Re: Galileo's Paradox and the Project of the Reals

Tony Orlow wrote:
MoeBlee wrote:
Tony Orlow wrote:
Mike Kelly wrote:
Uh, no it doesn't. The axiom of infinity doesn't order anything.

It declares existence of an ordered set.

It is a CONSEQUENCE of the axiom of infinity that there are certain
sets with certain orderings. The axiom itself just says that there
exists an successor inductive set. The axiom itself does not declare as
to any orderings of such sets.

Is an inductive successor set ordered?

Do you mean:

For every successor inductive set, does there exist an ordering of it?

First, what KIND of ordering? Second, depending on the KIND of
ordering, I might or might not know the answer to the question.

If so, then you said nothing
different from what I said.

We are not saying the same thing. The axiom of infinity declares the
existence of a successor inductive set. The axiom of infinity does not
itself mention any ordering.

And nothing you've said refutes that the axiom of infinity can be
stated in the pure language of set theory whose only non-logical
primitive is 'e'.


Except that it does use a form of successor function, unless you have
some other interpretation of y'.

AGAIN, for the hundredth time, the successor operation is definable in
terms of 'e'.

http://mathworld.wolfram.com/AxiomofInfinity.html

Oh please, get a decent set theory textbook already!

Question : do you *seriously* think we're lying to you when we say it
can be done?

I just want to point it out when you employ successor or '<' in your
derivation. You're not using just 'e'.

We're using syntax for which there is an algorithm to revert any
formula (including those using 'successor' and those using '<') to the
pure language of set theory whose only non-logical primitive is 'e'.
You just don't get it, since you've never read a book on the subject.


You don't get the fact that your derivation of successor from 'e' relies
on the assumption of the successor function.

WRONG! We keep telling you, and you would see for YOURSELF if you just
read a textbook.

Please state the axiom of
infinity in English, and see how y' translates in your head.

The axiom of infinity is not in English and is not determined by "how
it translates in my head".

The axiom of infinty is usually given in a language for set theory that
is extended through definitions (including the defintion of 'successor
of') but ALL formulas of set theory (including the axiom of infinity)
can be formalized in the pure (no definitions used) language of set
theory with its one primitive predicate symbol 'e'.

Of course, you will never see that, since you won't read a textbook
that shows that it is true.

OK, I'm reviewing. Whoops, can't find the definition of ordinal
subtraction. Can you help?

Right. It doesn't exist. {0,1,2..}=omega, and {1,2,3...}=omega, with one
fewer element.

No. card({0,1,2..}) = omega = card({1,2,3...}).

What you posted is not the case. It is not the case that {1,2,3...} =
omega.

Huh! Somehow I thought omega was the union of all finite naturals.

And 0 is a member of omega, while {1 2 3 ...} doesn't have 0 as a
member, so {1 2 3 ...} is not omega.

Meanwhile, you still haven't given a coherent definition of ordinal
subtraction. (Actually, we can define ordinal subtraction (cf. Enderton
'Elements Of Set Theory'), but it's nothing like your ignorant
misunderstandings of set theory.)

So, you are saying that for xeR omega-x<omega? That's bull.

I'm not saying "xeR omega-x<omega", especially since I don't even
recognize it as a rendering of anything that is well formed. Try to say
what you want to say in recognizably coherent formulas.

MoeBlee

.

Relevant Pages

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