The Set of Everything
Toby Smith
exist. Can anyone provide me with a proof?
cheers
John Jones
cannot
> exist. Can anyone provide me with a proof?
I challenge your presumption to a greater meaning to be found because of a
'mathmatical' treatment or proof.
..What proof would you have in mind?
Lets find out..
..'the set of everything'. Doesn't that presume a limitless creation of
objects? We create objects, or define them, when we want. 'The set of
everything' is a set of an act, the set of the 'act of creating objects',
and that is not a set of everything. So the 'set of everything' is not a
set.
But how do you expect that a mathmatical proof 'of' that, explains itself
better? except 'by' that?
JJ
Toby Smith <chocolat...@gmx.net> wrote in message
news:U%Lha.142$ZK2...@newsfep3-gui.server.ntli.net...
Toby Smith
the problem, as part of a course I am on.
TS
"John Jones" <scooby...@btopenworld.com> wrote in message
news:b683k9$oj6$1...@sparta.btinternet.com...
John Jones
cannot
> exist. Can anyone provide me with a proof?
I challenge your presumption to a greater meaning to be found because of a
'mathmatical' treatment or proof.
..What proof would you have in mind?
Lets find out..
..'the set of everything'. Doesn't that presume a limitless creation of
objects? We create objects, or define them, when we want. 'The set of
everything' is a set of an act, the set of the 'act of creating objects',
and that is not a set of everything. So the 'set of everything' is not a
set.
But how do you expect that a mathmatical proof 'of' that, explains itself
better? except 'by' that?
JJ
Toby Smith <chocolat...@gmx.net> wrote in message
news:U%Lha.142$ZK2...@newsfep3-gui.server.ntli.net...
John Jones
over its own intuitional grounds.
Its that grim.
JJ
Toby Smith <chocolat...@gmx.net> wrote in message
news:8HMha.151$ZK2...@newsfep3-gui.server.ntli.net...
Barb Knox
<chocolat...@gmx.net> wrote:
Clearly, if V is the set of everything then V is itself a member of V.
Now, most common formulations of set theory have an "axiom of foundation",
which says that NO set is an element of itself.
Without this, or something like it, you end up with Russell's paradox:
Let P be the set of all sets that are NOT members of themself. Then is P
a member of itself? By definition, it is if it isn't and it isn't if it
is.
BTW, there are other ways around Russell's paradox. For example,
"non-well-founded" set theories do allow a set to be a member of itself,
if that's what you really need.
--
---------------------------
| BBB b \ barbara minus knox at iname stop com
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
Philipp Frank
Toby Smith schrieb im Artikel <U%Lha.142$ZK2...@newsfep3-gui.server.ntli.net>:
> I'm having trouble following why the mathematical set of everything cannot
> exist. Can anyone provide me with a proof?
> cheers
You'll have to work in some axiom system for a rigorous proof, and the
most common (the one every mathematician uses) is ZFC (for Zermelo,
Fraenkel and the axiom of Choise). It's easy to prove it here:
If a is aset, define r(a) as the set of all x in a such that x is not a
member of itself (ZFC doesn't forbid sets being elements of
themselves, but we're just interested in the one which are not here).
This is a set by the axiom of comprehension. Now suppose that r(a) is an
element of a. That means we have r(a)\notin r(a) if and only if r(a)\in
r(a). Contradiction, so r(a) is not an element of a, so a could not
possibly be the set of all sets, and since it's arbitrary, we're done.
Hope that was what you were looking for
Cheers
Philipp
G. Frege
wrote:
>
> I'm having trouble following why the mathematical set of everything cannot
> [or rather does not] exist. Can anyone provide me with a proof?
>
Depends on the particular set theory your are working in.
For ZF this is quite easy to see. Given some set U, with separation we could
build the set R = {x in U : ~(x in x)}. If we would consider now the question R
in R? we would get: R in R iff R in U & ~(R in R). Now assume R in U, then we
would have a contradiction: R in R iff ~(R in R). Hence (with propositional
logic) ~(R in U). With other words, there can't be a "set of everything".
F.
Saikat Guha
> I'm having trouble following why the mathematical set of everything cannot
> exist. Can anyone provide me with a proof?
