Proper class.Proper class ?
zuhair
If A is a proper class, and B is a proper class , and A.B={ x| x in A,
x in B } ,
is A.B a set , or a proper class, or ( a set xor a proper class ) ,
or ( ~a set /\ ~ a proper class) ?
Zuhair
Jonathan Hoyle
I believe it's easy enough to show both possibilities:
Let A and B be proper classes as you have described, and let . be the
class intersection operator as you have defined them. For this
example, let us let 0 be the empty set, U be the class of successor
ordinals, while V be the class of limit ordinals.
When A=U, B=U, then A.B=U, a proper class
When A=U, B=V, then A.B=0, a set
Hope that helps,
Jonathan Hoyle
Aatu Koskensilta
It depends on A and B.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
zuhair
Ok, let me frame my question in a more proper way.
Let A be a proper class , Let B be a proper class : A.B = { x| x in A,
x in B } =/={ } and A.B=/=A and A.B=/=B. . What is A.B? is it a set , a
proper class, or ( a set xor a proper class) or (~ a set and ~ a proper
class).
In other words for proper classes A and B , that are not disjoint , nor
one of them subclass's the other , what is A.B ? is it a set, a proper
class, (a set xor a proper class), or (~ a set and ~ a proper class ).
Zuhair
george
zuhair wrote:
> Ok, let me frame my question in a more proper way.
>
> Let A be a proper class , Let B be a proper class : A.B = { x| x in A,
> x in B } =/={ } and A.B=/=A and A.B=/=B. . What is A.B?
It STILL *depends* on A and B. You are NOT going to
get a general law here.
> is it a set ,
sometimes.
> a proper class,
sometimes.
> or ( a set xor a proper class)
Under some simple class theories, ALWAYS, because
EVERYthing is a class, and is in turn proper xor a set.
> or (~ a set and ~ a proper class).
Again, under simpler more introductory class theories,
everything is a class (and therefore either a set or proper),
so this can never happen.
> In other words for proper classes A and B , that are not disjoint,
If their intersection is a set, then they might as well be disjoint,
for purposes of this argument. The intersection is "small" compared
to the two parent classes. Obviously, if you start with 2 disjoint
proper
classes A and B, you can just union a set c (also disjoint) to both of
them,
and wind up with 2 non-disjoint proper classes whose intersection is
that set. A u c . B u c = c. So clearly, the result of intersecting
2 proper
classes can be a set.
> nor one of them subclass's the other , what is A.B ? is it a set, a proper
> class,
Well, there are contexts in which those are the only two
possibilities, so, yes, always, it could be a set (as described
above), but if it isn't, then it must be a proper class. E.g.,
The intersection of the set of multiples of 2 with the set of multiples
of 3 would be the set of multiples of 6 (those may look more like
sets than proper classes to you, but there is a context in which
all infinite classes of finite numbers are proper).
zuhair
Ok, let me further define A and B in an exact way.
A is the class of all sets that subsets their power sets.
A= { x|x is a subset of P(x) }
B is the class of all sets that are bijectable to set S.
B={ x| x is bijectable to S }
It is clear that both A and B are proper classes.
A.B = {x| x is a subset of P(x) /\ x is bijectable to S }
Now what is A.B? is it a set? a proper class? either? neither?
Zuhair
Jonathan Hoyle
> one of them subclass's the other , what is A.B ? is it a set, a proper
> class, (a set xor a proper class), or (~ a set and ~ a proper class ).
Still easy. Let C be any proper class, and let x, y be any distinct
elements not in C.
If A = C U { x }, B = C U { y }, then A.B = C, a proper class. (Note
that A and B are not disjoint, nor is one a subclass of the other.)
Now, let C1 and C2 be any two disjoint proper classes, and x any
element not in C1 or C2.
If A = C1 U { x }, B = C2 U { x }, then A.B = { x }, a set. (Note that
A and B are not disjoint, nor is one a subclass of the other.)
Hope that helps.
zuhair
still my question about what is A.B? when A is the set of all sets that
subsets their power sets, and B is the set of all sets that has
bijection to S is till now not solved.
Zuhair
zuhair
for further clarification see the following:
Charlie-Boo
> Under some simple class theories, ALWAYS, because
> EVERYthing is a class
How about the classes that are not elements of themselves - is that a
class? Likewise for those loons who say everything is a set - how
about the sets that are not elements of themself - is that a set?
C-B
(author of CBL which answers all of these questions formally and
automatically)
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
> for further clarification see the following:
>
> A is the class of all sets that subsets their power sets.
>
> A= { x|x is a subset of P(x) }
>
> B is the class of all sets that are bijectable to set S.
>
> B={ x| x is bijectable to S }
>
> It is clear that both A and B are proper classes.
Not necessarily. Suppose S = {}.
And of course, unless one is in some system which allows proper classes,
one cannot have proper classes,
>
> A.B = {x| x is a subset of P(x) /\ x is bijectable to S }
If S is bijectable to some ordinal, isn't A.B just the set whose only
member is that ordinal?
zuhair
Virgil wrote:
> In article <1163118984....@m7g2000cwm.googlegroups.com>,
> "zuhair" <zalj...@yahoo.com> wrote:
>
> > for further clarification see the following:
> >
> > A is the class of all sets that subsets their power sets.
> >
> > A= { x|x is a subset of P(x) }
> >
> > B is the class of all sets that are bijectable to set S.
> >
> > B={ x| x is bijectable to S }
> >
> > It is clear that both A and B are proper classes.
>
>
> Not necessarily. Suppose S = {}.
Yea, you are right! but my question is when S=/={}.
so I should rephrase B={x|x is bijectable to S, S=/={ } }
Zuhair
zuhair
Virgil wrote:
> In article <1163118984....@m7g2000cwm.googlegroups.com>,
> "zuhair" <zalj...@yahoo.com> wrote:
>
> > for further clarification see the following:
> >
> > A is the class of all sets that subsets their power sets.
> >
> > A= { x|x is a subset of P(x) }
> >
> > B is the class of all sets that are bijectable to set S.
> >
> > B={ x| x is bijectable to S }
> >
> > It is clear that both A and B are proper classes.
>
>
> Not necessarily. Suppose S = {}.
Yea, you are right! but my question is when S=/={}.
so I should rephrase B={x|x is bijectable to S, S=/={ } }
Zuhair
Jonathan Hoyle
> subsets their power sets, and B is the set of all sets that has
> bijection to S is till now not solved.
First of all, the class of all sets is a proper class, not a set.
Secondly, what do you mean by the predicate "that subsets their
powersets"? And what is this S that you speak of in the second
sentence? You have undefined (or at least badly defined) terms here.
In any case, I believe the question you asked was answered definitely:
The intersection of two non-disjoint proper classes, where one is not a
subclass of the other, can either result in a set or another proper
class.
Jonathan Hoyle
Eastman Kodak
zuhair
S is any set.
Second I was speaking of the class of all sets that are subsets of
their power sets.
or in other language , the class of all transitive sets.
x is a transitive set <-> x:Amex,Anem ->nex
You can google for "transitive set" to see what it mean.
Every transitive set is a subset of its power set.
I think that the class of all transitive sets is a Proper class.
Now Let S be { { } , { {} } , { { {} } } }
Let A = { x| x is a transitive set }
Let B = { x| x has bijection to S}
Now what is A.B? is it a set , a proper class, both, neither
My question is very clear from the begining and I am astonished that
you didn't understand what I meant.
Anyhow it is not answered yet?
Zuhair
Jonathan Hoyle
>
> Second I was speaking of the class of all sets that are subsets of
> their power sets.
>
> or in other language , the class of all transitive sets.
>
> x is a transitive set <-> x:Amex,Anem ->nex
>
> You can google for "transitive set" to see what it mean.
>
> Every transitive set is a subset of its power set.
Ah, I see what you are getting at. Yes, I know what a transitive set
is, but thank you for explaining what you mean.
> I think that the class of all transitive sets is a Proper class.
It is.
> Now Let S be { { } , { {} } , { { {} } } }
>
> Let A = { x| x is a transitive set }
> Let B = { x| x has bijection to S}
>
> Now what is A.B? is it a set , a proper class, both, neither
If I understand this correctly, S is a three element set, so B is
essentially the class of all sets with cardinality 3 (which includes
S). We also note that since S is transitive, S is thus also an
element of A. Therefore A.B includes S as an element.
In fact, A.B is appears to be nothing more than the class of all
transitive sets of cardinality 3. Assuming no urelements in your Set
Theory, I believe A.B should may be a set. Another member of A.B would
be the set S1 = { {}, {{}}, {{}, {{}}} }. Without urelements, there
really can't be too many others to push A.B into proper classdom (or so
it seems to me).
> My question is very clear from the begining and I am astonished that
> you didn't understand what I meant.
Actually not. Your question from the beginning had to do with
arbitrary proper classes. But I see your question now.
> Anyhow it is not answered yet?
It should be now.
Jonathan Hoyle
Eastman Kodak
zuhair
Ok, you answered it when S has cardinality 3, but I want the answer for
S were S is any set other than { } . would A.B be a set .
More specifically A.B is the class of all transitive sets that has
bijection to S, were S is a set other than { }. Is A.B a set for each
S.
Zuhair
Ross A. Finlayson
Classes are non-sets. Non-sets in a set theory are non-sense.
There can be only one proper class or none, in a set theory it would be
a set.
There is no class of classes in set theory with classes.
There is no universe in ZF, ZF is inconsistent.
Ross
zuhair
ZFC is inconsistent because it doesn't have a universe, that's why I am
trying to develop a set theorum having a universe.
In reality the whole idea of axiomatization is silly really. Since e
is not defined and a set is not defined then the whole set theory
mounts to nothing. And those people seems to think that they can
produce things from nothing, which is philosophically absurd.
Zuhair
Ross A. Finlayson
Actually, to produce things from nothing is the requirement.
That's because there is obviously not nothing. There is existence, E,
as I recently was illustrating, in a technical way. The point of
deaxiomatization and deriving truths from that process, reverse
mathematics of a sort, is to attempt to gain insight from the most
primitive truths and how they could be at all.
It resolves basically to what looks like a paradox, on reexamination
always true because its opposite is the same.
That can be stated in the terms of mathematical logic, conveniently.
Ross
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
Anny system must have primitives, i.e., things not defined within the
system, to start with or one reall is making something out of nothing.
> And those people seems to think that they can
> produce things from nothing, which is philosophically absurd.
It is those who think they can produce something without even axioms and
primitives who trying to produce something from nothing.
Ross A. Finlayson
Virgil
ZF can't produce anything without everything.
Ross
george
Charlie-Boo wrote:
> george wrote:
>
> > Under some simple class theories, ALWAYS, because
> > EVERYthing is a class
>
> How about the classes that are not elements of themselves -
That depends. Some theories have a "foundation" axiom that prevents
that. Other theories allow it.
> is that a class?
There is a class of all SETS that are not elements of themselves.
No proper class is an element of ANY class, so, no, there is not a
class of all classes. That would be a hyperclass. Yes, this regress
is silly.
> Likewise for those loons who say everything is a set - how
> about the sets that are not elements of themself - is that a set?
There is no set of all&only such sets; however, there is a class of
them.
There are no "loons" alleging that everything is a set. That is
alleged
by certain THEORETICAL TREATMENTS of this stuff (like ZFC).
People's opinions are NOT involved.
> C-B
> (author of CBL which answers all of these questions formally and
> automatically)
Every individual set or class theory also answers them.
The problem is that different theories answer them in different ways.
Yours is just another raindrop in the hurricane.
Jonathan Hoyle
Yes, since that is how you defined S.
> ...but I want the answer for S were S is any set other than { } .
> would A.B be a set .
>
> More specifically A.B is the class of all transitive sets that has
> bijection to S, were S is a set other than { }. Is A.B a set for each
Given your definitions of B, the only pertinent thing is the
cardinality of S. For example, in the above case, you could define S =
{ 0, 1, 2 } or as S = { 100, googleplex, aleph_omega }, it doesn't
matter, as long as |S| = 3. If S is a set, it has a defined
cardinality |S|. Let us say for some arbitrary set S, |S| is some
cardinal number k.
By your definition of B, B is therefore the set of all sets with
cardinality k. (Note that B is not a proper class here, it is a set.)
Thus, A.B becomes nothing more than the set of all transitive sets with
cardinality k. Again, A.B is a set.
Hope that helps,
Jonathan Hoyle
Eastman Kodak
Jonathan Hoyle
Incorrect. Classes include sets.
> Non-sets in a set theory are non-sense.
Again, untrue. There are Set theories that define the existence of
proper classes.
> There can be only one proper class or none, in a set theory it would be
> a set.
??? The sentence makes no sense and is self-contradictory. A proper
class is, by definition, not a set. And why should there be only one?
If you remove a single element from a proper class, it still remains a
proper class.
> There is no class of classes in set theory with classes.
