Re: infinitely many nn's = infinite nn's?

The Ghost In The Machine wrote:
I'm frankly not sure either of these are true, but
primarily for technical reasons. Basically, aleph_null
is not a member of N, more or less by accident.

I disagreed,
Wrong; that is ENTIRELY by design.

By who's design?

Peano's, eventually, though of course he had predecessors.

I suppose one could make a case that one
defined aleph_null = card(N),

YES, THAT IS IN FACT the definition.

in which case I'll concede the point.

Thank you.

The proof is simple enough:
Bad explanations have always been part of Phil's problem and
crap like this is NOT helping.

Phil isn't listening to you either, last I looked. :-)

I thank you very kindly; sincerely, humor is an
constructive response to insult and I am always
grateful when I do NOT hit back as hard as I hit.


assume aleph_null is in N;

NO, do NOT do that.
Assume, since the axioms already say so,
that a number is natural iff it is 0 or the successor
of a natural. Note that aleph-0 is NOT the successor
OF ANYthing (and obviously is not 0). Therefore it
is not a natural number. QED.

Well, personally, I'd approach it using the usual proof by
contradiction; assuming succ(n) = aleph_null,
then aleph_null is in N

Only if n is natural. The point is NOT that aleph_null
is not the successor of any natural: the point is that aleph_null
is not the successor OF ANYTHING, PERIOD.

and, assuming that N includes 0, card(N) = aleph_null + 1,

NO, IT DOESN'T! You are NOT doing correct cardinal
arithmetic here! Cardinally, aleph_null + 1 *==* aleph_null!
If you take any set of size aleph_null and add one more element
to it, the bigger element STILL HAS CARDINALITY aleph_null!

which contradicts aleph_null = card(N).

Except it doesn't, since cardinally, aleph_null + 1 = aleph_null
= card(N).

since card(N) = aleph_null one has the interesting issue
that aleph_null + 1 is also in N, (by virtue of axiom 2);
ergo, contradiction if one does it right.

Yes, that is the relevant contradiction; aleph_null + 1 would
have to be one of the things "counted" by aleph_null and
you would basically wind up with aleph_null>aleph_null+1.

(There's a technical issue or two there.)

Indeed, but it doesn't arise if you do it the other way.
You have to start with the right definition of a limit ordinal.

"Aleph-0" is from at least TWO REALMS HIGHER UP
than the naturals in any case.

Why two? Beth_0 = card(N), aleph_0 = beth_0.
One could make
a case either way AFAICT.

You have LEFT OUT the FIRST level up from the
naturals: you left out w (omega)! The limit ordinal of
the naturals was an ORDINAL *before* it was a cardinal
number! We define the cardinal numbers as *initial* ORDINAL
numbers. *w* is the FIRST level up. Aleph_0 (cardinal) is
the SECOND level up.

You can't even mention the
two of them in the same breath without violating Occam's
Razor. All ANYone needs to say about the naturals for
purposes of THIS discussion is that "the class of ALL of
them",

The class of all of what? The naturals?

Of course.

Say, what is a class in this context anyway?

Sigh. DEPENDS. We haven't ESTABLISHED
the context. One thing that will hold context-free
is that every set is a class.

Is it the same as a set?

No, but every set is a class.
SOME of this will hold EVEN if N is NOT a set.
N is indisputably a class, and every set is a class.

I wish I knew why Phil wants to distinguish between
potential and actual infinities. While there are contexts
where that might make sense (Mandelbrot sets),

No, really, there aren't. You concede entirely too much.

A Mandelbrot set is the set of all numbers in C such that
the sequence z_0 = z, z_{i+1} = z_i^2 + z does not tend to
infinity. Sounds pretty potential to me...

OK, you win, I lose, on that one.
But you had to go to analysis & limits to get there
(you are at 2nd-order, with infinite sequences of
reals). THIS IS A *FIRST*-order discussion!
DON'T go there again! You can talk to me about
countable sets of naturals and NOTHING bigger!

But "does not tend to infinity" is AMBIGUOUS in a
FASCINATING way here. If you were just to consider
real trig functions and give somebody 2 SAT math
questions saying "evaluate the limit :"
ln |x| as x->0
ln |cot x| as x-> 0, the classical answer would be
that both of these limits simply do not exist.
But the latter approaches both +inf and -inf
(from opposite sides), while the former approaches
-inf from both sides and this seems to cry out for
an "actual" (if not a "real") answer!
My point being that since -inf is not a real number, restricting
yourself to that context would allow you to say that the limit
does not exist, but the behavior in question is IDENTIFIABLY,
DESCRIBABLY, MORE regular than the dichotomous behavior.
You really DO want to be ABLE to say that ONE function
APPROACHES -inf, while the other function, for from approaching
NOthing, approaches TWO things (approaches an ordered pair,
since there are ALWAYS going to be two sides from which
you could approach any one real). Arguably, the ONLY
case in which you should be ALLOWED to say "the limit does
not exist" is the case of something like cos(1/x) as x->0, where
you get oscillation on BOTH sides no matter HOW close you get.

So there is some legitimacy for "potential" as a technical
term IF you use it carefully and IF we are talking about
infinite sequences OF RATIONALS (or reals or complexs)
AS OPPOSED to naturals. But at first-order, this is just
not reasonably entertainable.

NO infinity
is used in the definition of the natural numbers.

Certainly not explicitly. However, Phil apparently sees
a potential infinity in the fifth Peano axiom.

There is more than one way to phrase that axiom.
It quantifies over infinitely many properties but that
is NOT the point. That is a LINGUISTIC quantification
via an axiom-SCHEMA. It doesn't even COUNT. It is
quantifying OUTSIDE the framework of the theory.

Darned if I know precisely where, though it does use a double
quantifier and an additional quantifier on a property
or set, as opposed to a variable:

Only in the 2nd-order version.
You are CONFUSING PHIL when you talk like this!
BE PRECISE, dammit!


(AS)(S subset of N . 1 in S . (An)(n in S => succ(n) in S)
=> (S = N) )


That is JUST WRONG.
More to the point, it is set theoretical.
As soon as you say "in" and "subset", you NEED A LOT
MORE than just these 5 axioms. You ALSO NEED some
axioms saying what SET means (and what "in" and "subset"
mean). If you want to avoid importing 10 axioms (including
at least 1 more INFINITARY axiom-SCHEMA, of Replacement)
from ZFC, then you have to use 2nd-order logic for this.
You also have to start at 0 instead of 1, JUST BECAUSE I
SAID SO (well,not JUST because; I do have some good
reasons, the most important of them being that you do
actually want to be ABLE TO SAY that x+0=x and x*0=0,
SINCE THOSE ARE THE BASE-CASES of the recursions).

Apparently N is finite

No, it isn't; not even to Phil
is anything THAT ridiculous "apparent".

or potentially infinite,

Yes, that is apparently the adjective Phil wants to use for
any set of numbers that both a) doesn't have an upper bound
and b) doesn't have any actually infinite *members*.
But you can't let him. It violates Occam's razor. You do not
NEED to introduce "potential infinity" for those sets.
You can simply say about them that their NUMBER OF
elements is ACTUALLY infinite, but that each INDIVIDUAL
element IS finite. You DON'T NEED a third term.

But since
there ARE an infinite number of finite naturals, no infinity
NEEDS to be used to call an infinite number of finite things
into ABSTRACT existence.

As long as one can guarantee that the generator of such things
does not generate copies. In Peano, this is covered by
Axiom 2.

Right; that is one of the ones we have to keep, even in the
little 3-axiom version.

.

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