> cheers
Every set has more subsets than elements. (From Cantor's theorem.)
But the "set of everything" would have all its subsets as elements,
and so would not have more subsets than elements. So there is no such
set. (QED)
Toby Smith
Excuse me? when did I say this?
But thanks to Barb Knox and Philipp Frank who have provided me with some
understandable reasoning.
TS
"John Jones" <scooby...@btopenworld.com> wrote in message
news:b686vo$aoj$1...@hercules.btinternet.com...
Mechanic
"Philipp Frank" <Philip...@web.de> wrote in message
news:gt296b...@192.168.0.2...
This is circular proof in the sense that you already define the set of
everything to be an antinomy and obviously you conclude a contradiction. But
the question was whether such a set can exists at all. Am I missing
something to we're living in a circle?
George Greene
: I'm having trouble following why the mathematical set of everything cannot
: exist. Can anyone provide me with a proof?
: cheers
Every set has more subsets than members.
--
---
"It's difficult ... you need to be united to have any
strength, but internal issues have to be addressed."
--- E. Ray Lewis, on liberalism in America
karl malbrain
No, you needn't miss anything. As I posted elsewhere in our group, PI
is now ONE, while the natural logarithms are put to bed, so to speak.
Do we live in a circle: yes, we do. THere is a great circle around
each and every one of us, but I can only speak for Stockton, CA in the
95204 area code.
Try re-fastening your SECONDARY CARRIER FACILITY. You'll find plenty
of MISSED SPECS in the NASA affair.
MALBRAIN, range-safety-officer, 1973, Houston, TX.
karl malbrain
karl malbrain, 735 N Tuxedo, Stockton, CA 95204
s...@sig.below (Barb Knox) wrote in message news:<see-310303...@192.168.1.2>...
karl malbrain
Now, that's a circle for you. OK, let's switch, again. This time the
CIRCLES are SQUARES, and the natural logarithms get to pass ZERO.
Malbrain
|-|erc
"George Greene" <gre...@eagle.cs.unc.edu> wrote
> "Toby Smith" <chocolat...@gmx.net> writes:
>
> : I'm having trouble following why the mathematical set of everything cannot
> : exist. Can anyone provide me with a proof?
> : cheers
>
> Every set has more subsets than members.
is that for finite sets?
if you could show the subsets are countable then the set exists.
EVERYTHING = {a,b,c....{a,b}{a,c}{b,c}....}
would imply
EVERYTHING = {a,b,c... {a,b} {a,c} {b,c}... {{a,b},{a,c}} {{a,b}{b,c}}...}
keeps growing might not be countable.
Herc
Nick Argall
You can have a set theory which is inconsistent (allows Russel's paradox) or
a set theory with is incomplete (forbids Russel's paradox). Mathematicians
dislike inconsistency more than they dislike incompleteness, and the rules
change that forbids Russel's paradox also forbids the Set of Everything.
Nick
karl malbrain
> "George Greene" <gre...@eagle.cs.unc.edu> wrote
> > "Toby Smith" <chocolat...@gmx.net> writes:
> >
> > : I'm having trouble following why the mathematical set of everything cannot
> > : exist. Can anyone provide me with a proof?
> > : cheers
> >
> > Every set has more subsets than members.
Try making simple COMBINITORICS a prerequisite from high-school.
Otherwise, read on:
An engineer by any other name
Legislature to decide if computer programmers can legally use the
title
By R.G. RATCLIFFE
Copyright 2003 Houston Chronicle Austin Bureau
AUSTIN -- One of the oddest battles of the 78th Legislature is pitting
Texas' licensed professional engineers against the high-tech
industry's software dudes.
At issue is just who in Texas can call himself an engineer.
"It's one of the silliest issues we're having to deal with this
session, but it's also one of the most important," said Steven Kester,
legislative director of the American Electronics Association, an
organization of computer companies.
Texas has one of the nation's strictest engineering practices acts and
limits the title of engineer to those people who have studied
engineering and passed a licensing exam.
And that law puts most of the "engineers" in the high-tech industry
out of the field. Kester said the restriction threatens high-tech
growth in Texas.
But Ken Rigsbee, chairman of the Texas Society of Professional
Engineers legislative committee, said the restriction is needed to
protect the public.