Not in traditional set theories, but there are some non-well founded
theories that do include the class of classes.
> There is no universe in ZF.
What do you mean by this statement? You repeat it often enough, you
ought to finally make sense of it.
> ZF is inconsistent.
Which axiom in ZF do you think introduces the inconsistency? You make
this claim often but (as usual) never manage to prove it.
Jonathan Hoyle
Eastman Kodak
zuhair
Jonathan Hoyle wrote:
> > Ok, you answered it when S has cardinality 3...
>
> Yes, since that is how you defined S.
>
> > ...but I want the answer for S were S is any set other than { } .
> > would A.B be a set .
> >
> > More specifically A.B is the class of all transitive sets that has
> > bijection to S, were S is a set other than { }. Is A.B a set for each
>
> Given your definitions of B, the only pertinent thing is the
> cardinality of S. For example, in the above case, you could define S =
> { 0, 1, 2 } or as S = { 100, googleplex, aleph_omega }, it doesn't
> matter, as long as |S| = 3. If S is a set, it has a defined
> cardinality |S|. Let us say for some arbitrary set S, |S| is some
> cardinal number k.
>
> By your definition of B, B is therefore the set of all sets with
> cardinality k. (Note that B is not a proper class here, it is a set.)
No, this is not true, you cannot say that A.B defined as the set of all
sets bijectable to S(were S hac cardinality k) is a set, A.B is not a
set, Moe Blee have presented a proof against A.B being a set. in simple
words UA.B would be the set of all sets, which is not a set in ZFC,
therefore A.B is not a set in ZFC.
You are mistaken, however you can say that the set of all transitive
sets is not a proper set, and if so A.B will of course be a set (
separation).
Zuhair
Ross A. Finlayson
If the universe exists, for example the one we inhabit, where functions
between objects are objects, then that's an example that infinite sets
are equivalent.
That's argued against by saying the physical universe is not the
mathematical universe, conversely every mathematical construct is only
an abstraction to the extent that it is invented by the mind, in the
physical universe.
The universe is infinite because, as example, two free massy point
particles have infinitely differentiable motion.
You're correct about some classes being sets, as they are so defined.
While that may be so:
a) you can't quantify over elements of a proper class,
b) there is nothing to contain proper classes because nothing can
contain a proper class,
c) there could only exist one or no proper classes in any theory,
d) proper classes are non-sets and there are none in any pure set
theory, and
e) classes can't contain classes, for then the same perceived
problems in regular set theory which brought about the device of
classes would apply.
About ZF and the universal quantifier requiring a universe and there
being none in ZF:
i) ZF requires a universe to quantify,
ii) there is no universe in ZF over which to quantify, thus
iii) there are no sets in ZF, which contradicts assertion that there
are, thus
iv) ZF is inconsistent.
Jon, I hope you might reply to the previous message about these notions
and well-ordering the reals and etcetera from some few weeks ago.
There is no set of ordinals in ZF, nor set of cardinals, nor set of
sets.
Ross
Jonathan Hoyle
> between objects are objects, then that's an example that infinite sets
> are equivalent.
The physical universe we live in is finite, and therefore not
equivalent to the universe of mathematical objects, which is infinite.
For example, the total number of elementary particles that make up the
universe is somewhere in the 10^80 - 10^90 range, which is vanishingly
tiny compared to the number of integers there are.
> That's argued against by saying the physical universe is not the
> mathematical universe, conversely every mathematical construct is only
> an abstraction to the extent that it is invented by the mind, in the
> physical universe.
Yup, you got it.
> The universe is infinite because, as example, two free massy point
> particles have infinitely differentiable motion.
How do you know that? (Even if so, that doesn't make the universe
infinite.)
> You're correct about some classes being sets, as they are so defined.
> While that may be so:
>
> a) you can't quantify over elements of a proper class,
I thought you could in NBG Set theory, but I could be wrong. But there
are other Set Theories which do define this.
> b) there is nothing to contain proper classes because nothing can
> contain a proper class,
Again, true in NBG but not in some others (eg, non-well founded
theories).
> c) there could only exist one or no proper classes in any theory,
This is false. The class of all sets, the class of all ordinals and
the class of all cardinals are each proper classes, yet they are each
distinct from one another.
> d) proper classes are non-sets and there are none in any pure set
> theory, and
Again, this is false. As I have repeatedly pointed out, NBG Set Theory
allows for proper classes.
> e) classes can't contain classes, for then the same perceived
> problems in regular set theory which brought about the device of
> classes would apply.
You can have well founded theories in which classes contain other
classes (NBG is not one of them), but to have a "class of all classes",
you need to go to a non-well founded theory, since (by definition) a
class of all classes must contain itself as a member.
> About ZF and the universal quantifier requiring a universe and there
> being none in ZF:
>
> i) ZF requires a universe to quantify,
True, although ZF can be applied to any number of different
"universes", countable or uncountable alike.
> ii) there is no universe in ZF over which to quantify, thus
Again, you parrot this sentence, and again, i have no idea what you are
talking about. Perhaps you are making the freshman mistake of
confusing the model with the theory? ZF can be used to model a
sufficiently rich universe. Or perhaps you are referring to the fact
that ZF does not address a "set of all sets"? That merely makes ZF
incomplete (which Godel proved in 1931), not inconsistent.
> iii) there are no sets in ZF, which contradicts assertion that there
> are, thus
Who says there are no sets in ZF?!? Just because ZF does not address
the class of all sets does imply that it addresses no sets. (This is
where your sloppy thinking has has undone you again.)
> iv) ZF is inconsistent.
A false conclusion following a series of false assumptions or false
deductions.
> Jon, I hope you might reply to the previous message about these notions
> and well-ordering the reals and etcetera from some few weeks ago.
Is there a particular date I should look to respond?
> There is no set of ordinals in ZF, nor set of cardinals, nor set of
> sets.
Again, you are confusing the model with the theory. You certainly can
have a model (universe) with those things, only the theory ZF does
speak of those particular entities. You can keep the same model
(universe) and merely enlarge the theory to NBG to encompass them. The
universe is not tied to the theory, nor is the theory tied to the
model. This is almost the entirity of your confusion.
> Ross
You might wish to familiarize yourself on NBG a bit more, here are a
couple of links:
Wikipedia:
http://en.wikipedia.org/wiki/Von_Neumann-Bernays-Gödel_set_theory
PlanetMath.org:
http://planetmath.org/encyclopedia/VonNeumannBernausGodelSetTheory.html
Also on non-well founded set theories:
Wikipedia: http://en.wikipedia.org/wiki/Non-well-founded_set_theory
Jonathan Hoyle
> sets bijectable to S(were S hac cardinality k) is a set, A.B is not a
> set, Moe Blee have presented a proof against A.B being a set. in simple
> words UA.B would be the set of all sets, which is not a set in ZFC,
> therefore A.B is not a set in ZFC.
I didn't see that proof, so maybe I am missing something here. Let's
use the example you gave above where |S| = 3. So A.B is the class of
all transitive sets of cardinality 3. Moe Blee's claim is that UA.B is
the class of all sets? I don't see how that is possible, since even
the transitive set 4 = { 0, 1, 2, 3 } = { {}, {{}}, {{},{{}}},
{{},{{}},{{},{{}}}} } is not in A.B. Are you sure Moe Blee was talking
about the same sets here.
I do agree, however, that B is a proper class, and I mispoke when I
suggested otherwise.
Jonathan Hoyle
Eastman Kodak
Ross A. Finlayson
NBG's not a pure set theory, there are only sets in pure set theories.
In pure set theories everything's a set, remarkably in no standard
"set" theory is everything a set, where "naive" set theory is pure set
theory. A set theory without a universe is not Cantorian.
I've heard that NBG, basically replacing axioms of ZFC with axiom
schema, is not the same thing as NBG with classes, that is to say, NBG
is not ZFC with classes. If von Neumann's, Bernays', and Goedel's
theory, by whoever put those three together, where Goedel apparently
tacked himself onto von Neumann and Bernays, actually does have these
notions of proper classes from the beginning, I was thinking they were
separate notions.
I see on those web pages that they think the main focus of NBG is to
have proper classes.
The planetmath exposition has basically there being axiom schema, or
alternatively proper classes.
I think it could be agreed that by default talk of "NBG" is talk of
"the mainstream variant of NBG with proper classes", and it's not a
pure set theory.
Proper classes don't have enough features to do much of anything with
them. They're basically a placeholder in vocabulary for where
otherwise saying, and requiring, "collection of sets" would invalidate
a variety of closely held standard set-theoretic notions, of
regular(ized) set theories.
Proper classes aren't sets. They seem just another manifestation of
incapacity to understand the infinite, counting upwards forever and
never getting anywhere. Counting on nothing, the numbers get higher.
The group noun game never gets anywhere, and Zeno stalls.
About models and theories, basically models are to theories as classes
are to sets. There is no maximal element of ZF to the model of ZF, ZF
has no model. ZF, in its entirety, has no model (ordinal large enough
to well-order all its elements, ie, its universe), and for any portion
to have a model, the entirety must have a model.
About the universe being infinite, each function between each function
between each function ad infinitum of however many particles you think
there are are objects and the universe is infinite.
Ross
Ross A. Finlayson
> > If the universe exists, for example the one we inhabit, where functions
> > between objects are objects, then that's an example that infinite sets
> > are equivalent.
>
> The physical universe we live in is finite, and therefore not
> equivalent to the universe of mathematical objects, which is infinite.
> For example, the total number of elementary particles that make up the
> universe is somewhere in the 10^80 - 10^90 range, which is vanishingly
> tiny compared to the number of integers there are.
>
> > That's argued against by saying the physical universe is not the
> > mathematical universe, conversely every mathematical construct is only
> > an abstraction to the extent that it is invented by the mind, in the
> > physical universe.
>
> Yup, you got it.
>
> > The universe is infinite because, as example, two free massy point
> > particles have infinitely differentiable motion.
>
> How do you know that? (Even if so, that doesn't make the universe
> infinite.)
>
In more about the universe and how it's infinite, big science astronomy
continually learns more, hopefully, about the universe, and as it does
and the numbers are crunched: it gets bigger. That is to say, the
data always points to a larger universe, and your 2^90 particles today
was 2^50 some few years ago.
Similarly, the more exactly subatomic particles are measured: the
smaller they appear to be. In the last fifteen years the measured
sizes of subatomic particles have decreased by several orders of
magnitude. That's not just smaller bounds on the measurements, that's
measurement of smaller quantities.
Now, as to why there is gauge invariance, that's a good question and I
don't have a very good answer for that.
This is where for about a century it has been realized that the
constituent elements of matter as particles are as well waves,
correspondingly with energy and the solitons and waves. There's no
separation in the dichotomy, they must be both.
In forming agglomerates as particles and fields as waves, certain
abstractions enable much tractability, for example in considering
particles as only points. While that is so, it is well-known that they
are not only that, and indeed in the large and small the behavior of
these otherwise idealizable entities can not be predicted solely by
those methods. The infinitesimal analysis is truly useful, and since
its inception has been about summing infinitesimals from zero to one
and calling it one.
I draw parallels between that notion of particle/wave duality and a
notion of the duality of real numbers as contiguous points on a line
and complete ordered field, although I mean the complete
super-Archimedean complete ordered field.
Then, with polydimensional perspective, modern classical
interpretations of the real real numbers shall offer otherwise
inaccessible predictions.
Ross
Jonathan Hoyle
> In pure set theories everything's a set, remarkably in no standard
> "set" theory is everything a set, where "naive" set theory is pure set
> theory.
I wouldn't be concerned about the difference in names between "class"
and "set". In most non-well founded theories, they use only the term
"set" and speak of "the set of all sets" rather than the "class of all
classes". It's merely a name. If you don't like the word "class",
replace it with "set" if you want. The main distinction (inasfar as
NBG is concerned) is that classes cannot be members of any other class
or set; in almost all other ways, classes and sets are equivalent.
> I've heard that NBG, basically replacing axioms of ZFC with axiom
> schema, is not the same thing as NBG with classes, that is to say, NBG
> is not ZFC with classes.
Actually, it essentially is. The finite axiom schema supplied by NBG
is proven to be equivalent to ZFC's, so they can be thought of as the
same. All treatments I have seen of NBG have included proper classes
(as that is one of the things that makes NBG so interesting), but I
suppose it can be bypassed. In any case, everything I have said about
NBG still stands.
> I see on those web pages that they think the main focus of NBG is to
> have proper classes.
>
> The planetmath exposition has basically there being axiom schema, or
> alternatively proper classes.
>
> I think it could be agreed that by default talk of "NBG" is talk of
> "the mainstream variant of NBG with proper classes", and it's not a
> pure set theory.
I don't know what you mean by a "pure set theory". You seem to be
contradicting yourself. On one hand, you complain that there is no set
of all sets. So, I give you NBG in which there is a class of all sets.
