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MoeBlee
Posted: Fri Aug 25, 2006 10:47 pm
Guest
Tony Orlow wrote:
Quote:
MoeBlee wrote:
Tony Orlow wrote:
Given a set x, can we always determine card(x)?

I'm guessing, but I'm pretty sure that there is not an algorithm that
will tell you the ordinal index of the aleph that is the cardinality of
any given set.

Good guess, except I have no idea what your driving at.

Of course you don't, because you are ignorant of even of the basics of
your own question. With the axiom of choice (which is one method),
every set has a cardinality. That cardinality is a cardinal number. And
that cardinal number is an aleph indexed by an ordinal. So I answered
your quesion: No, I don't think there is a procedure to determine which
cardinal is the cardinality of a given set. But in my first answer I
gave an even sharper formulation by pining the question down to the
ordinal index. Of course, for some sets, we can prove that a certain
cardinal is the cardinality that set; but I don't think there can be a
general procedure to produce an answer for any set that might be
submitted to test.

Quote:
First you need to define the set in set theory and prove that it exists
in set theory. Then, set theory would not be inconsistent for there not
being an algorithm to determine the ordinal index of the aleph that is
the cardinality of any given set.

So, you're saying that identifying a set with no cardinality doesn't
make set theory inconsistent? I think I saw a different opinion yesterday.

You INSIST on twisting just about every answer given you.

If it is a theorem that every set has a cardinality, then it would be
inconsistent if it were also a theorem that there exists a set without
a cardinality. But the fact that we don't have a procedure to always
announce what SPECIFIC cardinality a set has does not contradict any
theorem of set theory.

Quote:
Again, you're just yapping without regard for the specific definitions
of such things as 'consistent'.

Consistent: Adj. Without contradiction.

Yes, and your remarks were without sense of that definition, as
SPECIFICALLY you've been told about a million times that a theory is
inconsistent iff there is a contradiction in the theory, which means
that there is sentence and its negation that are both members of the
theory.

MoeBlee
MoeBlee
Posted: Fri Aug 25, 2006 10:51 pm
Guest
Tony Orlow wrote:
Quote:
So, is it technically incorrect to say 1 = 3/3 = 1.000... = 0.999...? I
never thought so.

Those are notations that can be understood precisely only in context of
the particular treatment in which they occur. I said that IN A SYSTEM
that constructs the reals as either Dedekind cuts or as equivalence
classes of Cauchy sequences, then the naturals and rationals are
isomorphically embedded but are not actual subsets of the reals. (That
is BASIC undergraduate mathematics. I mean, even I know that, and I
don't even have an education in mathematics! It's appalling how
ignorant you are WHILE you so categorically opine.) If the system and
construction are of a different sort, then we'd have to evaluate upon
the specifics of that system and construction. But, again, as I said,
for contexts that do not need to be so pedantic, such notation is
understood well enough that the equations you mentioned do of course
hold.

By the way, a while ago I posted a very rigorous explication of decimal
notation in a post to you.

Quote:
Because it's more a philosophical issue or an issue of terminology
outside the system. The purpose of set theory is not to address the
question of what is and is not a number. Rather, among the purposes of
set theory is to axiomatize and construct the various number systems
that are of interest.

So, how do you know when you're doing math, and when you've strayed into
other territory?

That's a question for philosophy. I have my own views on what
mathematics is, but I don't hold that my own concept of mathematics is
definitive.

Quote:
you don't see scientists accepting transfinitology either.
Of course, you have a survey of scientists to support your claim.
Uh, yes, right here. Why don't you survey thise that object, regarding
what they do?

That's quite an unscientific survey method of yours.

Yo, man, it's empirical evidence. Ask around. Life's an experiment.

It's anecdotal evidence only. Check it out. A scientific experiment is
not just any life experience.

Quote:
Numbers are among the primary concern of mathematics. Set theory
axiomatizes and constructs number systems.

Which are the other primary concerns, if any? (how did we define
"number" again?)

I don't define 'number' in a formal theory. But in informal discussion
about mathematics I don't demur from using the word 'number' in its
ordinary dictionary senses.

Other concerns of mathematics are sets, relations, spaces, geometries,
topologies, algebras, for example, and more.

Quote:
I'll take as an "I dunno".