Rigsbee said state restrictions on who can call themselves engineers
were set up decades ago after someone misengineered a heating pipe
system at the New London Junior-Senior High School.
An explosion of natural gas in the pipe system killed 300 students and
teachers in 1937.
Rigsbee said the licensed professional engineers of Texas have been
protecting their title from encroachment ever since. There are 49,000
state-licensed professional engineers.
Rigsbee said the high-tech problem mostly involves computer
programmers whom the industry likes to call computer engineers.
Rigsbee said the industry holds out its products as having been
"engineered." And he said there is a belief that the computer
companies are in a better position to win contracts if they can say
they have 150 engineers on staff instead of 150 programmers.
"What we have a problem with is a graduate of a two-year computer
programming school or some technicians ... holding themselves out as
engineers when they clearly are not," Rigsbee said.
The computer industry had been happy to function under an exemption in
state law that allowed a company to call in-house personnel whatever
it wanted to so long as the engineering title was not held out to the
public.
But the Texas Board of Professional Engineers sent cease-and-desist
letters to some high-tech industry specialists who used the title of
engineer in correspondence.
That led to a request to former Attorney General John Cornyn to
clarify the issue. Cornyn last July said the matter is simple when it
comes to state law.
"The Texas Engineering Practice Act ... does not allow an in-house
employee of a private corporation, though classified internally as an
`engineer' or under another engineering title, to use the title
`engineer' on business cards, cover letters or other forms of
correspondence that are made available to the public," Cornyn said.
Boom. In a single sentence, the computer programming engineers of
Texas became software dudes.
Actually, while software programmers make up the bulk of the high-tech
industry's engineers, the industry also uses the title for electrical
and mechanical engineers not licensed by the state. Texas Instruments
also has "customer support engineers."
"Texas is becoming a laughingstock of the global high-technology
community," said Steve Taylor, director of corporate affairs for
Applied Materials.
Taylor said there are about 100,000 high-tech personnel in Texas who
have "engineer" in their title, but they are not licensed by the
state.
"They risk fines of up to $3,000 a day for handing out business cards
to a supplier or even dropping it in a fish bowl at a restaurant for a
chance at a free lunch," Taylor said.
AEA's Kester said electronics professionals from around the country
are called engineers within their firms and in the industry. Suddenly,
he said, they are now required to carry one set of business cards for
Texas and another for the other 49 states.
"It's a matter of professional pride," Kester said. "They've built up
a lot of experience and earned the title of engineer in their
industry."
Kester said the electronics industry has made changing the state law a
top priority because it is making it difficult to recruit employees
from other states and around the world.
"We run the risk of not having them move here," Kester said. "That
puts us at a significant disadvantage."
Legislation to loosen the title requirements is being carried by Sen.
Rodney Ellis, D-Houston, and Rep. Warren Chisum, R-Pampa.
George Greene
: "George Greene" <gre...@eagle.cs.unc.edu> wrote
: > "Toby Smith" <chocolat...@gmx.net> writes:
: >
: > : I'm having trouble following why the mathematical set of everything cannot
: > : exist. Can anyone provide me with a proof?
: > : cheers
: >
: > Every set has more subsets than members.
:
: is that for finite sets?
Yes, and for all other sets as well.
: if you could show the subsets are countable then the set exists.
If the original alleged universe was countably infinite, then each
individual subset is countable, but the class of all of them
is not countable.
: EVERYTHING = {a,b,c....{a,b}{a,c}{b,c}....}
:
: would imply
: EVERYTHING = {a,b,c... {a,b} {a,c} {b,c}... {{a,b},{a,c}} {{a,b}{b,c}}...}
:
: keeps growing might not be countable.
Indeed, it isn't countable.
Charlie-Boo
> "Toby Smith" <chocolat...@gmx.net> writes:
>
> : I'm having trouble following why the mathematical set of everything cannot
> : exist. Can anyone provide me with a proof?
> : cheers
>
> Every set has more subsets than members.
The fact that a (non-empty) set and its subsets are of different
cardinality doesn't mean you can't have a set with everything. You
simply cannot compare its elements to its subsets. The existance of
this set per se is not a problem. But without it, you do have
problems. If there is no universal set, then either you have no empty
set or you cannot in general take the complement of a set. You would
have to recheck all of your proofs!