Then you complain because the word "class" is used instead of "set".
So then I mention the existence of non-well founded theories which do
have a set of all sets. But you do not seem to like that either.
What exactly are you asking for? Do you want a set of all sets or not?
If so, you must embrace a non-well founded theory, since such a set
must include itself as a member. If not, ZFC is just fine for you.
Either way, I am at a complete loss as to why you are complaining.
> Proper classes don't have enough features to do much of anything with
> them. They're basically a placeholder in vocabulary for where
> otherwise saying, and requiring, "collection of sets" would invalidate
> a variety of closely held standard set-theoretic notions, of
> regular(ized) set theories.
What specific features of sets that you want are missing in proper
classes?
> Proper classes aren't sets. They seem just another manifestation of
> incapacity to understand the infinite, counting upwards forever and
> never getting anywhere. Counting on nothing, the numbers get higher.
> The group noun game never gets anywhere, and Zeno stalls.
This is just simply untrue. Within the same universe, a proper class
in one theory can easily become a set in another. Consider the
universe of V_kappa for some sufficiently large (strongly inaccessible)
kappa as a model for NBG. Then all subsets of V_kappa with cardinality
less than kappa are "sets" in this model, whereas those with
cardinality kappa are "proper classes". kappa itself is the class of
ordinals in this model.
But since kappa is arbitrary, we can choose another strongly
inaccessible kappa2 > kappa and look at V_kappa2. Now in this model,
all those sets of cardinality kappa which were proper classes in the
former theory are now simply sets in this theory. Now only those
collections of size kappa_2 are proper classes.
Any model of NBG where X is a proper class can be widened so that X is
a set, and some other (larger) Y becomes a proper class.
> About models and theories, basically models are to theories as classes
> are to sets.
Not even a little bit.
> There is no maximal element of ZF to the model of ZF, ZF
> has no model.
Huh? This sentence makes absolutely no sense. Please rephrase, and
when you do so, try to be a bit more rigorous in your thinking. Your
sloppiness is your greatest undoing here.
> ZF, in its entirety, has no model (ordinal large enough
> to well-order all its elements, ie, its universe), and for any portion
> to have a model, the entirety must have a model.
I am again at a complete loss. Most of this makes no sense to me. And
what does make sense is clearly (and provably) wrong.
> About the universe being infinite, each function between each function
> between each function ad infinitum of however many particles you think
> there are are objects and the universe is infinite.
I don't understand what you mean by "each function between each
function between each function...". You apparently have some
non-rigorous, fuzzy, and ill-defined philosophy here which breaks down
with minimal inspection. Please try again, and try to be precise with
your statements.
Jonathan Hoyle
Eastman Kodak
zuhair
Sorry, there was a mistake, I meant B as defined as the set of all sets
that have cardinality k, is not a set, since this would mean that B is
the set of all sets that are bijective to S when S has cardinality k.
It has been proved that such a set is not in ZFC. In short UB ( the set
union of B) would be the set of all sets, since you can replace any
member of a member of B by any set v, without this member of B being
excluded from B. And since the set of all sets is not a set in ZFC,
then B is not a set. But it can be a proper class, this will safe it
from being inconsistent.
zuhair
Jonathan Hoyle
> that have cardinality k, is not a set, since this would mean that B is
> the set of all sets that are bijective to S when S has cardinality k.
Yes, you are correct. Moe Blee's proof is a good one. It is simple
enough to show just using k = 1. Each set containing a cardinal number
is certainly a set. But the collection of all these sets is not a set,
as its union would be the class of cardinal numbers.
> It has been proved that such a set is not in ZFC. In short UB ( the set
> union of B) would be the set of all sets, since you can replace any
> member of a member of B by any set v, without this member of B being
> excluded from B. And since the set of all sets is not a set in ZFC,
> then B is not a set. But it can be a proper class, this will safe it
> from being inconsistent.
While A and B are each proper classes, it seems to me that A.B should
still be a set, certainly for finite S anyway. I'm not sure why the
class of a transitive sets with cardinality omega should not be a set
either.
Jonathan Hoyle
Eastman Kodak
zuhair
It is not about whether you are sure or not, it is about proof. what is
the proof that for every S:S=/={}, y={x|x is a transitive set /\ x is
bijectable to S ) -> y is a set.
This is important.
Zuhair
Jonathan Hoyle
> the proof that for every S:S=/={}, y={x|x is a transitive set /\ x is
> bijectable to S ) -> y is a set.
There are various ways to show that a given class is a set. The most
obvious way is to show its cardinality. Cardinal numbers are sets, and
a set equinumerous to some cardinal number must therefore be a set as
well. Proper classes in NBG are "too big" for standard cardinal
numbers.
I haven't thought it through carefully (it is 1:30AM my time), but I
think the following would be a reasonable sketch of such a proof:
By your definitions earlier, A.B is the class of all transitive sets
with cardinality k, where k = |S| for some predefined S. Assuming no
urelements, each element of A.B must therefore a subset of V_k. (If
not, then the set would have to have cardinality > k by transitivity).
This makes A.B a subset of V_k+1. Thus |A.B| <= |V_k+1|, and therefore
a set.
It's been a while since I did Set Theory as a student, so someone might
want to verify this thinking. But it certainly seems reasonable to me.
> This is important.
Why exactlty is this important?
zuhair
In another thread I have attempted to rdefine cardinality without
choice, by stating that
card(S) = { x|x is transtive /\ x is bijectable to S }.
In other words a cardinality is the equivalence class of transitive
sets under equivalence relation bijection.
This will bridge the gap of Von Neumann's definition of cardinality,
since the later cannot describe the cardinality of non well orderable
sets like P(w) were w=Omega., this new definition f cardinality can
describe such sets, since the power set of every transitive set is a
transitive set, and using the fact that for every two sets x and y, if
x is bijectable to y then P(x) is bijectable to P(y), this
bijectability do not require choice. and so this definition of
cardianlity will overcome the shortcomings of both Von Neumann's
definition of cardinality and Frege's definition of cardinality.
Zuhair
Jonathan Hoyle
> without choice, by stating that
>
> card(S) = { x|x is transtive /\ x is bijectable to S }.
I'm not sure what this definition is buying you.
With Choice, all sets have cardinality, and the cardinal numbers can be
well ordered.
Without Choice, all sets can be mapped to an Aleph, but those "numbers"
can't be well-ordered.
With your definition of cardinality, you can well order the cardinals
again, but then not all sets will have cardinality. That is to say,
without AC, you can have a set S which is not bijectable to any
transitive set.
What do you do with the "cardinality" of those sets?
Jonathan Hoyle
Eastman Kodak
zuhair
I think you misunderstood me.
All of Von Neumann's cardinals are ordinals, but the converse is of
course not true.
The problem is wihtout AC, P(w) which is not a well orderable set ,
cannot be bijected to any of Von Neumann's ordinals, because a not well
orderable set have no bijection with a well ordered set, and since
every Von Neumann's cardinal is an ordinal and every ordinal is a well
ordered set, then no not well orderable set can have cardinality as
defined by Von Neumann, so P(w), P(P(w), etc........ all have no Von
Neumann cardinalities without AC.
But using my definition of cardinals as the equivalence class of
transitive sets bijectable to a set, will bridge this gap in Von
Neumann's.
Since P(w), P(P(w),..etc can all be bijected to equivalence classes of
transitive sets bijectable to them.
The trick, is that the Power set of EVERY transitive set is a
transitive set. Therefore we can always find cardinalities for sets
that are not well orderable.
Cardinality as defined by my way are more general than cardinality
defined by Von Neumann given that the axiom of choice is not in action.
You see not all transitive sets are well orderable.
Also this definition avoids the problems Frege's definition of
cardinality entails, those of weather the equivalence class of all sets
bijectable to a set, is a set or a proper set, and even if it is a
proper sets, proper sets are heavy entities, not having all the merits
sets has.
That's why I insisted from the beginning on knowing weather the sets of
all transitive sets bijectable to a predefined set S, is a set or not.
Since you say it is a set. Then the definition I have given is the best
definition of cardinality.
Without the axiom of choice, many sets that Von Neumann's definition of
cardinality do not cover, this definition cover, and Thus it is bette
than Von Neumann's , if AC is dropped.
Zuhair
Charlie-Boo
george wrote:
> Charlie-Boo wrote:
> > george wrote:
> >
> > > Under some simple class theories, ALWAYS, because
> > > EVERYthing is a class
> >
> > How about the classes that are not elements of themselves -
>
> That depends. Some theories have a "foundation" axiom that prevents
> that. Other theories allow it.
It is inconsistent to allow it, as it is inconsistent to say that
everything is a set or everything is a class.
> > is that a class?
>
> There is a class of all SETS that are not elements of themselves.
> No proper class is an element of ANY class, so, no, there is not a
> class of all classes. That would be a hyperclass. Yes, this regress
> is silly.
Then how can you say that everything is a class?
> > Likewise for those loons who say everything is a set - how
> > about the sets that are not elements of themself - is that a set?
>
> There is no set of all&only such sets; however, there is a class of
> them.
> There are no "loons" alleging that everything is a set. That is
> alleged
> by certain THEORETICAL TREATMENTS of this stuff (like ZFC).
> People's opinions are NOT involved.
They believe it but I know better because the sets that are not
elements of themselves is not a set, so it is inconsistent to say that
everything is a set, or everything is a class. That is loony.
C-B
zuhair
Charlie-Boo wrote:
> george wrote:
> > Charlie-Boo wrote:
> > > george wrote:
> > >
> > > > Under some simple class theories, ALWAYS, because
> > > > EVERYthing is a class
> > >
> > > How about the classes that are not elements of themselves -
> >
> > That depends. Some theories have a "foundation" axiom that prevents
> > that. Other theories allow it.
>
> It is inconsistent to allow it, as it is inconsistent to say that
> everything is a set or everything is a class.
Hey, do you agree with me that every set is in itself.
Zuhair
Math1723
> >
> > That depends. Some theories have a "foundation" axiom that prevents
> > that. Other theories allow it.
>
> It is inconsistent to allow it, as it is inconsistent to say that
> everything is a set or everything is a class.
Neither needs be inconsistent. There are set theories which do allow
sets/classes to contain themselves as members. There are also theories
which a universal set/class is defined. Inconsistency is avoided in
such theories by denying the Axiom of Separation.
> > > is that a class?
> >
> > There is a class of all SETS that are not elements of themselves.
> > No proper class is an element of ANY class, so, no, there is not a
> > class of all classes. That would be a hyperclass. Yes, this regress
> > is silly.
You appear to be speaking of NBG Set Theory. Yes, in this theory, sets
and classes contain only sets (or urelements). Proper classes cannot
be members of any class (or set). However, NBG Set Theory is a
Well-Founded theory; there are non-Well Founded theories which do not
have this limitation.
Charlie-Boo
zuhair wrote:
> Charlie-Boo wrote:
> > george wrote:
> > > Charlie-Boo wrote:
> > > > george wrote:
> > > >
> > > > > Under some simple class theories, ALWAYS, because
> > > > > EVERYthing is a class
> > > >
> > > > How about the classes that are not elements of themselves -
> > >
> > > That depends. Some theories have a "foundation" axiom that prevents
> > > that. Other theories allow it.
> >
> > It is inconsistent to allow it, as it is inconsistent to say that
> > everything is a set or everything is a class.
>
> Hey, do you agree with me that every set is in itself.
Then there is no empty set. No, not by the normal (intuitive)
definition of a set.
But regardless of whether sets can contain themselves or not, or
whether all do, saying that the set/class of sets/classes that don't
contain themselves is a set/class is inconsistent. Statements like:
"everything is a set"
"everything is a class"
"every set contains itself"
are just stupid.
The fact of the matter is, a set can contain itself. There is nothing
logically wrong with a set containing itself. Have you ever
experienced a program that outputs only itself? There is no problem
there, and a set can likewise be defined so what matches the definition
happens to be a single definition, namely itself.
Define f(x) = {x(x)}. Then f(f) = {f(f)} is a set that contains
itself. Look up QUINE ATOM on Google - see the first reference.
A great deal of what is written on Logic, Set Theory and Computer
Science is stupid nonsense that professors get published because the
editors and referees are more stupid professors. It's true, but few
people seem to realize it.
C-B
Charlie-Boo
Math1723 wrote:
> > > > How about the classes that are not elements of themselves -
> > >
> > > That depends. Some theories have a "foundation" axiom that prevents
> > > that. Other theories allow it.
> >
> > It is inconsistent to allow it, as it is inconsistent to say that
> > everything is a set or everything is a class.
>
> Neither needs be inconsistent. There are set theories which do allow
> sets/classes to contain themselves as members. There are also theories
> which a universal set/class is defined. Inconsistency is avoided in
> such theories by denying the Axiom of Separation.
That's stupid professor-speak. If everything is a set, then is
{x|~(xex)} a set?