Then you'll, AS USUAL, take incorrectly.

MoeBlee
Tony Orlow
Posted: Fri Aug 25, 2006 10:51 pm
Guest
MoeBlee wrote:
Quote:
Tony Orlow wrote:
MoeBlee wrote:
Tony Orlow wrote:
Given a set x, can we always determine card(x)?
I'm guessing, but I'm pretty sure that there is not an algorithm that
will tell you the ordinal index of the aleph that is the cardinality of
any given set.
Good guess, except I have no idea what your driving at.

Of course you don't, because you are ignorant of even of the basics of
your own question. With the axiom of choice (which is one method),
every set has a cardinality. That cardinality is a cardinal number. And
that cardinal number is an aleph indexed by an ordinal. So I answered
your quesion: No, I don't think there is a procedure to determine which
cardinal is the cardinality of a given set. But in my first answer I
gave an even sharper formulation by pining the question down to the
ordinal index. Of course, for some sets, we can prove that a certain
cardinal is the cardinality that set; but I don't think there can be a
general procedure to produce an answer for any set that might be
submitted to test.

First you need to define the set in set theory and prove that it exists
in set theory. Then, set theory would not be inconsistent for there not
being an algorithm to determine the ordinal index of the aleph that is
the cardinality of any given set.
So, you're saying that identifying a set with no cardinality doesn't
make set theory inconsistent? I think I saw a different opinion yesterday.

You INSIST on twisting just about every answer given you.

If it is a theorem that every set has a cardinality, then it would be
inconsistent if it were also a theorem that there exists a set without
a cardinality. But the fact that we don't have a procedure to always
announce what SPECIFIC cardinality a set has does not contradict any
theorem of set theory.

However, if we have a set which we can prove does NOT have any
cardinality within the system, then there's a hole, yes?

Quote:

Again, you're just yapping without regard for the specific definitions
of such things as 'consistent'.
Consistent: Adj. Without contradiction.

Yes, and your remarks were without sense of that definition, as
SPECIFICALLY you've been told about a million times that a theory is
inconsistent iff there is a contradiction in the theory, which means
that there is sentence and its negation that are both members of the
theory.

So, is "every set has a cardinality, either finite or an indexed aleph"
a theorem or not? You suggest that the axiom of choice makes it so. I
say that the set of bit positions required by the binary naturals does
not have such a cardinality.

Quote:

MoeBlee
MoeBlee
Posted: Fri Aug 25, 2006 10:51 pm
Guest
Tony Orlow wrote:
Quote:

Logic determines truth. Induction is more than just a form of proof.

Logic determines VALIDITY. And induciton is indeed more than just a
form of proof. But your concept of induction is uninformed,
superficial, and confused.

MoeBlee
MoeBlee
Posted: Fri Aug 25, 2006 10:51 pm
Guest
MoeBlee wrote:
Quote:
With the axiom of choice (which is one method),
every set has a cardinality. That cardinality is a cardinal number. And
that cardinal number is an aleph indexed by an ordinal.

CORRECTION: That should be, in ZFC, every set has a cardinality. That
cardinality is either a natural number or an aleph (and, of course,
every aleph is indexed by an ordinal).
Dik T. Winter
Posted: Fri Aug 25, 2006 10:51 pm
Guest
In article <44ef3b88@news2.lightlink.com> Tony Orlow <tony@lightlink.com> writes:
Quote:
Dik T. Winter wrote:
....
There is no such specific natural number. It is when we have them
all, but as there is no largest number, this can not be achieved by
taking them one by one.

The set of all naturals numbers consists of only natural numbers. There
is NO natural number where the count becomes infinite. So there is no
point in the set, even if you COULD get to the "last" one, where any
infinite set has been achieved.

And there is no point in the set where you have the complete set. Yes.
Indeed. So what?

So, if there is no point in the set which can even remotely be
considered infinitely far from the beginning, what makes it actually
infinite?

What the set of all finite numbers makes infinite? The axiom of infinitys
states that the set of all finite (natural) numbers does exist. From that
is is easy to prove that that set is not finite, hence infinite.

Quote:
If no element of the set can be an infinite number of steps
from the start, you may not be able to find an end.

And indeed, when you go step by step you will not get at the end.

Quote:
But does that mean
it's "greater than" every finite, or only "greater than or equal"?