Actually, every attempt to be as general as possible (Logic, Theory of
Computation, Set Theory, Mathematics, et. al.) includes the empty and
the universal object. The former is the wff FALSE that is never
provable, a program that consists of only HALT NO, or {}. The latter
is the wff TRUE, the program that is HALT YES, and the universal set.
Beyond that, read up on my axiomitization of computer science, and
you'll see how it all fits together.
Charlie Volkstorf
Cambridge, MA
Chris Menzel
> In article <U%Lha.142$ZK2...@newsfep3-gui.server.ntli.net>, "Toby Smith"
> <chocolat...@gmx.net> wrote:
>
> > I'm having trouble following why the mathematical set of everything cannot
> > exist. Can anyone provide me with a proof?
> > cheers
>
> Clearly, if V is the set of everything then V is itself a member of V.
> Now, most common formulations of set theory have an "axiom of foundation",
> which says that NO set is an element of itself.
>
> Without this, or something like it, you end up with Russell's paradox:
> Let P be the set of all sets that are NOT members of themself. Then is P
> a member of itself? By definition, it is if it isn't and it isn't if it
> is.
The presence or absence of foundation is irrelevant to Russell's
Paradox. (In fact, foundation was not even included in Zermelo's
original axiomatization of set theory.) Note in particular that the
axiom does not prevent you from defining P or asking whether or not P is
a member of itself -- the mere coherence of that question is all that's
necessary to set the paradox a-going. The culprit behind the paradox is
rather the principle that lets you define P in the first place, viz.,
the principle (usually called "naive comprehension") that for any
specifiable condition, there is a set of things satisfying that
condition. Plug in the condition "x is not a member of x" and out pops
P ex nihilo and we're off to the races. This is the principle that gets
clobbered in modern set theories and hence which blocks the familiar
routes to paradox.
Chris Menzel
Saikat Guha
> George Greene <gre...@eagle.cs.unc.edu> wrote in message news:<xesof3q...@eagle.cs.unc.edu>...
> > "Toby Smith" <chocolat...@gmx.net> writes:
> >
> > : I'm having trouble following why the mathematical set of everything cannot
> > : exist. Can anyone provide me with a proof?
> > : cheers
> >
> > Every set has more subsets than members.
>
> The fact that a (non-empty) set and its subsets are of different
> cardinality doesn't mean you can't have a set with everything. You
> simply cannot compare its elements to its subsets.
Eh? How is that supposed to work? If they have different
cardinalities, won't one be bigger than the other?
(There might be a class that has no power set, and whose subsets
thereby have no cardinality. Usually those things are called proper
classes, not sets. Maybe that's what you have in mind.)
Barb Knox
ch...@aol.com (Charlie-Boo) wrote:
> George Greene <gre...@eagle.cs.unc.edu> wrote in message
news:<xesof3q...@eagle.cs.unc.edu>...
> > "Toby Smith" <chocolat...@gmx.net> writes:
> >
> > : I'm having trouble following why the mathematical set of everything
cannot
> > : exist. Can anyone provide me with a proof?
> > : cheers
> >
> > Every set has more subsets than members.
>
> The fact that a (non-empty) set and its subsets are of different
> cardinality doesn't mean you can't have a set with everything. You
> simply cannot compare its elements to its subsets.
Huh? Every one of its subsets IS one of its elements, by most definitions
of "everything". Even if you are using some idiosyncratic definition of
"everything" here, I'm sure the OP wasn't.
And BTW, the fact that card(x) < card(2^x) is true for the empty set too,
so the "(non-empty)" qualification isn't needed.
[snip]
Charlie-Boo
> > > : exist. Can anyone provide me with a proof?
> > > : cheers
> > >
> > > Every set has more subsets than members.
> >
> > The fact that a (non-empty) set and its subsets are of different
> > cardinality doesn't mean you can't have a set with everything. You
> > simply cannot compare its elements to its subsets.
>
> Eh? How is that supposed to work? If they have different
> cardinalities, won't one be bigger than the other?
>
> (There might be a class that has no power set, and whose subsets
> thereby have no cardinality. Usually those things are called proper
> classes, not sets. Maybe that's what you have in mind.)