C-B
george
Charlie-Boo wrote:
> > Charlie-Boo wrote:
> > > How about the classes that are not elements of themselves -
I replied,
> > That depends. Some theories have a "foundation" axiom that prevents
> > that. Other theories allow it.
> It is inconsistent to allow it,
It is inconsistent to have a class of ALL&ONLY those classes
that are not elements of themselves. But if you don't have foundation
(and allow some classes to be members of themselves), then you
CAN have a class containing MOST of the classes that don't
contain themselves (it just doesn't contain itself), AND a class having
ALL of the classes that don't contain themselves, PLUS one that
does (itself).
> as it is inconsistent to say that
> everything is a set or everything is a class.
THAT is NOT inconsistent, dumbass. ZFC says that
everything is a set. NBG says that everything is a class.
Both of them are consistent.
> > There is a class of all SETS that are not elements of themselves.
> > No proper class is an element of ANY class, so, no, there is not a
> > class of all classes. That would be a hyperclass. Yes, this regress
> > is silly.
>
> Then how can you say that everything is a class?
I *didn't* say it. NBG says it. Or rather, presupposes it.
The class of all classes does not exist in 1st-order NBG,
but all its elements (all the classes) do; one of these elements
is the class of all sets. In 1st-order ZFC, the class of
all sets doesn't exist either.
> They believe it but I know better because the sets that are not
> elements of themselves is not a set,
And, in ZFC, IT ALSO DOESN'T EXIST, so there is no inconsistency.
In NBG, this class DOES exist. There is no inconsistency there either.
> so it is inconsistent to say that
> everything is a set, or everything is a class.
I don't mean the same thing as you do by "everything".
Or rather, standard classical first-order logic doesn't mean the
same thing that you do by "everything". The standard classical
semantics for "everything" means "everything THAT IS IN MY
DOMAIN OF DISCOURSE". Everything ELSE is simply not known
about at all.
> That is loony.
Maybe, but it has some nice theoretical properties; that's
why it is the standard.
> > > C-B
> > > (author of CBL which answers all of these questions formally and
> > > automatically)
I repeat,
zuhair
Charlie-Boo wrote:
> zuhair wrote:
> > Charlie-Boo wrote:
> > > george wrote:
> > > > Charlie-Boo wrote:
> > > > > george wrote:
> > > > >
> > > > > > Under some simple class theories, ALWAYS, because
> > > > > > EVERYthing is a class
> > > > >
> > > > > How about the classes that are not elements of themselves -
> > > >
> > > > That depends. Some theories have a "foundation" axiom that prevents
> > > > that. Other theories allow it.
> > >
> > > It is inconsistent to allow it, as it is inconsistent to say that
> > > everything is a set or everything is a class.
> >
> > Hey, do you agree with me that every set is in itself.
>
> Then there is no empty set.
Yes, that is the fact. There is not empty set.
Every set is in itself. No set is not in itself.
In fact what the word set mean is "what is in itself".
that's the true definition of "set" that eluded mathematicians and
logicians all of these years.
the reason becaue they assumed the converse approach.
a set that is not in itself is not a set, perhaps it is a ur-element.
Zuhair
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
> a set that is not in itself is not a set
> Zuhair
Not in reasonable set theories.
Math1723
> > > everything is a set or everything is a class.
> >
> > Neither needs be inconsistent. There are set theories which do allow
> > sets/classes to contain themselves as members. There are also theories
> > which a universal set/class is defined. Inconsistency is avoided in
> > such theories by denying the Axiom of Separation.
>
> That's stupid professor-speak. If everything is a set, then is
> {x|~(xex)} a set?
Perhaps I misunderstood you? When you said "everything is a set", I
thought you meant "the set of all sets is a set". There are Non-Well
Founded Set Theories that do indeed allow for such sets to exist. It
is not inconsistent to allow it, it merely requires a different
axiomatic system. Obviously, ZFC does not allow for it, as "the set of
all sets", if it existed, would violate the Axiom of Regularity.
Russel's Paradox is avoided by limitations imposed by the Axiom of
Separation. You cannot call {x|~(xex)} a set under even non-Well
Founded theories (at least as i undertsnad them).
If uyou need to read up on this, there are a number of web sites that
discuss Anti-Foundation set Theories. Wikipedia has a good of one as
any starting point, so here'it is:
Charlie-Boo
> Charlie-Boo wrote:
> > > Charlie-Boo wrote:
> > > > How about the classes that are not elements of themselves -
> > > > is that a class?
>
> I replied,
> > > That depends. Some theories have a "foundation" axiom that prevents
> > > that. Other theories allow it.
> It is inconsistent to have a class of ALL&ONLY those classes
> that are not elements of themselves.
Then the systems that allow it are inconsistent.
If everything is a set, then are the sets that do not contain
themselves a set? Answer: No, that leads to an inconsistency. If
everything is a class, then are the classes that do not contain
themselves a class? Answer: No, that leads to an inconsistency.
So it is inconsistent to say that everything is a set or that
everything is a class. It doesn't matter that your professors told
you otherwise. Simple logic shows that they are wrong.
There is no reason to refer to various set axioms. The above is true
for any axioms, because it is based on simple general logic.
"In questions of science, the authority of a thousand is not worth the
humble reasoning of a single individual." - Galileo Galilei
> But if you don't have foundation then you
> CAN have a class having
> ALL of the classes that don't contain themselves
You are simply being inconsistent in your statements.
The truth of the matter is, there is no reason to believe that a set
can't contain itself. A computer program can output only itself, and
likewise a set can contain only itself. People who think of sets as
being paper bags have a problem with this fact. But those who realize
that a set is just a definition of whatever we want to include, and
that definition can easily be constucted to contain its own definition
(by use of self-reference, as I showed.)
C-B
Charlie-Boo
> Charlie-Boo wrote:
> > zuhair wrote:
> > > Charlie-Boo wrote:
> > > > george wrote:
> > > > > Charlie-Boo wrote:
> > > > > > george wrote:
> > > > > >
> > > > > > > Under some simple class theories, ALWAYS, because
> > > > > > > EVERYthing is a class
> > > > > >
> > > > > > How about the classes that are not elements of themselves -
> > > > >
> > > > > That depends. Some theories have a "foundation" axiom that prevents
> > > > > that. Other theories allow it.
> > > >
> > > > It is inconsistent to allow it, as it is inconsistent to say that
> > > > everything is a set or everything is a class.
> > >
> > > Hey, do you agree with me that every set is in itself.
> >
> > Then there is no empty set.
>
> Yes, that is the fact. There is not empty set.
>
> Every set is in itself. No set is not in itself.
>
> In fact what the word set mean is "what is in itself".
>
> that's the true definition of "set" that eluded mathematicians and
> logicians all of these years.
Then mathematics would no longer work. How could I look for
{x|x^2+3*x-4=0}? It is meant to contain only one or two numbers, but
you say that it also contains itself.
Then we have no control over what is in a set. We can no longer just
give a condition and there is a set containing just its elements.
Your "Set Theory" has little or nothing to do with how sets function.
C-B
Charlie-Boo
> You cannot call {x|~(xex)} a set under even non-Well
> Founded theories (at least as i undertsnad them).
That's why it is stupid to say that everything is a set. {x|~(xex)}
is not a set.
> If uyou need to read up on this, there are a number of web sites that
> discuss Anti-Foundation set Theories. Wikipedia has a good of one as
> any starting point, so here'it is:
>
> http://en.wikipedia.org/wiki/Non-well-founded_set_theory
"In questions of science, the authority of a thousand is not worth the
humble reasoning of a single individual." - Galileo Galilei
C-B
zuhair
OF course it contains itself, no doubt. you are confusing determination
with membership.
yet it is true that this set is detemined by two numbers, but it
contain itself.
Zuhair
>
> Then we have no control over what is in a set. We can no longer just
> give a condition and there is a set containing just its elements.
>
> Your "Set Theory" has little or nothing to do with how sets function.
On the contrary have everything to do with how sets function. in
addition it has a universe that it can quantify on. it superseed
mathematics.
Zuhair
Jonathan Hoyle
>
> If everything is a set, then are the sets that do not contain
> themselves a set? Answer: No, that leads to an inconsistency. If
> everything is a class, then are the classes that do not contain
> themselves a class? Answer: No, that leads to an inconsistency.
>
> So it is inconsistent to say that everything is a set or that
> everything is a class. It doesn't matter that your professors told
> you otherwise. Simple logic shows that they are wrong.
>
> There is no reason to refer to various set axioms. The above is true
> for any axioms, because it is based on simple general logic.
Technically though, even "simple general logic" is axiomatic. You could
theoretically have a weaker logical framework in which it is impossible
to generate a contradiction from this. However, such systems would be
less interesting since it would be difficult to generate any theorems
at all. (I would imagine that a logical system without, say, modus
ponens as an axiom or theorem, would indeed be dull and unexciting.)
> The truth of the matter is, there is no reason to believe that a set
> can't contain itself.
In Anti-Foundational systems, some sets certainly can contain
themselves as members. But not every set does, as Zuhair is attempting
to make.
I think Zuhair's problem is failing to comprehend the distinction
between the concepts: "contains as a subset" and "contains as an
element". To illustrate this point, let D = the Set of Dogs.
Obviously, D cannot contain itself because D contains only dogs, not
sets. If you tried to add D to itself, it would no longer truly be
"the Set of Dogs", since it would then contain a non-Dog element.
This distinction is apparently lost on Zuhair, where his universe
contains only sets, without urelements like dogs. Even in mathematics,
the vast majority of sets are treated as collections of urelements, not
as sets of sets: "a set of points on a line", "a set of linear
equations", "a set of solutions to a differential equation", etc.
Although each of these examples could be forced into being sets of sets
without urelements, with some effort, but it is completely irrelevant
to anyone outside of Set Theory.
As long as Zuhair is unable to understand the difference between the
container ("set") and what's inside of it (its "elements"), he will be
barking up this tree to no avail.
Jonathan Hoyle
Eastman Kodak
zuhair
Sure. you got it ha.
To illustrate this point, let D = the Set of Dogs.
> Obviously, D cannot contain itself because D contains only dogs, not
> sets. If you tried to add D to itself, it would no longer truly be
> "the Set of Dogs", since it would then contain a non-Dog element.
what an example. from the begining Johanth. you shouldn't say a set of
dogs, there is no such a thing. there is "dogs"which is different from
a set of dogs, even I don't like this word, a set of dogs, it should be
rather told as the set of dogs and the set itself. a set of dogs
without the set itself being a member of itself is just meaningless as
saying that a dog is not a dog.
Anyhow. there is no point from all of this. since the majority here do
not prefer sets to be in themselfs. then my game has only one player in
it. and so it is useless.
>
> This distinction is apparently lost on Zuhair, where his universe
> contains only sets, without urelements like dogs. Even in mathematics,
> the vast majority of sets are treated as collections of urelements, not
> as sets of sets: "a set of points on a line", "a set of linear
> equations", "a set of solutions to a differential equation", etc.
> Although each of these examples could be forced into being sets of sets
> without urelements, with some effort, but it is completely irrelevant
> to anyone outside of Set Theory.
>
> As long as Zuhair is unable to understand the difference between the
> container ("set") and what's inside of it (its "elements"), he will be
> barking up this tree to no avail.
barking up? thanks for this nice expression.
>
> Jonathan Hoyle
> Eastman Kodak
Jonathan Hoyle
> > Obviously, D cannot contain itself because D contains only dogs, not
> > sets. If you tried to add D to itself, it would no longer truly be
> > "the Set of Dogs", since it would then contain a non-Dog element.
>
> what an example. from the begining Johanth. you shouldn't say a set of
> dogs, there is no such a thing.
There's no such thing as "a set of dogs" ???
> there is "dogs"which is different from a set of dogs...
How is that different? It seems to me that any number of dogs can be
viewed as a set of dogs.
> ...even I don't like this word, a set of dogs...
Of course you don't like it. After all, it undermines your entire
philosophy, so why would you like it? Your liking it or not, however,
has no baring on the matter.
> it should be rather told as the set of dogs and the set itself.
But then that wouldn't be a set of dogs, now would it? A set of dogs
contains only dogs, no sets.
> ...a set of dogs without the set itself being a member of itself is just meaningless as
> saying that a dog is not a dog.
How is that the same? A set doesn't have to be a member of itself to
be a set.
zuhair
I think you should cary on the discussion to a much deeper level than
this.
YOu think that dogs are not sets. fare enough, what is a set? so that
dogs are not sets.
This is a problem u know.
Lets take y={x|P(x)} were P(x)<-> x is a dog.
Now P is the collection of properties that defines a dog.
My question is a dog a dog.
U say no. a dog is not a dog.
you mean there is the objective dog which is formed from cells , flesh
and blood, who we hear him barking ,etc,etc....
and there is the subjective dog, which is an abstraction in our mind of
objective dogs, and that this abstraction is not the objective dog,
this abstraction is the set of objective dogs.