What is the difference? Assuming you mean "aleph-0" when you write "it",
it is easily proven that:
aleph-0 is greater than or equal to each natural
gives the theorem:
aleph-0 is greater than each natural.

Because: suppose aleph-0 >= each n in N. Now suppose in addition that it
is equal to some particular n. Well, n + 1 is in N, and so aleph-0
should also be larger or equal to n + 1. Hence it can not be equal to n.
So it is not equal to any n at all. And so aleph-0 > each n in N.

Quote:
Indeed. The set of all natural numbers is just sufficient.

No, it is far too great. If you have a countably infinite number of bit
positions, then you have an uncountably infinite set of strings. Where
bit positions are indexed by the naturals, the naturals are the power
set of the number of bit positions,

Wrong. This is plain nonsense. Suppose there are three bit positions.
The set of naturals {1, 2, 3} is sufficent to index them. In what way
is {1, 2, 3} the power set of the number of bit positions (3)?

Quote:
No one is obligated to accept the theory at all. Whether it is proven
to be "correct" or not, as I have no idea what "correct" in this context
means. Is Euclidean geometry "correct"? Is hyperbolic geometry "correct"?
Is elliptic geometry "correct"?

Ah, now you bring up a prime example. Euclid set down laws for flat 2D
geometry, and questioning those axioms led to new shapes for space.
Accrdingly, the axioms of set theory might work together to describe a
system, but it is not impossible that entirely other systems might arise
from different starting assumptions.

And indeed, I never did state the opposite. But if you want to get at a
new system, provide axioms, definitions, and whatever. Tell us what
axioms to retain and what axioms to reject. And if there are axioms to
be rejected, come up with alternatives.

Quote:
But what is the case is that if you accept the axioms, you also have
to accept what follows from the axioms.

Yes, I understand that, and much to the consternation of some, I don't.

Yes, much consternation, I can understand that. So apparently you are
accepting the axioms, but not what follows from the axioms. What kind
of logic are you using?

Quote:
Rusin had the gall to tell me that if I don't accept that there are an
infinite number of finite naturals, then I will join JSH and others on
his "kill list". I don't claim that my conclusions are derived purely
from ZFC or NBG, but that there are more fundamental concerns which
contradict both, and that some other prioritization of principles needs
to happen. Proper subsets are smaller. The addition of a single element
needs to be reflected in the size of the set. Infinite values are larger
than finite values. Things like that.

Well, you are stating such. So provide a framework. Either within the
accepted axioms, or without it. But if you want to remain within the
accepted axioms, you should also accept what follows from these axioms.
And if you want to go outside, provide your set of axioms.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
MoeBlee
Posted: Fri Aug 25, 2006 10:51 pm
Guest
Tony Orlow wrote:
Quote:
If it is a theorem that every set has a cardinality, then it would be
inconsistent if it were also a theorem that there exists a set without
a cardinality. But the fact that we don't have a procedure to always
announce what SPECIFIC cardinality a set has does not contradict any
theorem of set theory.

However, if we have a set which we can prove does NOT have any
cardinality within the system, then there's a hole, yes?

If, using only first order logic with identity, you can prove from ZFC
that there exists a set that does not have a cardinality, then you will
have proven that ZFC (and just ZF for that matter, since ZFC is
relatively consistent with ZF) is inconsistent.

Quote:
So, is "every set has a cardinality, either finite or an indexed aleph"
a theorem or not?

It is a theorem of ZFC. And every aleph is indexed, so you don't need
to mention that for this purpose.

And by the way, that reminds me that I made a mistake in not qualifying
that only INFINITE sets have an aleph as their cardinality. Finite sets
have a natural number as their cardinality and infinite sets have an
aleph as their cardinality.

And if I mentioned ZC before, it should have been ZFC, since (as far as
I know, we also need the axiom schema of replacement to have all the
ordinals we need).

So, yes, it is a theorem of ZFC that every finite set has a natural
number as its cardinality and every infinite set has an aleph as its
cardinality.

Quote:
You suggest that the axiom of choice makes it so.

And I should have mentioned that the axiom schema of replacement also
is need (as far as I know) in addition to the Z axioms that are not
derivable from the axiom schema of replacement.