If I asked you to characterize the set of inputs for which the program
"HALT YES" halts yes, how would you describe it? Or the values of x
for which the wff x=x is true (or provable)?
The empty set and the universal set not only exist (are sets), they're
also recursive.
If the set of "everything" is taken to its logical conclusion, it
simply has no cardinality. Any attempt to assign it a cardinality is
easily refuted!
Charlie Volkstorf
Cambridge, MA
http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/20021008.1/1
http://www.arxiv.org/html/cs.lo/0003071
George Greene
: The fact that a (non-empty) set and its subsets are of different
: cardinality doesn't mean you can't have a set with everything.
In the first place, a denumerable set and its denumerable
subsets are NOT of different cardinalities; each denumerable
subset is the SAME cardinality (denumerable, DUH) as the whole
set. The fact that the power-set of the original set is of
BIGGER cardinality (not just different: BIGGER) means that
the original set CANNOT have "everything". In particular,
it must not contain some of its own subsets as members.
Some set theories will allow you to prove the absence of
particular subsets as members (the set itself, or its Russell
subset, e.g.). But none of that is really the point.
The point is that there are simply MORE subsets, and since
each of these subsets is 1 thing in "every" thing, the
original set did NOT contain everything.
: You simply cannot compare its elements to its subsets.
That is irrelevant. Regardles of what you can or cannot comapre,
the subsets still EXIST, therefore, each of them is 1 THING that
"every" THING would have to include. They would therefore all
have to be members. But they can't all be. End of story.
: The existance of this set
WHAT set? The set of everything?
: per se is not a problem.
It is so, too! It does not contain all of its subsets.
Therefore it does not exist.
: But without it, you do have problems.
: If there is no universal set, then either you have no empty
: set or you cannot in general take the complement of a set. You would
: have to recheck all of your proofs!
If the proofs are in ZFC then you never NEED to take the complement,
in general of a set.
: Actually, every attempt to be as general as possible (Logic, Theory of
: Computation, Set Theory, Mathematics, et. al.) includes the empty and
: the universal object.
A universal object, yes -- there is a CLASS of all sets.
But THAT is NOT a "set of" EVERYthing. There CANNOT BE
"a set of everything". There cannot be a class of everything,
either.
: The former is the wff FALSE that is never
: provable, a program that consists of only HALT NO, or {}. The latter
: is the wff TRUE, the program that is HALT YES, and the universal set.
:
: Beyond that, read up on my axiomitization of computer science, and
: you'll see how it all fits together.
DAMN, you're an idiot.
Charlie-Boo
> : If there is no universal set, then either you have no empty
> : set or you cannot in general take the complement of a set. You would
> : have to recheck all of your proofs!
>
> If the proofs are in ZFC then you never NEED to take the complement,
> in general of a set.
I sure remember a lot of books talking about and using the complement
of a set. You're saying that's a useless concept?
> DAMN, you're an idiot.
I think your problem is that you're confusing the concept of a
universal set with its construction. Regardless of what you consider
to be a "thing", if you look at the wff TRUE or the program that says
only HALT YES, in each case every value that you supply is accepted
(the wff is true and provable, the program halts yes). THAT is the
concept of a universal set: a procedure that always accepts its input.
If you're having a problem formulating what you want to consider to
be in the universal set, that doesn't change the above facts.
Fundamental to any system is the question of the characteristics of
its universal set. It if is recursively enumerable, then every
recursive set is also recursively enumerable.
Don't worry. A lot of people are unclear on this point. Standard
thought is that "every recursive set is recursively enumerable", but
if you follow through the logic involved, you will see that it depends
on whether or not the universal set is r.e. The unwritten assumption
is that in typical systems it is. That is what Peano really proved.
karl malbrain
> > : If there is no universal set, then either you have no empty
> > : set or you cannot in general take the complement of a set. You would
> > : have to recheck all of your proofs!
> >
> > If the proofs are in ZFC then you never NEED to take the complement,
> > in general of a set.
>
> I sure remember a lot of books talking about and using the complement
> of a set. You're saying that's a useless concept?
No, we gave you the >Invert operator for this.
> > DAMN, you're an idiot.