This platonisim exists in your mind.
to me a dog is a dog weather objective or subjective, sets and
predicate should have nothing to do with this subjective\objective
distinction of your.
therefore a dog satisfies P(x)<->x is a dog, since a dog is a dog.
Therefore the set of dogs is { Johanthan's dog, Ross's dog, Zuhair'
dog,.......,a dog }.
You see the point. the above is a set of dogs.
to say that a dog is not a ur-element. Ok, fare enough. I will tell you
that there are no sets who'es only members are Ur elements.
Zuhair
MoeBlee
> Lets take y={x|P(x)} were P(x)<-> x is a dog.
>
> Now P is the collection of properties that defines a dog.
No, it's not. 'P' is a formula.
Moeblee
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
> Jonathan Hoyle wrote:
If Zuhair barks up trees, whether right ones or wrong ones, does that
make him a member of that set of dogs?
zuhair
Anyhow forget all of that.
I had a simple question.
when two proper classes interesect. when do that intersection is a set?
and when do that intersection is a proper? I mean is there general rule
by which we can know the answer?
Zuhair
Jonathan Hoyle
> and when do that intersection is a proper? I mean is there general rule
> by which we can know the answer?
Sometimes it's a proper class, sometimes it's a set, and sometimes it's
empty (hence a set). There is no general rule; however, anyone can
give you very simple examples of these:
Suppose A and B are proper classes, and let C be the intersection of A
and B.
1. If A is the class of all sets of cardinality 1, and B is the class
of all sets of cardinality 2, then C is the empty set. (Obviously any
two disjoint proper classes for A and B would work here.)
2. If A is the class of all sets of cardinality 1, and B is the class
of all sets of cardinality 2 unioned with the set of integers, then C
is the set of integers.
3. If A is the class of all sets with cardinality <= 2, and B is the
class of all sets with cardinality >= 2, then C is the (proper) class
of all sets with cardinality 2.
george
> george wrote:
> > It is inconsistent to have a class of ALL&ONLY those classes
> > that are not elements of themselves.
Charlie-Boo wrote:
> Then the systems that allow it are inconsistent.
NO "system" allows it: IT IS INCONSISTENT!
> If everything is a set, then are the sets that do not contain
> themselves a set?
Your "if" part is IRrelevant. EVEN if everything is NOT a set,
it is STILL the case that having a set of all sets that do not
contain themselves is inconsistent.
> Answer: No, that leads to an inconsistency.
The fact that you cannot
have a set of all&only the sets containing themselves has NOTHING
whatever to do with whether "everything" is a set.
> If everything is a class, then are the classes that do not contain
> themselves a class?
Of course not; no proper class is a member of any class.
The SETS that do not contain themselves ARE a class, though.
> Answer: No, that leads to an inconsistency.
In the usual class theory, you don't get anywhere NEAR that,
since NO proper class (REGARDLESS of whether it contains
itself) is a member of any class.
> So it is inconsistent to say that everything is a set or that
> everything is a class.
That simply has nothing to do with anything.
ZFC and NBG both say that and they are both consistent.
> It doesn't matter that your professors told
> you otherwise. Simple logic shows that they are wrong.
Simple logic shows nothing of the kind, but you wouldn't
know that, since, as you have just proved, You Can't Do simple logic.
> There is no reason to refer to various set axioms.
Of course there is; you can't have sets without them.
> The above is true
> for any axioms,
It is obviously NOT true for axiom-sets THAT DON'T
EVEN HAVE any sets AT ALL.
> because it is based on simple general logic.
No, it isn't, and YOU DON'T know what "simple general logic"
MEANS. There is a 2nd-order tautology relating to Russell's
Paradox that *is* purely logical, but it has nothing directly
to do with belonging to a set; it holds for binary relations
in general, of which a bleonging to b is only one.
> > But if you don't have foundation then you
> > CAN have a class having
> > ALL of the classes that don't contain themselves
>
> You are simply being inconsistent in your statements.
No, I'm not.
> The truth of the matter is, there is no reason to believe that a set
> can't contain itself.
WE KNOW THAT, DUMBASS.
WE'VE HEARD of Aczel and of set theories with ANTI-
foundation axioms. You, unfortunately, have not.
> A computer program can output only itself, and
> likewise a set can contain only itself.
Of course. But they can't do it in the context of an axiom
that SAYS they can't. You are of course equally free to
have an axiom that says they can, or to say nothing about
the issue one way or the other. But the point is, ZFC says something.
> People who think of sets as
> being paper bags have a problem with this fact. But those who realize
> that a set is just a definition of whatever we want to include, and
> that definition can easily be constucted to contain its own definition
> (by use of self-reference, as I showed.)
YOU haven't showed JACK. LITERALLY MILLIONS
of other people "showed" this before you did.
Google "Aczel" and "foundation" if you would like
a clue.
zuhair
All of this in not an answer to my question
My question is the following: IS there a general rule by which we can
know that the interesection of any two proper classes is a set or a
proper class.
and if there such a thing then what it is.
Your examples only proves that this intersection can be a set or a
proper class. I know that, my question was when C would be a set and
when C would be a proper class.
Zuhair
MoeBlee
> Jonathan Hoyle wrote:
> > > when two proper classes interesect. when do that intersection is a set?
> > > and when do that intersection is a proper? I mean is there general rule
> > > by which we can know the answer?
> >
> > Sometimes it's a proper class, sometimes it's a set, and sometimes it's
> > empty (hence a set). There is no general rule; however, anyone can
> > give you very simple examples of these:
> >
> > Suppose A and B are proper classes, and let C be the intersection of A
> > and B.
> >
> > 1. If A is the class of all sets of cardinality 1, and B is the class
> > of all sets of cardinality 2, then C is the empty set. (Obviously any
> > two disjoint proper classes for A and B would work here.)
> >
> > 2. If A is the class of all sets of cardinality 1, and B is the class
> > of all sets of cardinality 2 unioned with the set of integers, then C
> > is the set of integers.
> >
> > 3. If A is the class of all sets with cardinality <= 2, and B is the
> > class of all sets with cardinality >= 2, then C is the (proper) class
> > of all sets with cardinality 2.
>
> All of this in not an answer to my question
> My question is the following: IS there a general rule by which we can
> know that the interesection of any two proper classes is a set or a
> proper class.
He said, "There is no general rule". Don't you read what is written to
you?
MoeBlee
zuhair
Let Johanth. answer. anyhow if there is no general rule ,there is not
need to give examples.
>
> MoeBlee
MoeBlee
> Let Johanth. answer.
He did!
MoeBlee
Jonathan Hoyle
> > you?
>
> Let Johanth. answer. anyhow if there is no general rule ,there is not
> need to give examples.
As MoeBlee pointed out, I did answer that already: There is no general
rule.
It would be like asking "Is there a general rule for when the
intersection of two sets has cardinality 13?" There are uncountably
many ways for this to happen, and uncountably many ways for it not to
happen, and there is no simplistic rule which covers all possible
cases.
Is that clear enough?
Regards,
Jonathan Hoyle
Charlie-Boo
Do you need to have an empty set? Should that be possible?
Your idea is all fucked up - it eliminates the empty set and all of the
other familiar sets that we know and use. The idea of set axioms is to
allow us to define sets as we have grown to use them. It is well known
that the empty set is very useful (it is like 0 is in solving equations
in algebra.) It is one of the two trivial sets, with special
properties given by Rice.
You idea actually makes no sense. That is rare in life. Usually a
person has some sort of purpose and it makes sense to pursue one's
purposes in life. However, what is the purpose here? Making every set
contain itself does not serve any purpose that Mathematicians have ever
had. It is pure, unadulterated academic (acanemic) bullshit.
C-B
zuhair
no it cannot be possible you no that, since every set is a member of
itself, then there cannot be an empty set, this is clear.
>
> Your idea is all fucked up - it eliminates the empty set and all of the
> other familiar sets that we know and use.
True, but you can find equivalent sets to these sets you know, for
example the empty set is somehow equivalent to the singlton set in this
theory Q={Q}, since it can be regarded as empty of members other than
itself.
The idea of set axioms is to
> allow us to define sets as we have grown to use them.
It depends on the goal of the axioms, what if the set as we have grown
to use them are not enough to explain universal sets like the set of
all sets, the set of all ordinals, the set of all well founded
sets,etc...
It is well known
> that the empty set is very useful (it is like 0 is in solving equations
> in algebra.) It is one of the two trivial sets, with special
> properties given by Rice.
yes I know, but you can have its equivalent doing exactly the same uses
as your empty set.
>
> You idea actually makes no sense. That is rare in life. Usually a
> person has some sort of purpose and it makes sense to pursue one's
> purposes in life.
However, what is the purpose here? Making every set
> contain itself does not serve any purpose that Mathematicians have ever
> had.
what are U saying, this is wrong. The purpose I made it very clear. my
purpose it to understand universal sets, we cannot do that without
saying that every set is in itself.
You say why you don't go ask yourself, what raised Russell's paradox.
You will see that what raised it is the idea that there can be sets
that are not in themselfs,i.e all of their members are different from
them. getting rid of this property will remove this paradox whatsoever,
and enable us to traverse this unknown world of universal sets, that
ZFC cannot deal with it without encountering Russell's paradox.
It is difficult to construct a set theory that keep the sets we are
used to know ( ie the sets that are not in themselfs) and at the same
time have a universal set, without inflicting some restrictions on
separation. To have a full theory in which separation and unrestriced
comprehension exist and that can deal with these universal sets, then
we should say that all sets are in themselfs.
It is pure, unadulterated academic (acanemic) bullshit.
You are speaking of you misunderstanding of the matter, not of my
intentions.
Zuhair
Charlie-Boo
Jonathan Hoyle wrote:
> > Classes are non-sets.
>
> Incorrect. Classes include sets.
>
> > Non-sets in a set theory are non-sense.
>
> Again, untrue. There are Set theories that define the existence of
> proper classes.
So? Classes are a total waste of time. It just postpones the paradox.
What about the class of all classes that don't contain themselves?
The Russell Paradox is simply imposing too many (3 particular)
requirements on a system. Each is ok by itself, but they are
collectively incompatible.
Let your base of computing be YES(a,b). So M represents set YES(M,a)
and we ask if various expressions are representable. (In the Theory of
Computation, YES means "TM a halts yes on input b." and representable
means r.e. In Set Theory it means there is a set that contains just
the things for which the epression holds.)
1. YES(x,y)
2. P(x) => ~P(x)
3. P(x,y) => P(x,x)
where P(x) means P is representable in your base of computing (YES.)
Can you formalize Russell's Paradox, the Liar Paradox, and some Theory
of Computation theorems using the above?
C-B
> > There can be only one proper class or none, in a set theory it would be
> > a set.
>
> ??? The sentence makes no sense and is self-contradictory. A proper
> class is, by definition, not a set. And why should there be only one?
> If you remove a single element from a proper class, it still remains a
> proper class.
>
> > There is no class of classes in set theory with classes.
>
> Not in traditional set theories, but there are some non-well founded
> theories that do include the class of classes.
>
> > There is no universe in ZF.
>
> What do you mean by this statement? You repeat it often enough, you
> ought to finally make sense of it.
>
> > ZF is inconsistent.
>
> Which axiom in ZF do you think introduces the inconsistency? You make
> this claim often but (as usual) never manage to prove it.
>
> Jonathan Hoyle
> Eastman Kodak
MoeBlee
> Jonathan Hoyle wrote:
> > > Classes are non-sets.
> >
> > Incorrect. Classes include sets.
> >
> > > Non-sets in a set theory are non-sense.
> >
> > Again, untrue. There are Set theories that define the existence of
> > proper classes.
>
> So? Classes are a total waste of time. It just postpones the paradox.
> What about the class of all classes that don't contain themselves?
In theories such as NBG, there is no class x such that for all classes
y, y in x iff y not in y.
MoeBlee
Jonathan Hoyle
> What about the class of all classes that don't contain themselves?
It doesn't "just postpone the paradox" because proper classes are not
like sets and have different behavior. For example, although you can
have sets containing other sets, you cannot have proper classes
containing other proper classes. Classes can contain only sets (and
urelements), but never other proper classes. They are not a waste of
time, but they have some limitations in exchange for their power.
Jonathan Hoyle
Eastman Kodak
Charlie-Boo
Then why not change "class" to "set" and drop the notion of a class?
C-B
> MoeBlee
Jonathan Hoyle
Because classes have different behavior than sets. For example, any
set can be a member of a class, whereas it's not true that any class
can be a member of a set. All sets are classes, but not all classes
are sets. Sets are members of sets and proper classes, but proper
classes are not members of any set or proper class.