Quote:
I
say that the set of bit positions required by the binary naturals does
not have such a cardinality.

First, define, from the language of ZFC, every term you use in that
formulation and every term you use to prove that formulation. Second,
show an argument using only first order logic with identity and that
proceeds only from axioms and already established theorems of ZFC. Then
we'll talk about it.

MoeBlee
Dik T. Winter
Posted: Fri Aug 25, 2006 10:51 pm
Guest
In article <44ef36ae$1@news2.lightlink.com> Tony Orlow <tony@lightlink.com> writes:
Quote:
Dik T. Winter wrote:
....
So you can compute all solutions of a polynomial equation even of
higher than fourth degree in finite time? I doubt that.

I did not state that. I said that the numbers were computable, where
I use the mathematical sense of computable.

Such that one can specify which finite number of iterations will get one
within a specific finite range of accuracy, gven a specific method of
approximation? It's a limit concept, really, yes?

Computer scientist you are? How wrong you are. The very first definition
is that a number is computable if there is a Turing machine so you can give
it a specific integer n and it will calculate all digits from the first
to the n-th. There is no limit concept involved at all. All algebraic
numbers are computable, as are a host of non-algebraic numbers (e, pi).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
Posted: Fri Aug 25, 2006 10:51 pm
Guest
In article <virgil-B3927D.12230325082006@news.usenetmonster.com> Virgil <virgil@comcast.net> writes:
Quote:
In article <1156501289.435365.119480@75g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
....
The problem is not indexing but "indexing without covering".
That is easy to prove impossible for any finite natural number in unary
representation. And, alas, there are only finite natural numbers.

I have no idea what "indexing without covering" means, and until it has
a clear definition, I will continue to state that indexing all of them
in any way indexes all of them.

What I have been able to find was that with indexing of digit WM means
that every digit position can be indexed by a natural number. With
covering WM means that all positions to a certain one can be indexed
by a natural number. What WM is asserting is that when a number can
be totally indexed, it also can be totally covered. Quantifier dislexia
disguised.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
Posted: Fri Aug 25, 2006 10:51 pm
Guest
In article <44ef3da9@news2.lightlink.com> Tony Orlow <tony@lightlink.com> writes:
Quote:
Dik T. Winter wrote:
.....
Why do you need an axiom for that? Why is it
not derivable logically?

Because without the axiom of infinity the set of naturals need not exist,
and indeed, you can build a completely logical system with the negation
of the axiom of infinity and with all other axioms remaining. It is
similar to the parallel axiom in geometry.

But without an axiom of infinity, it is demonstrable that, given the
axiom of internal infinity (continuity), x<z -> x<y<z, that any finite
interval includes an infinite number of points. Start with the line, and
identify points. There's infinity.

Your axiom uses things that are not defined. What is the *meaning* of
"x<z"?

Quote:
You both do not have an linearisation of the reals. You both are doing
other things.

Describe those "other things". How are they "other"?

You are expanding the reals in some undefined matter.

Quote:
The second proof of the uncountability of the reals is not about reals?
That's news. What was it about, then, in your opinion?

Sorry, I already explained that in previous articles to which you have
replied. But if you do not read the articles you reply to there is
something seriously wrong in the discussion.

It is about digital representation, which is the same as power set, even
in other number bases, which means more than two states of inclusion per
member.

Darn. Try to read. Cantor's proof is not about reals, it is neither
about digital representations. It is about none of the things you are
mentioning. But nevertheless you maintain that it is news that it is
not about the reals, while you read what I wrote?

Quote:
Larger than any finite. The set of naturals is as large as, but no
larger than, every natural.

That is not a definition, because it makes no sense. "The set of naturals
is as large as every natural"?

It is no larger than all naturals

That is something completely different again.

Quote:
From that: "The set of naturals is as
large as 1", "The set of naturals is as large as 2". What is the meaning
of these statements?

That is when you substitute "every", meaning "each", for "all". Careful.

Yes, you should be careful in what you mean, and not use a word that has
multiple meanings so that you can be misunderstood. So I will refrase:
Quote:
Larger than any finite. The set of naturals is as large as, but no
larger than, all naturals.
Is that what you intended? In that case you just stated a tautology.


