Re: infinity
- From: "Jonathan Hoyle" <jonhoyle@xxxxxxx>
- Date: 22 Sep 2005 08:31:37 -0700
>> When I am talking about ordinal numbers, I talk
>> on meta-mathematic level. So I have ZFC as
>> meta-mathematic - because I wanna to speak
>> about On. The large numbers are out of my
>> interes now - in fact in standard analysis the
>> most useful cardinalities are all less then
>> Aleph_5 - and what I do is finitistic construction
>> of real numbers, so I work only with very small
>> ordinals. Do you working with
>> strongly-inaccessible ones? Do you?
When you speak of On (the class of all ordinals), I am assuming that
you were including *all* these ordinals, including the inaccessibles on
up. If this is not the case, then you can simply define your universe
of discourse as all the ordinals and cardinals below the first
inaccessible theta and be done with it. Do I typically need to work
with strongly inaccessible cardinals? No, not usually; however, I
certainly would be if I were dealing with the entire Class of Ordinals
as you claim to be doing. I think that some of the most interesting
work going on in Set Theory has to do with these large cardinals and
their implications on the rest of mathematics. You do not appear to be
contributing to this work though. Instead of On, it sounds like you
want to simply stop at sets in theta_0.
As for your claiming that the most useful cardinalities are below
Aleph_5, that is certainly not true (not at least without the
additional assumption of GCH). The cardinality of the reals, c, is not
required to be below Aleph_5. In fact, c could theoretically be
greater than the first weakly inaccessible. 2^c, the cardinality of
all functions on the real line, is greater still. 2^2^c, the
cardinality of the power set of all such functions, is staggeringly
larger. Unless you have a proof of GCH in your back pocket that you've
forgotten to mention, a limitation to Aleph_5 seems both arbitrary and
unnecessary.
>> GREAT! You are right. N belongs to N+1, but
>> there is also strictly one natural number,
>> which is N-1. In standard On this number
>> does not exist.
I don't see how you get around the fact that this is provably false.
What you call N, the set of natural numbers (which is usually called
omega when dealing with ordinals), is a limit ordinal, not a successor
ordinal. The ordinal omega, by definition, contains only finite
natural numbers, and by the Axiom of Choice, is provably the smallest
infinite ordinal. From the Axiom of Infinity, we know that if k is in
omega, then so is k+1. Therefore, if your N-1 were in N, then (N-1)+1
is also in N, which implies N is in N, which violates the Axiom of
Foundation. You clearly do not seem to be working within ZFC, and so I
ask again, what framework are you using?
>> But how many steps you get, if you have ONE
>> for each finite natural except natural 5 and 7...
Removing the integers 5 and 7 from omega still gives you a set with
cardinality aleph_0, if that is what you are asking.
>> where there will be only zero steps?
I don't understand what you mean here.
>> In standard math you obtain also Omega steps,
>> in theory of One-compressions you will get
>> natural number N-2. (Or Omega-2 if you wanna
>> to use this notation, but |N-2| < |N|)
I am not familiar with the "theory of One-compressions", but you appear
to be postulating the existence of infinite cardinalities between the
finite and aleph_0. This is provably false by the Axiom of Choice
(even by its weaker cousin the Axiom of Countable Choice). Without the
Axiom of Choice, you can have other set which do not contain a
denumerable subset, but its cardinality is not comparable to Aleph_0,
so you cannot claim this inequality. Furthermore, even without AC, N \
{ 5, 7 } still has cardinality Aleph_0.
>> In fact I believe, that whole modern Set theory is
>> controversy from the begining. I can proof it also
>> - but I am not able to proof it with finite formula
>> yet. All mathemations know, that Set Theory is
>> dis-proof-able, but they just believe in it. I don´t.
>> Do you?
I am not sure how to respond to this as it awkward both grammatically
and conceptually. I am not sure what it means to prove Set Theory is a
contraversy, with or without a finite formula. Do I "believe" in Set
Theory? I am functionally a Formalist with regard to this, and so I
believe that any given axiomatic system, ZFC, ZF + AD, whatever, play
the game by the rules and look at the interesting results.
The Platonist part of me does show up from time to time on certain
matters. For example, I believe that the statement "the 1 trillionth
digit of pi is 0" must be either true or false, regardless of what is
provable from any given axiomatic system. Certain undecidables seem
compelling to me, such as the Axiom of Choice (why shouldn't I be able
to pick out an element from each of a collection of sets?). On the
other hand, the Formalist in me comes out with the Parallel
Postulate...it's neither true nor false...if you assume it's true, you
get Euclidean geometry, if you don't, you get one of the non-Euclidean
geometries. Each is just as valid, neither being any more "true" than
the other.
In general though, my "belief" in Set Theory should be completely
independent of what is provable from it. If I wish to include a
certain "belief" in my proof, I suppose I could add it as an axiom to
the system and see if a contradiction arises. If one does, then I need
to rethink my assumptions. You might want to consider a similar
approach.
Regards,
Jonathan Hoyle
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