MoeBlee
It's not a matter of dropping notions. It's a matter of adopting
different axioms. Yes, there is an advantage in simplicity in not
having to qualify as to whether certain statements pertain to classes
in general or only to sets. On the other hand, having the existence of
proper classes makes certain other propostions easier to express since
we don't have to go out of the theory into the meta-theory to talk
about formulas but rather can talk directly about the classes.
Moreover, and in that vein, having proper classes permits a finite
axiomatization that is not possible for Z set theories.
Those and other ramifications of having proper classes or not having
proper classes are so commonly discussed in the basic introductory
literature of set theory that I don't know why you didn't already know
the answer to your question.
MoeBlee
Charlie-Boo
MoeBlee wrote:
> Charlie-Boo wrote:
> > MoeBlee wrote:
> > > Charlie-Boo wrote:
> > > > Jonathan Hoyle wrote:
> > > > > > Classes are non-sets.
> > > > >
> > > > > Incorrect. Classes include sets.
> > > > >
> > > > > > Non-sets in a set theory are non-sense.
> > > > >
> > > > > Again, untrue. There are Set theories that define the existence of
> > > > > proper classes.
> > > >
> > > > So? Classes are a total waste of time. It just postpones the paradox.
> > > > What about the class of all classes that don't contain themselves?
> > >
> > > In theories such as NBG, there is no class x such that for all classes
> > > y, y in x iff y not in y.
> >
> > Then why not change "class" to "set" and drop the notion of a class?
> >
> > C-B
>
> It's not a matter of dropping notions. It's a matter of adopting
> different axioms. Yes, there is an advantage in simplicity in not
> having to qualify as to whether certain statements pertain to classes
> in general or only to sets. On the other hand, having the existence of
> proper classes makes certain other propostions easier to express since
> we don't have to go out of the theory into the meta-theory to talk
> about formulas but rather can talk directly about the classes.
Examples?
> Moreover, and in that vein, having proper classes permits a finite
> axiomatization that is not possible for Z set theories.
>
> Those and other ramifications of having proper classes or not having
> proper classes are so commonly discussed in the basic introductory
> literature of set theory that I don't know why you didn't already know
> the answer to your question.
Because the books are wrong.
C-B
> MoeBlee
Charlie-Boo
Jonathan Hoyle wrote:
> > Then why not change "class" to "set" and drop the notion of a class?
>
> Because classes have different behavior than sets.
Then why not include cows - they're even more differenter?
C-B
Jonathan Hoyle
> > proper classes are so commonly discussed in the basic introductory
> > literature of set theory that I don't know why you didn't already know
> > the answer to your question.
>
> Because the books are wrong.
If you believe the books - being written by logicians and
mathematicians - are wrong, why did you think you'd get a different
answer from a newsgroup populated by logicians and mathematicians?
MoeBlee
> > It's not a matter of dropping notions. It's a matter of adopting
> > different axioms. Yes, there is an advantage in simplicity in not
> > having to qualify as to whether certain statements pertain to classes
> > in general or only to sets. On the other hand, having the existence of
> > proper classes makes certain other propostions easier to express since
> > we don't have to go out of the theory into the meta-theory to talk
> > about formulas but rather can talk directly about the classes.
>
> Examples?
What is the purpose of your question? To understand more concretely my
point? If so, just read from the first chapter of many a basic textbook
in set theory, since it is not very interesting for me to regurgitate
such basic information. But I will at least give you one salient
example to get you started: (1) Replacement is a schema in Z but a
single sentence in NBG.
> > Moreover, and in that vein, having proper classes permits a finite
> > axiomatization that is not possible for Z set theories.
> >
> > Those and other ramifications of having proper classes or not having
> > proper classes are so commonly discussed in the basic introductory
> > literature of set theory that I don't know why you didn't already know
> > the answer to your question.
>
> Because the books are wrong.
Since you are unfamiliar even with the basic relative advantages and
disadvantages of Z and NBG, it seems you never read these books that
you say are wrong.
MoeBlee
MoeBlee
Charlie-Boo wrote:
> Jonathan Hoyle wrote:
> > > Then why not change "class" to "set" and drop the notion of a class?
> >
> > Because classes have different behavior than sets.
>
> Then why not include cows - they're even more differenter?
Udderly idiotic.
MoeBlee
Charlie-Boo
Why do you believe that all authors of books on logic and all
participants in this newsgroup are competent, honest logicians and
mathematicians?
You need to be more of a scientist and think less about personalities
and titles. You could start by substantiating (self-contained) the
assertion that there is any sense in defining "classes".
BTW Are you a teacher of some sort (someone whose income relies on
publishing)?
"In questions of science, the authority of a thousand is not worth the
humble reasoning of a single individual." - Galileo Galilei
C-B
Jonathan Hoyle
> >
> > If you believe the books - being written by logicians and
> > mathematicians - are wrong, why did you think you'd get a different
> > answer from a newsgroup populated by logicians and mathematicians?
>
> Why do you believe that all authors of books on logic and all
> participants in this newsgroup are competent, honest logicians and
> mathematicians?
I don't have to believe that they are all competent and honest. I only
need to believe that some of them are, and observe that they all pretty
much agree on this point. Besides, I can and have read the proofs for
myself.
You act as if this were a contraversial topic. It is not, and has not
been for over a century. Although there will always be crackpots (such
as yourself), the professional mathematical and logical community
agrees. You can doubt the comptency and honesty of someone who claims
"2+2=4", but if the proof is valid, their personal competency and
honesty is irrelevant.
> You need to be more of a scientist and think less about personalities
> and titles. You could start by substantiating (self-contained) the
> assertion that there is any sense in defining "classes".
Correct. And it has been proven that NBG Set Theory (in which classes
are defined) is equi-consistent with ZFC, So it is valid.
> BTW Are you a teacher of some sort (someone whose income relies on
> publishing)?
No, although it would be irrelevant if I were.
Jonathan Hoyle
Eastman Kodak
Charlie-Boo
Jonathan Hoyle wrote:
> > > > Because the books are wrong.
> > >
> > > If you believe the books - being written by logicians and
> > > mathematicians - are wrong, why did you think you'd get a different
> > > answer from a newsgroup populated by logicians and mathematicians?
> >
> > Why do you believe that all authors of books on logic and all
> > participants in this newsgroup are competent, honest logicians and
> > mathematicians?
>
> I don't have to believe that they are all competent and honest. I only
> need to believe that some of them are, and observe that they all pretty
> much agree on this point.
Then why ask why I should get a different answer from this newsgroup if
your statement refers to only some of its participants?
How does their agreeing make them not wrong?
> You act as if this were a contraversial topic.
How did I act like that? I didn't say anything about popular
opinion.
> It is not, and has not been for over a century.
On what basis can you make that claim?
If there is no controversy, then why are there so many alternatives
(ZF, ZFC, NF, NFU, MK, NBG, etc.)? That is quite the opposite of your
claim of it not being controversial.
It is ironic that you would refer to "basic introductory literature
of set theory" when essentially all such discussions talk about the
various opposing opinions as to how to axiomatize set theory - a
direct contradiction to your claim that there is no controversy.
> Although there will always be crackpots (such as yourself),
I beg your pardon? What is that supposed to mean?
> the professional mathematical and logical community agrees.
What specifically demonstrates that?
> You can doubt the comptency and honesty of someone who claims "2+2=4",
When did I say that? What does 2+2=4 have to do with anything?
> but if the proof is valid, their personal competency and honesty is irrelevant.
What proof? Your suggestion, "Why did you think you'd get a
different answer from a newsgroup populated by logicians and
mathematicians?" doesn't refer to any particular proof, but rather
the collective thoughts of a group of people, which certainly would be
affected by incompetence and dishonesty.
> > You need to be more of a scientist and think less about personalities
> > and titles. You could start by substantiating (self-contained) the
> > assertion that there is any sense in defining "classes".
>
> Correct.
Yet you again refer to popular opinion - is that a basis for a
scientific conclusion? Trying to use popular opinion to discredit
someone's technical assertions is pure politics, a corruption of the
scientific method.
> And it has been proven that NBG Set Theory (in which classes
> are defined) is equi-consistent with ZFC, So it is valid.
Valid in what sense? Simply declaring that two things are equivalent
doesn't prove that either of them is of value.
> > BTW Are you a teacher of some sort (someone whose income relies on
> > publishing)?
>
> No, although it would be irrelevant if I were.
It is relevant to my own observations concerning corruption in the
publishing business.
BTW If anybody is interested in why logicians have had such a hard time
coming up with an axiomatization of set theory that they can agree on,
it is because they are misusing the principle of axiomatization.
An axiomatic system is a reflection of truth. It formalizes what we
already know. For example, we all know there is a zero and a successor
function. Peano's axioms formalize that fact. However, logicians
use an axiomatic system to make arbitrary declarations about sets that
are inconsistent from one axiomatization to the next. When you simply
declare that a set can or cannot be a member of itself, you are not
providing a basis for that assertion. Sure, axioms are given and not
proven. But if the system is to be useful, the axioms should be
consistent with what is known, not arbitrary declarations (varying from
one system to the next) because the author can't decide what is true
and what isn't.
Another problem is that any traditional axiomatization will only define
an aleph-0 number of sets, because there are only that many possible
expressions. To truly encompass the intuitive notion of sets (of which
there are more than aleph-0), we must refer to a higher order object.
For example, the axioms could refer to "functions" and have a
symbol for function, which represents a higher order number of
possibilities not limited to the aleph-0 number of expressions.
You have still not justified the use of "classes". They are just a
kludge introduced to avoid problems with definitions of sets. (What do
they correspond to? Nothing.) But you have the same problems with
classes: Do the classes that are not elements of themselves form a
class? Then they make more arbitrary declarations, about "classes"
(and now they're talking about something totally contrived, having no
intuitive counterpart) and they are back to grasping at straws, rather
than really determining the truth about sets.
C-B
> Jonathan Hoyle
> Eastman Kodak
Jonathan Hoyle
> > need to believe that some of them are, and observe that they all pretty
> > much agree on this point.
>
> Then why ask why I should get a different answer from this newsgroup if
> your statement refers to only some of its participants?
They don't refer to just some of them. All professional mathematicians
agree on this point.
> How does their agreeing make them not wrong?
Because they all have read the proofs.
> > You act as if this were a contraversial topic.
>
> How did I act like that? I didn't say anything about popular
> opinion.
>
> > It is not, and has not been for over a century.
>
> On what basis can you make that claim?
On the basis that the entirity of the professional mathematical and
logical community agree.
> If there is no controversy, then why are there so many alternatives
> (ZF, ZFC, NF, NFU, MK, NBG, etc.)? That is quite the opposite of your
> claim of it not being controversial.
You seem to think that because there are multiple axiomatic systems fo
set theory that therefore there must be some contraversy. Nothing can
be further from the truth. Professional logicians are may work in any
one of them. Just as there exists both Euclidean and Non-Euclidean
Geometries, mathematicians may happily work in either. Different
axioms yield different results. That is not contraversial in the
least.
> It is ironic that you would refer to "basic introductory literature
> of set theory" when essentially all such discussions talk about the
> various opposing opinions as to how to axiomatize set theory - a
> direct contradiction to your claim that there is no controversy.
Never once have I alluded to "various opposing opinions as to how to
axiomatize set theory". That different axiomatic systems exist is
true. However, there are "no opposing opinions" about them. Logicians
and mathematicians agree that they are each valid in their own system.
> > Although there will always be crackpots (such as yourself),
>
> I beg your pardon? What is that supposed to mean?
That you are a crank. For more detailed information on what that
means, go to http://www.crank.net and look up the section on
Mathematical cranks.
> > the professional mathematical and logical community agrees.
>
> What specifically demonstrates that?
All of their publications for over a century.
> Yet you again refer to popular opinion - is that a basis for a
> scientific conclusion?
I am not referring to popular opinion. I am referring to mathematical
proofs that have been confirmed by the professional mathematical
community en masse.
<remaining rant snipped>
Jonathan Hoyle
Eastman Kodak
Charlie-Boo
> > > I don't have to believe that they are all competent and honest. I only
> > > need to believe that some of them are, and observe that they all pretty
> > > much agree on this point.
> >
> > Then why ask why I should get a different answer from this newsgroup if
> > your statement refers to only some of its participants?
>
> They don't refer to just some of them. All professional mathematicians
> agree on this point.
How do you know what all of them believe?
If, as you say, you don't have to believe they are all competent and
honest, then they might as well be stupid liars, so their opinion is of
limited value, isn't it?
> > How does their agreeing make them not wrong?
>
> Because they all have read the proofs.
But can't a person read a proof and then draw erroneous conclusions
about it?
What prompted the development of ZFC in the first place?
> > > You act as if this were a contraversial topic.
> >
> > How did I act like that? I didn't say anything about popular
> > opinion.
> >
> > > It is not, and has not been for over a century.
> >
> > On what basis can you make that claim?
>
> On the basis that the entirity of the professional mathematical and
> logical community agree.
How do you know that they all agree?
