Re: Orlow cardinality question

Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> David Kastrup said:
>> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:
>>
>> > David Kastrup said:
>> >> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:
>> >>
>> >> > Actually, I have no problem with the naturals going on forever,
>> >> > as long as they achieve infinite values, which I have no problem
>> >> > with either.
>> >>
>> >> Unfortunately, the naturals have a problem achieving infinite
>> >> values. They can't. While every possible limit gets exceeded by
>> >> some naturals, there is no single natural that could exceed all
>> >> limits.
>> >
>> > Then one can say the same about set size, which is incremented in
>> > exactly the same steps as the maximal value.
>>
>> Rubbish. The set of natural numbers is defined by five axioms, not by
>> somebody putting numbers into a bag. The size of the sets of naturals
>> is not "incremented", it just is.
> it's defined recursively with a base case and each element defined with respect
> to its predecessor. It's incremented.
>>
>> > The set of all finite naturals is finite, indeterminate as its size
>> > may be.
>>
>> Stomping your feet and sulking is not going to make it so.
> No, the restriction of finiteness does that all on its own.
>>
>> >> > I really don't see what problems that introduces. If you want to
>> >> > say the series of finite naturals has no end because the end is
>> >> > not identifiable, then that is okay, but it is of the "boundless"
>> >> > variety, and not infinite.
>> >>
>> >> "Infinite" as a set measure is defined as the ability to place a
>> >> set into bijection with a proper subset. The successor relation
>> >> does just that with the naturals. So the set is infinite.
>>
>> > That's one way to look at it. So? What does having infinite whole
>> > numbers break in your world?
>>
>> The fifth Peano axiom. Any set containing 0 and including for each of
>> its numbers its successors, contains _all_ elements of the naturals.
>> Since the successor operation will give a finite number from a finite
>> number, the induction axiom strictly rules out infinite numbers in the
>> set of naturals.
> Not after the infinite number of increments required to produce the infinite
> set.

The set is not "produced". There is no time involved.

>>
>> Now if you want to, you can call the set size of the naturals a
>> "number", or you can call it nonexistent. But what you _can't_ call
>> it is a natural number. That name is already taken for the set of
>> numbers defined by the five Peano axioms, and the size of this set
>> can't be a member of this set itself, because it does not obey the
>> axioms.
> Yes it does. I have already shown how the size of the set IS the maximal
> element. Same thing, a natural number, which is used to count things, like
> natural numbers. Natural numbers are things used to count things, like natural
> numbers, which are things used to count things, like natural numbers.........

So you admit that your theory is fundamentally inconsistent.
N is a natural number according to you, but N+1 is not, in
direct contradiction with the definition of the natural numbers.

I wonder if you think N-1 is a natural number or not?
Given your totally arbitrary system, who knows?

<snip>

>>
>> But whatever such numbers are, however useful or
>> enlightening they might be: one thing they aren't by definition:
>> members of the set of naturals.
> The size of the set of naturals is in the set.

And therefore N+1 is in the set.

>>
>> The axioms for that set leave no place for such a number designating
>> the size of the set _as_ _a_ _member_ of the set.
> Poppy***!

Explain how N can be in the set of natural numbers but N+1
is not? The axioms clearly state that if x is a natural number,
then x+1 is a natural number. It does not get much simpler
than that.

Stephen
.

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