>
> I think your problem is that you're confusing the concept of a
> universal set with its construction.
That's why he swears at you. ALL SETS ARE SPECIFIED BY THEIR
CONSTRUCTORS, and specifications are finite. They STEM from infinity,
which is where we ALL start from, without limits, as kids with
parents.
> Regardless of what you consider
> to be a "thing",
Things have seven parts. It's not a thing without each and every one.
MALBRAIN.
George Greene
: > : If there is no universal set, then either you have no empty
: > : set or you cannot in general take the complement of a set. You would
: > : have to recheck all of your proofs!
: >
: > If the proofs are in ZFC then you never NEED to take the complement,
: > in general of a set.
:
: I sure remember a lot of books talking about and using the complement
: of a set. You're saying that's a useless concept?
I'm saying it's irrelevant to set theory with infinite objects.
Most of those things are inside finitary contexts.
Most TM-paradigms constrain their input to be finite, even
when that produces an infinite number of different possible inputs.
ch...@aol.com (Charlie-Boo) writes:
: I think your problem is that you're confusing the concept of a
: universal set with its construction.
No, I'm not. YOUR problem is that you are confusing
"constructions" in general with things that might actually
matter.
: Regardless of what you consider to be a "thing",
By definition, EVERY thing is "a thing".
:if you look at the wff TRUE or the program that says
: only HALT YES, in each case every value that you supply is accepted
: (the wff is true and provable, the program halts yes). THAT is the
: concept of a universal set: a procedure that always accepts its input.
That does NOT qualify as even a "construction" or "representation"
of a "set of everything", IF "everything" INCLUDES things that CAN'T BE
"input to a program".
: If you're having a problem formulating what you want to consider to
: be in the universal set, that doesn't change the above facts.
The above facts simply have no relevance to "everything".
They all presume a smaller universe than that. If you're having a
problem formulating what you want to consider to be in the
universal set, then you are too stupid to understand the meaning
of "everything".
: Fundamental to any system is the question of the characteristics of
: its universal set.
Exactly, but a "set of everything", BY DEFINITION, would include
EVERY system and EVERY one of ALL their universal sets AS MEMBERS!
NOTHING that *you* call a "system" is anywhere NEAR BIG enough
to encompass this! ALL your systems PREsuppose some LIMITED universe
that OBVIOUSLY does NOT contain EVERYthing!
: It if is recursively enumerable, then every
: recursive set is also recursively enumerable.
Every recursive set is recursively enumerable BY DEFINITION.
NO "If" required. The recursively enumerable sets are a proper
superclass of the recursive sets. A set is recursive iffDF it
1) is recursively enumerable, AND 2) has a complement that is
also recursively enumerable. This complement is computed with
respect to a universe with a FINITARY definition; it is NOT
any sort of "complement In General".
: Don't worry. A lot of people are unclear on this point. Standard
: thought is that "every recursive set is recursively enumerable", but
: if you follow through the logic involved, you will see that it depends
: on whether or not the universal set is r.e.
The universal set is REGULAR, dumbass.
: The unwritten assumption
: is that in typical systems it is.
It is not unwritten.
It is explicitly written in most texts treating the issue
___
that the input language is \ *
/__
i.e., the class of all FINITE strings over some FINITE
alphabet sigma.
Needless to say, THAT class is NOT "everything".
--
John Jones
meaning to us, above and beyond the intuitional grounds upon which it is
based?
What is the absoluteness of the message of a mathmatical truth?news:D_Vha.202$VZ4....@newsfep2-win.server.ntli.net...
mitch
Toby Smith wrote:
> I'm having trouble following why the mathematical set of everything cannot
> exist. Can anyone provide me with a proof?
> cheers
It is extremely important to understand that the classical treatment of
language interpretation parameterizes the universal quantifier over
collections. Moreover, the existential quantifier is interpreted only with
respect to its status as a derivative concept relative to the universal
quantifier. These are the underlying assumptions of construction that allow
the self-inconsistency of a specific syntactic form to be extended to a
metaphysical assertion of reality.
Think about the problem differently. If a set is a collection taken as an
object, how are we to understand a collection as indecomposable? What object
can be dually interpreted as a collection?