I'm a Mathematician and I don't agree.
> > If there is no controversy, then why are there so many alternatives
> > (ZF, ZFC, NF, NFU, MK, NBG, etc.)? That is quite the opposite of your
> > claim of it not being controversial.
>
> You seem to think that because there are multiple axiomatic systems fo
> set theory that therefore there must be some contraversy. Nothing can
> be further from the truth.
If there is controversy as to what the solution should be, then people
would propose alternate solutions to what has been published,
wouldn't they? How is the current situation any different from that?
> Professional logicians are may work in any
> one of them. Just as there exists both Euclidean and Non-Euclidean
> Geometries,
Those are different theories, not different axiomatizations of the same
theory.
> > It is ironic that you would refer to "basic introductory literature
> > of set theory" when essentially all such discussions talk about the
> > various opposing opinions as to how to axiomatize set theory - a
> > direct contradiction to your claim that there is no controversy.
>
> Never once have I alluded to "various opposing opinions as to how to
> axiomatize set theory".
Who said you did?
> That different axiomatic systems exist is
> true. However, there are "no opposing opinions" about them. Logicians
> and mathematicians agree that they are each valid in their own system.
If the problem is adequately solved, why do they need to come up with
additional solutions?
> > > Although there will always be crackpots (such as yourself),
> >
> > I beg your pardon? What is that supposed to mean?
>
> That you are a crank. For more detailed information on what that
> means, go to http://www.crank.net and look up the section on
> Mathematical cranks.
What have you seen that makes you invoke that notion?
> > > the professional mathematical and logical community agrees.
> >
> > What specifically demonstrates that?
>
> All of their publications for over a century.
Have you read them all?
> > Yet you again refer to popular opinion - is that a basis for a
> > scientific conclusion?
>
> I am not referring to popular opinion. I am referring to mathematical
> proofs that have been confirmed by the professional mathematical
> community en masse.
Isn't that just an expression of their opinion? Isn't it possible
(in principle) that someone might disagree?
"In philosophical discussion, the merest hint of dogmatic certainty as
to finality of statement is an exhibition of folly." - Alfred North
Whitehead
Do you believe Whitehead?
> <remaining rant snipped>
You wrote, "And it has been proven that NBG Set Theory (in which
classes are defined) is equi-consistent with ZFC, So it is valid."
I replied, "Valid in what sense? Simply declaring that two things
are equivalent doesn't prove that either of them is of value."
How is that a rant? Is asking a question a rant?
Does simply declaring that two things are equivalent in fact prove that
either of them is of value?
Jonathan Hoyle
> Jonathan Hoyle wrote:
> > > > I don't have to believe that they are all competent and honest. I only
> > > > need to believe that some of them are, and observe that they all pretty
> > > > much agree on this point.
> > >
> > > Then why ask why I should get a different answer from this newsgroup if
> > > your statement refers to only some of its participants?
> >
> > They don't refer to just some of them. All professional mathematicians
> > agree on this point.
>
> How do you know what all of them believe?
Because mathematical proofs are not subject to opinion. A
mathematician believes in what has been proven. There may be some who
will agree or disagree on whaat has not been proven. For example,
although the majority believe that Goldbach's Conjecture is true, I
imagine there are those who do not. But all agree that a proof one way
or the other will decide the matter entirely.
> If, as you say, you don't have to believe they are all competent and
> honest, then they might as well be stupid liars, so their opinion is of
> limited value, isn't it?
>
> > > How does their agreeing make them not wrong?
> >
> > Because they all have read the proofs.
>
> But can't a person read a proof and then draw erroneous conclusions
> about it?
An untrained person, sure. But mathematicians are trained to read and
understand proofs. That is not to say that one can make a mistake. In
1994, Wiles' initial proof of Fermat's Last Theorem had a couple of
minor mistakes in it, which he subsequently rectified, thanks to the
help of other mathematicians. One person can, and will, make a
mistake. But the community as a whole is there to help peer review
proofs to prevent them.
Note that Wiles' FLT proof is many orders of magnitude more difficult
than the things we aare talking about here. And undergrad
upperclassman can follow the proofs we are discussing here.
> What prompted the development of ZFC in the first place?
The desire to remove the non-rigorous thinking of sets that led to
paradoxes. By axiomitizing the assumptions specifically (and different
theories can have different axioms), rigor is once again restored.
> > > > You act as if this were a contraversial topic.
> > >
> > > How did I act like that? I didn't say anything about popular
> > > opinion.
> > >
> > > > It is not, and has not been for over a century.
> > >
> > > On what basis can you make that claim?
> >
> > On the basis that the entirity of the professional mathematical and
> > logical community agree.
>
> How do you know that they all agree?
Because I can read.
> I'm a Mathematician and I don't agree.
On what basis do you claim to be a mathematician? I find it hard to
believe that a mathmatician can make the most obvious errors in
thinking which you have.
> > You seem to think that because there are multiple axiomatic systems fo
> > set theory that therefore there must be some contraversy. Nothing can
> > be further from the truth.
>
> If there is controversy as to what the solution should be, then people
> would propose alternate solutions to what has been published,
> wouldn't they? How is the current situation any different from that?
You need a refresher course in logic. Here, let me help you using your
own statements:
Let A = "there is controversy as to what the solution should be"
Let B = "people would propose alternate solutions to what has been
published"
You appear to be trying to use B to prove A. But it is a logical
fallacy to argue "If A then B" and "B" therefore "A". A is simply
false. The fact that B is true says nothing about A.
> > Professional logicians are may work in any
> > one of them. Just as there exists both Euclidean and Non-Euclidean
> > Geometries,
>
> Those are different theories, not different axiomatizations of the same
> theory.
Untrue. Although ZFC and NBG are equi-consistent, they are not the
same theories. For example, in NBG, the class of all ordinals exists,
whereas in ZFC it does not. It appears you are confusing the "model"
with the "theory".
> > That different axiomatic systems exist is
> > true. However, there are "no opposing opinions" about them. Logicians
> > and mathematicians agree that they are each valid in their own system.
>
> If the problem is adequately solved, why do they need to come up with
> additional solutions?
You claim to be a mathematician and yet ask such a question? There are
many reasons to find additional solutions to the same problem, as
different proofs may hilight other areas in an instructive way.
Besides, proving that NBG is equi-consistent to ZFC is not just
"another solution to the same problem". It also yields results in a
metalogic framework when comparing different axiomatic systems.
> > > > Although there will always be crackpots (such as yourself),
> > >
> > > I beg your pardon? What is that supposed to mean?
> >
> > That you are a crank. For more detailed information on what that
> > means, go to http://www.crank.net and look up the section on
> > Mathematical cranks.
>
> What have you seen that makes you invoke that notion?
A previous post of your when you stated "the books are wrong" when
referring to the standard introductory texts on logic. One of the
hallmarks of a crank is to believe that the entirity of the
professional community is wrong, and only he is right.
A typical crankish response at this point might be for you to comment
on how Newton and Einstein were once thought incorrect, and that they
were later vindicated, and thus by extension, so will you be.
> > > > the professional mathematical and logical community agrees.
> > >
> > > What specifically demonstrates that?
> >
> > All of their publications for over a century.
>
> Have you read them all?
No, nor is it necessary to. As stated above, a mathematical proof is
not subject to opinion; it does not matter how many places it is
printed.
If you believe a particular proof in Set Theory is incorrect, please
state which theorem you have in mind, and where you believe the error
occurs. This is ultimately the final issue. (Vague rants about how
the textbooks are incorrect, notwithstanding.)
> > > Yet you again refer to popular opinion - is that a basis for a
> > > scientific conclusion?
> >
> > I am not referring to popular opinion. I am referring to mathematical
> > proofs that have been confirmed by the professional mathematical
> > community en masse.
>
> Isn't that just an expression of their opinion? Isn't it possible
> (in principle) that someone might disagree?
Please give me an example of a mathematical proof which is under
disagreement by the mathematical community.
> "In philosophical discussion, the merest hint of dogmatic certainty as
> to finality of statement is an exhibition of folly." - Alfred North
> Whitehead
>
> Do you believe Whitehead?
I do. And it might even be relevant if we were having a philosophical
discussion. As it is, this is a mathematical one. Furthermore, it is
mathematical proof, not dogma, which you appear to be arguing against.
Again, I must insist: what theorem(s) do you think are incorrect and
why? Failing to answer would make you failing to observe Whitehead's
warning, as you would be following your own dogma.
> You wrote, "And it has been proven that NBG Set Theory (in which
> classes are defined) is equi-consistent with ZFC, So it is valid."
>
> I replied, "Valid in what sense? Simply declaring that two things
> are equivalent doesn't prove that either of them is of value."
This is true. However, I was not attempting to use that argument as a
proof of ZFC's consistency. rather it is a response to your previous
statement "You could start by substantiating (self-contained) the
assertion that there is any sense in defining 'classes'." ZFC has no
notion of classes, whereas NBG does. By demonstrating the
equi-consistency between ZFC (which has no classes) and NBG (which
does), this is proof that the inclusion of classes by themselves does
no harm.
Let's cut to the chase: perhaps you believe that ZFC is inconsistent?
If so, which of the ZFC axioms do you think introduces the
inconsistency? Here is a link to a list of the ZFC axioms, in case you
do not remember them:
http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html
If you are not saying ZFC is inconsistent, what exactly are you whining
about? Please be specific.
Jonathan Hoyle
Eastman Kodak
Jonathan Hoyle
so I will respond to it here.
> BTW If anybody is interested in why logicians have had such a hard time
> coming up with an axiomatization of set theory that they can agree on,
> it is because they are misusing the principle of axiomatization.
Firstly, logicians have not had a hard time at all. Secondly, they are
not in disagreement. Different theories with different axioms yield
different results. As long as the system is not inconsistent, there
are a number of models for most of the common theories.
Axiomization is not a "principle" which can be "misused". You simply
list your full set of assumptions (in the language of logic) and see
what theorems result. Some axiomizations may yield interesting
results, whereas others are not so interesting. In any case, you
merely get what you ask for.
> An axiomatic system is a reflection of truth.
It can be, but it doesn't have to be. For example, I "believe" in the
full Axiom of Choice from a Platonist prepective, although there are
others who do not. Regardless, we all can happily examine results from
ZFC or ZF+~AC or any number of differing systems. We simply agree that
if these are our starting assumptions, those are the results we get.
No value judgement necessary.
> It formalizes what we already know.
Usually not. By convention, we typically attempt to minimize the set
of axioms we use to remove redundancies. For example, we may already
"know" that 2+2=4, but it is pointless to make "2+2=4" one of our
axioms. Instead, we choose more basic fundamental principles as
axioms, and let 2+2=4 be a theorem.
> For example, we all know there is a zero and a successor
> function. Peano's axioms formalize that fact. However, logicians
> use an axiomatic system to make arbitrary declarations about sets that
> are inconsistent from one axiomatization to the next. When you simply
> declare that a set can or cannot be a member of itself, you are not
> providing a basis for that assertion.
If there were a provable "basis", then we would let that be a theorem,
not an axiom.
> Sure, axioms are given and not proven.
That is the point.
> But if the system is to be useful, the axioms should be
> consistent with what is known, not arbitrary declarations (varying from
> one system to the next) because the author can't decide what is true
> and what isn't.
Typically, you do want to choose axioms which are intuitively obvious.
But our intuition is not flawless, and we often make errors when we
rely upon it more than a mathematical proof. Set Theory is the perfect
example of that. Naive Set Theory, using only our intuition and prior
to axiomization, led to a number of inconsistencies. It is only when
mathematicians more rigorously questioned their assumptions by listing
them axiomatically, did Set Theory become rigorous.
Besides, what does it mean to be "true" or "false"? Is Euclid's
Parallel Postulate true or false? It seems neither to me: if you
assume it's true, you get Euclidean Geometry; if you assume it's false,
you get one of the Non-Euclidean Geometries? What about the Continuum
Hypothesis? There are models in which it is true, and models in which
it is false. What about the Axiom of Choice? Or the Axiom of
Foundation? If you have a strong opinion one way or the other, then
perhaps you should study the results of an axiomatic system with your
assumptions. But that doesn't take away from the other systems.
> Another problem is that any traditional axiomatization will only define
> an aleph-0 number of sets, because there are only that many possible
> expressions.
I am not sure I understand this part. The real line contains 2^aleph-0
points, and each point can be in a singleton set, and thus at least
2^aleph-0 sets. The power set of the reals makes it even larger:
2^2^aleph-0. And so it goes.
Are you making reference to Skolem's Theorem here that there are always
countaable models? Although true, that doesn't in any way discount the
existence of uncountable models.
> You have still not justified the use of "classes".
Classes do not need justification. Classes do not exist in ZFC, but do
in NBG. If by justification, perhaps you want a proof that they do not
introduce an inconsistency? Well, as stated in another post, NBG is
proven to be equi-consistent with ZFC (which has no classes). That is
not to say NBG is consistent; it merely means that it was ZFC, not the
added classes, which caused the problem.
