definition has to admit 0,1,2….

I say you can’t, but I haven’t proven that.

BTW, if you add an appropriate formulation of the Axiom of Foundation
to ZF00, I believe you are able to do this. But I’d like to do this
without just the blanket assumption that all sets are well-founded.

Here is an example of diagonalization

123
456
789

Diag = 159

AntiDiag = 260   <<<<<<<NEW SEQUENCE NOT ON THE LIST!

YOU ALL THINK THIS WORKS ON THE LIST OF COMPUTABLE REALS!

DON’T YOU!!!

Gee it works for 159, must work in the infinite case too, who cares if there’s
no new digit sequence that can be formed.

You’re all DIM!  How can you form a new digit sequence when they’re all
computed up to infinite length?  

Or as George Greene puts it, they’re all computed up to ALL (infinite) FINITE lengths.

And as George Greene puts it there’s a new digit sequence at some FINITE point.

Well I can’t see it.

Herc
—
the nonexistence of a box that contains the numbers of all the boxes
that don’t contain their own box number implies higher infinities.
– Cantor’s Proof (the holy grail of paradise in mathematics)

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The voting paradox …is a situation noted by the Marquis de Condorcet
in the late 18th century, in which collective preferences can be
cyclic …even if the preferences of individual voters are not. This
is paradoxical, because it means that majority wishes can be in
conflict with each other. When this occurs, it is because the
conflicting majorities are each made up of different groups of
individuals. For example, suppose we have three candidates, A, B and
C, and that there are three voters with preferences as follows
(candidates being listed in decreasing order of preference):

Voter 1: A B C
Voter 2: B C A
Voter 3: C A B

If C is chosen as the winner, it can be argued that B should win
instead, since two voters (1 and 2) prefer B to C and only one voter
(3) prefers C to B. However, by the same argument A is preferred to B,
and C is preferred to A, by a margin of two to one on each occasion.
The requirement of majority rule then provides no clear winner.

Also, if an election were held with the above three voters as the only
participants, nobody would win under majority rule, as it would result
in a three way tie with each candidate getting one vote. However,
Condorcet’s paradox illustrates that the person who can reduce
alternatives can essentially guide the election. For example, if Voter
1 and Voter 2 choose their preferred candidates (A and B
respectively), and if Voter 3 was willing to drop his vote for C, then
Voter 3 can choose between either A or B – and become the agenda-
setter.

When a Condorcet method is used to determine an election, a voting
paradox among the ballots can mean that the election has no Condorcet
winner. The several variants of the Condorcet method differ on how
they resolve such ambiguities when they arise to determine a winner.
Note that there is no fair and deterministic resolution to this
trivial example because each candidate is in an exactly symmetrical
situation.

The phrase "Voter’s Paradox" is sometimes used for the rational choice
theory prediction that voter turnout should be 0.

http://en.wikipedia.org/wiki/Voting_paradox
http://en.wikipedia.org/wiki/Voting_system#Foundations_of_voting_theory
http://www.google.com/search?hl=en&q=define%3ATransitivity

Hi,

Could anyone help me with this question (from Bostock’s Intermediate
Logic)

"The theory of groups can be presented as having in its vocabulary
just identity and a single two-place function f(x,y) which we write as
‘x.y’. The usual laws for identity apply, and in addition these three
axioms:"

(A1) Axyz(x.(y.z) = (x.y).z)

(A2) AxyEz(x = z.y)

(A3) AxyEz(x = y.z)

Prove:

a = a.c   |=   c = c.c

(Hint: Use Ez(c = z.a))

Thanks for any help,
Mitch.