It is possible to interpret the universal quantifier with respect to a
reflexive order relation. The language primitive must be a strict transitive
order, and, an axiom explicitly assuming "almost universality" must relate a
primitive membership predicate to that order relation. A congruence relation
derived from these two primitives (incorrectly referred to as identity in
first-order logic with identity) provides for a reflexive case juxtaposed
against the strict order relation. This exclusive disjunction distinguishes
between terms referring to proper parts (the strict order relation corresponds
to the mereological "proper part" predicate) and terms referring to the class
universe (the set of all sets).
If you look at the class universe as some sort of primitive topology, the
proper parts are topologies with the subspace topology. The axiom schema of
separation ensures that these topologies are totally separated. The power set
axiom ensures that all separations correspond to proper subspaces of the class
universe. Hereditary definition ensures that all proper subspaces of the
universe correspond to dichotomies (labelings of term referents into two
categories) by virtue of characteristic functions. That is, all of the proper
subspaces of the class universe are totally disconnected.
With this formulation, the language can support reference to a greatest
class. The assertion of "almost universality" ensures that "proper classes"
different from the class universe cannot be referred to within the theory.
So, complements of the hereditarily defined classes are disallowed.
Consequently, the class universe is a connected topology for which every
proper subspace is totally disconnected.
That describes an indecomposable object that reflects the semantics of a
collection.
Do not confuse interpretation of a particular instantiation of
self-inconsistent syntax with a metaphysical reality.
Good luck.
:-)
mitch
George Greene
:
: > I'm having trouble following why the mathematical set of everything cannot
: > exist. Can anyone provide me with a proof?
: > cheers
Yes.
Every set has more subsets than members.
mitch <mit...@rcnNOSPAM.com> writes:
: It is extremely important to understand that the classical treatment of
: language interpretation parameterizes the universal quantifier over
: collections.
Right.
But that doesn't change anything. What you there call a "collection"
is precisely what everybody else is going to understand, throughout
this discussion, as a "set" (even though classical set theories think
it is a proper class).
: Moreover, the existential quantifier is interpreted only with
: respect to its status as a derivative concept relative to the universal
: quantifier.
That is just plain not true: the duality is COMPLETE, ABSOLUTE,
and symmetric. EITHER quantifier could be taken as foundational
(and the other defined in terms of it); duality is just LIKE that.
: These are the underlying assumptions of construction that allow
: the self-inconsistency of a specific syntactic form to be extended to a
: metaphysical assertion of reality.
No extension is required.
That every set (or class, or collection) has MORE sub-
sets (or classes, or collections) than members is as hard
a metaphysical reality as you will ever confront.
: Think about the problem differently. If a set is a collection taken as an
: object, how are we to understand a collection as indecomposable?
You AREN'T, dumbass!
IF it is a collection, then it CANNOT be decomposable!
The individual THINGS THAT *GOT* collected into it, the
things that MAKE it a collection AS OPPOSED to something
"decomposable", will ALWAYS be things INTO WHICH it CAN be
"decomposed"!
: What object can be dually interpreted as a collection?
Interpretation is not really relevant; the object either
IS a collection or it isn't. Sometimes, an interpretation
may manage to retain consistency while disagreeing
with the reality, but that doesn't change the reality.
: It is possible to interpret the universal quantifier with respect to a
: reflexive order relation.
No, it isn't.
What is the domain of this relation?
The quantifier IS NOT *universal* unless it quantifies
over EVERYthing. Yes, it is possible to write a symbol,
CALL it a universal quantifier, and then RESTRICT what it
quantifies over to something less than the universe, but
that is just cheating.
: The language primitive must be a strict transitive order,
Orders are usually considered to be binary relations.
Again, the question must arise, what is the domain of this
relation?
: and, an axiom explicitly assuming "almost universality" must relate a
: primitive membership predicate to that order relation. A congruence relation
: derived from these two primitives (incorrectly referred to as identity in
: first-order logic with identity) provides for a reflexive case juxtaposed
: against the strict order relation.
Suppose we use first-order logic WITHOUT identity.
Is there a first-order formulation of these axioms?
If so, how have you gaine anything? You are still fundamentally
locked into an FOL framework. If not, how are you even coherent
at all?