> They are just a kludge introduced to avoid problems with definitions
> of sets. (What do they correspond to? Nothing.)
Incorrect. The definition of "set" is not altered in the slightest.
Proper classes simply allow you to speak of things that you were not
able to before. For example, in NBG you can speak of the class of all
ordinals, the class of all cardinals, or even the class of all sets
(you cannot do any such thing in ZFC).
> But you have the same problems with classes: Do the classes that
> are not elements of themselves form a class?
No class can be an element of itself. In exchange for the power that
proper classes give you, you have some limitations in their use (at
least within NBG). For example, classes (including sets) may contain
only sets or urelements. No class may contain a proper class as one of
its elements.
(As someone who claims to be a mathematician, how is it that you do not
know this?)
> Then they make more arbitrary declarations, about "classes"
> (and now they're talking about something totally contrived, having no
> intuitive counterpart) and they are back to grasping at straws, rather
> than really determining the truth about sets.
If your complaint is with regard to wanting sets to contain themselves
as members, there are set theories that certainly do allow this.
Neither ZFC or NBG allow this by the Axiom of Foundation (or Axiom of
Regularity as it is sometimes called); however, there are some Non-Well
Founded Set Theories that certainly do allow sets to contain themselves
as members. See this link for more information:
CBT...@gmail.com
Jonathan Hoyle wrote:
> Hi, Charlie. For some reason, I missed the bottom part of this post,
> so I will respond to it here.
>
> > BTW If anybody is interested in why logicians have had such a hard time
> > coming up with an axiomatization of set theory that they can agree on,
> > it is because they are misusing the principle of axiomatization.
>
> Firstly, logicians have not had a hard time at all. Secondly, they are
> not in disagreement. Different theories with different axioms yield
> different results. As long as the system is not inconsistent, there
> are a number of models for most of the common theories.
>
> Axiomization is not a "principle" which can be "misused".
Impossible to misuse? I'll have to remember that the next time someone
(you perhaps?) says I'm misusing it.
If all of the Mathematicians agree, then what is the consensus
(unanimity) regarding the answer to the question of whether the Axiom
of Choice should indeed be an axiom?
(The rest of this message is an assemblage of miscommunications.)
C-B
CBT...@gmail.com
Jonathan Hoyle wrote:
> Hi, Charlie. For some reason, I missed the bottom part of this post,
> so I will respond to it here.
>
> > BTW If anybody is interested in why logicians have had such a hard time
> > coming up with an axiomatization of set theory that they can agree on,
> > it is because they are misusing the principle of axiomatization.
>
> Firstly, logicians have not had a hard time at all. Secondly, they are
> not in disagreement. Different theories with different axioms yield
> different results. As long as the system is not inconsistent, there
> are a number of models for most of the common theories.
>
> Axiomization is not a "principle" which can be "misused".
Impossible to misuse? I'll have to remember that the next time someone
(you perhaps?) says I'm misusing it.
If all of the Mathematicians agree, then what is the consensus
(unanimity) regarding the answer to the question of whether the Axiom
of Choice should indeed be an axiom?
(The rest of this message is an assemblage of miscommunications.)
C-B
> You simply
Jonathan Hoyle
> (unanimity) regarding the answer to the question of whether the Axiom
> of Choice should indeed be an axiom?
That's easy:
When a result in mathematics you wish to show requires AC, then AC
should be called out as an axiom. [One common example is the
contruction of a non-measurable set.] In certain areas of mathematics,
it is conventional to assume AC right from the beginning, so no such
specific reference need be taken.
Likewise, when a result you wish to show requires ~AC, you likewise
must indicate that you are assuming one of the formulations of ~AC as
an axiom. Although this is less common, you will see this from time to
time. [The construction of an infinite Dedekind-finite set would be an
example.]
There's no debate here. Everyone agrees. It's kind of like baseball
with the DH rule. You can play with it or without it, you just need a
convention to follow with regard to it. You may have a preference one
way or the other, and that it certainly your choice. However, it does
not mean that all baseball games which follow a DH rule different from
your own, is somehow contraversial.
Jonathan Hoyle
Eastman Kodak
Charlie-Boo
If it were contraversial, how would it be different than the way it is
now?
Charlie-Boo
> When a result in mathematics you wish to show requires AC, then AC
> should be called out as an axiom. [One common example is the
> contruction of a non-measurable set.]
> when a result you wish to show requires ~AC, you likewise
> must indicate that you are assuming one of the formulations of ~AC
as
> an axiom. [The construction of an infinite Dedekind-finite set
would be an
> example.]
That is being inconsistent and shows the controversy - people cannot
decide on what Set Theory should be. There is the informal intuitive
notion of a set, but when people try to formalize it they get
paradoxes/inconsistencies in the system itself, as well as
inconsistency in the reactions as to what to do to try to fix it.
> There's no debate here. Everyone agrees.
Everyone agrees to be inconsistent? You are not agreeing with yourself
from one moment to the next. First you say assume AC for one theorem,
and then you say to assume ~AC for another theorem. So you end up with
a collection of theorems that have contradictory premises. Great!
Controversy: "A dispute, especially a public one, between sides
holding opposing views."
The opposing views are regarding whether AC is true, false, an axiom,
not an axiom, a special axiom, etc. For examples, see your words
above.
> it does not mean that games which follow a rule different from
> your own is somehow contraversial.
"1904/1908 Zermelo introduces axioms of set theory, explicitly
formulates AC and uses it to prove well-ordering theorem, thereby
rising a storm of controversy."
Concerning the Axiom of Choice, John L. Bell
"This fine book provides a full account of the various controversies.
It is interesting, comprehensive and exact."
Zermelo's Axiom of Choice: Its Origins, Development, and
Influence", Gregory H. Moore, American Mathematical Monthly 91,10
(Dec. 1984), pp 654-662 - Review by Robert Bunn, University of
British Columbia
Jonathan Hoyle
> > should be called out as an axiom. [One common example is the
> > contruction of a non-measurable set.]
>
> > when a result you wish to show requires ~AC, you likewise
> > must indicate that you are assuming one of the formulations of ~AC
> > as an axiom. [The construction of an infinite Dedekind-finite set
> > would be an example.]
>
> That is being inconsistent and shows the controversy - people cannot
> decide on what Set Theory should be.
Completely untrue. Do you think mathematicians are being
"inconsistent" and "show controversy" because both Euclidean and
Non-Euclidean geometries are taught? It's the same thing here, except
that it's the Parallel postulate, instead of the Axiom of Choice, which
can go either way. Are baseball players being "inconsistent" and "show
controversy" because there is both an American League and National
League which differ on the designated hitter?
> There is the informal intuitive
> notion of a set, but when people try to formalize it they get
> paradoxes/inconsistencies in the system itself, as well as
> inconsistency in the reactions as to what to do to try to fix it.
Just the opposite is true. Paradoxes arose when the "informal
intuitive notion of a set" was used, and were resolved only once the
system was formalized.
> > There's no debate here. Everyone agrees.
>
> Everyone agrees to be inconsistent? You are not agreeing with yourself
> from one moment to the next.
No, everyone agrees to call out the assumption being used.
> First you say assume AC for one theorem,
> and then you say to assume ~AC for another theorem. So you end up with
> a collection of theorems that have contradictory premises. Great!
Having different theorems resulting from different premises is not
inconsistent. It is basic Aristotilean Logic.
> Controversy: "A dispute, especially a public one, between sides
> holding opposing views."
>
> The opposing views are regarding whether AC is true, false, an axiom,
> not an axiom, a special axiom, etc. For examples, see your words
> above.
These are not "opposing views" any more than Euclidean and
Non-Euclidean Geometries are "opposing". they are equally valid
systems and can be studied in parallel (no pun intended).
> > it does not mean that games which follow a rule different from
> > your own is somehow contraversial.
>
> "1904/1908 Zermelo introduces axioms of set theory, explicitly
> formulates AC and uses it to prove well-ordering theorem, thereby
> rising a storm of controversy."
It is no longer 1908. The contraversy was resolved when it was proven
independent by Fraenkel. Just because something was controversial a
hundred years ago does not imply it is still so today.
You'll note that the "controversy" of whether or not the world is round
has also been resolved. (At least to those of us who wish to be in the
21st century. I guess you'll have to decide for yourself.)
Jonathan Hoyle
Eastman Kodak
Virgil
"Charlie-Boo" <shyma...@gmail.com> wrote:
> Jonathan Hoyle wrote:
>
> > When a result in mathematics you wish to show requires AC, then AC
> > should be called out as an axiom. [One common example is the
> > contruction of a non-measurable set.]
>
> > when a result you wish to show requires ~AC, you likewise
> > must indicate that you are assuming one of the formulations of ~AC
> as
> > an axiom. [The construction of an infinite Dedekind-finite set
> would be an
> > example.]
>
> That is being inconsistent and shows the controversy - people cannot
> decide on what Set Theory should be.
Producing different theorems for different set theories is no more
inconsistent than producing different theorems for different geometries.
Virgil
"Jonathan Hoyle" <jonh...@mac.com> wrote:
> > > When a result in mathematics you wish to show requires AC, then AC
> > > should be called out as an axiom. [One common example is the
> > > contruction of a non-measurable set.]
> >
> > > when a result you wish to show requires ~AC, you likewise
> > > must indicate that you are assuming one of the formulations of ~AC
> > > as an axiom. [The construction of an infinite Dedekind-finite set
> > > would be an example.]
> >
> > That is being inconsistent and shows the controversy - people cannot
> > decide on what Set Theory should be.
>
> Completely untrue. Do you think mathematicians are being
> "inconsistent" and "show controversy" because both Euclidean and
> Non-Euclidean geometries are taught? It's the same thing here, except
> that it's the Parallel postulate, instead of the Axiom of Choice, which
> can go either way. Are baseball players being "inconsistent" and "show
> controversy" because there is both an American League and National
> League which differ on the designated hitter?
>
> > There is the informal intuitive
> > notion of a set, but when people try to formalize it they get
> > paradoxes/inconsistencies in the system itself, as well as
> > inconsistency in the reactions as to what to do to try to fix it.
>
> Just the opposite is true. Paradoxes arose when the "informal
> intuitive notion of a set" was used, and were resolved only once the
> system was formalized.
Minor point, but at least one definition of "paradox allows true
statements to be paradoxical:
<quote>
Websters Concise Electronic Dictionary
4 sense(s) for paradox
1. paradox
(noun) [paradoxes]
statement that seems contrary to common sense yet is perhaps true
<end quote>
It is outright "contradictions" which the formalizations are designed to
eliminate. and have so far seemed to be successful in doing so.
> > The opposing views are regarding whether AC is true, false, an axiom,
> > not an axiom, a special axiom, etc. For examples, see your words
> > above.
>
> These are not "opposing views" any more than Euclidean and
> Non-Euclidean Geometries are "opposing". they are equally valid
> systems and can be studied in parallel (no pun intended).
>
> > > it does not mean that games which follow a rule different from
> > > your own is somehow contraversial.
> >
> > "1904/1908 Zermelo introduces axioms of set theory, explicitly
> > formulates AC and uses it to prove well-ordering theorem, thereby
> > rising a storm of controversy."
>
> It is no longer 1908. The contraversy was resolved when it was proven
> independent by Fraenkel. Just because something was controversial a
> hundred years ago does not imply it is still so today.
>
> You'll note that the "controversy" of whether or not the world is round
> has also been resolved. (At least to those of us who wish to be in the
> 21st century. I guess you'll have to decide for yourself.)
Is Hoyle now being spokesman for the Flat Earth Society?
Jonathan Hoyle
LOL, I guess I should be careful in the event one takes my remarks that
way. :-)
(I have to admit, of all cranks, the Flat Earth-ers have to be my
favorite...if for no other reason than I am curious as to how - in this
day and age - they are able to maintain its credibility even in their
own minds.)
MoeBlee
> That is being inconsistent and shows the controversy - people cannot
> decide on what Set Theory should be. There is the informal intuitive
> notion of a set, but when people try to formalize it they get
> paradoxes/inconsistencies in the system itself, as well as
> inconsistency in the reactions as to what to do to try to fix it.
Different people prefer different theories, some of those theories
contradicting one another. But that is vastly different with there
being any one of those theories itself being inconsistent.
You are utterly confused about all of this.
> Everyone agrees to be inconsistent? You are not agreeing with yourself
> from one moment to the next. First you say assume AC for one theorem,
> and then you say to assume ~AC for another theorem. So you end up with
> a collection of theorems that have contradictory premises. Great!
You're trolling, willfully obtuse, or stupid.
We do NOT adopt as a theory for working mathematics a "collection of
theorems" from two different axiomatizations that have negations of
each others' axioms. What we do is study separately such different
theories and compare their similarities and differences.
MoeBlee