Quote:
Then I don't know what proof you are talking about. When people say
"Cantor's second", they are generally referring to his second proof of
the uncountablility of the reals based on the diagonal argument, as
opposed to the first, based on an unreachable intermediate value.

But they are wrong. The proof was *not* about the uncountability of the
reals. The diagonal proof Cantor provided was not about that. It was
a proof about the things I outlined just above.

It was about power set and digital representation, which are identical.
It was about symbolic sets.

You finally did read it? If so, you really should improve your German.

Quote:
I thought it was clear that I was using a notion of infinite, like WM,
from a quantitative standpoint, rather than set-theoretic.

Without definition.

Greater than any finite. Simple enough?

So the cardinality of the naturuals is infinite?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Virgil
Posted: Sat Aug 26, 2006 2:18 am
Guest
In article <44ef9b7d@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:

Quote:
What is the difference between Dedekind infinite and
infinite?

A set is Dedekind infinite if it has an injection to a proper subset,
and finite otherwise.

A set is (non-Dedekind) finite if it has any bijection with any von
Nuemann natural, and is otherwise (non-Dedekind) infinite.

In ZFC and NBG, they are equivalent definitions.
Virgil
Posted: Sat Aug 26, 2006 2:25 am
Guest
In article <44ef9c12@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:

Quote:
MoeBlee wrote:
Tony Orlow wrote:
No, it all rests on the notions of identity and equality. As Leibniz
pointed out, when the properties of two objects are all exactly the
same, then they are the same object. So, when we say two numbers are
equal, that means all properties of the two are equal.

Ha! The fallacy of reversing implication right there! An example of
just about the most basic fallacy.

When you say "a=b", you say "A a A b P(a)=P(b)". The two are equivlent
statements, and therefore imply each other.

When one says of sets A and B that A = B, that is equivalent to
For all x (x \in A <==> x \in B). At least in ZF, ZFC and NBG.
Quote:


No, the indiscernibility of identicals does NOT imply the identity of
indiscernibles. You need both implications; you can't derive one from
the other. And, in first order logic, one direction can be posited only
in the semantics not in the axioms.

You prove two quantities equal by showing there is no difference, do you
not?

When one says of sets A and B that A = B, that is equivalent to
"For all x (x \in A <==> x \in B). At least in ZF, ZFC and NBG."

http://en.wikipedia.org/wiki/Axiom_of_extension
Virgil
Posted: Sat Aug 26, 2006 2:30 am
Guest
In article <44ef9c5a$1@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:

Quote:
MoeBlee wrote:
Tony Orlow wrote:
Yes, and the universe is consistent by definition, so math should be
consistently overall as well.

Unless the universe is a set of sentences, the notion of consistency
does not even apply.

The universe is governed by the properties of the elements within it,
which properties are statements true about those elements.

But as those "properties" noumenon and we can only observe phenomema in
the "real world". And they can never meet.
Virgil
Posted: Sat Aug 26, 2006 2:37 am
Guest
In article <44ef9ecc@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:



Quote:
There is NO system that will give you any kind of even minimal
mathematics without adopting axioms that are not true in all domains of
discourse.

I am not convinced of that.

TO is convinced only of the infallibility of his intuition.
Others, having seen the quagmires into which where his intuition has led
him, are even more convinced of its fallibility.
Quote:


If Ross and I each have our linearizations of the reals, then they
exist, independent of the axiom of choice (which might as well be called
the axiom of free will).

We prove existence of objects in a theory from axioms. You posit the
existence of objectw without any system of logic, primitives, or
axioms.

Logic is the primitive of argument. Logical truth is founded upon
quantity, and quantity upon geometry. Space is a priori. Smile

Only that empty one between TO's ears is a priori.

Quote:

Define 'finite'.

In what context? That word is, like many, one whose meaning depends on
the context in which it is used.
Virgil
Posted: Sat Aug 26, 2006 2:40 am
Guest
In article <44ef9f9c@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:

Quote:
MoeBlee wrote:
Tony Orlow wrote:
Given a set x, can we always determine card(x)?

I'm guessing, but I'm pretty sure that there is not an algorithm that
will tell you the ordinal index of the aleph that is the cardinality of
any given set.

Good guess, except I have no idea what your driving at.

He is driving at the fact that for infinite ordinals there are infinite
sets of different ones having the same cardinality.
 
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