Re: A consideration concerning the diagonal argument of G. Cantor
Jim Burns
herbzet wrote [to Dave Seaman]:
>
> To save typing, I'll call the sentence "There exists
> an infinite set with no denumerable subset" W (for weird).
> Now if we substitute the axiom of choice for W we get ZFC
> in which W is refuted. This is what suggested to me that
> ZF + ~C proves W. But you say that ZF + ~C does not
> decide W, if I understand you correctly. So we need
> a yet stronger axiom to add to ZF in order to decide W
> (in the affirmative).
Everyone in this sub-thread seems to agree that ZF alone
will not prove ~W. I expect that means someone has
proven it, but I can't seem to find it or it's reference.
I would appreciate it if someone were to point this
out to me.
To be honest, I am somewhat skeptical that ZF cannot
prove ~W without Choice, but I am also skeptical of my
ability to detect my use of the Axiom of Choice --
it just seems so obviously true. Am I using Choice
in the proofs sketched below?
There are two common definitions of "infinite set",
which are not equivalent without AC. (1) A Dedekind
infinite set has a bijection with a proper subset
of itself. (2) An infinite set does not have a
bijection with any of the finite ordinals.
It seems to me that I can prove "Every
[infinite/Dedekind infinite] set has a denumerable
subset." either way without using Choice.
If I'm wrong, where did I use AC?
Thanks,
Jim Burns
-------------------
-- Every Dedekind infinite set has a denumerable
subset. --
Let A be a Dedekind infinite set. There exists a
bijection f: A -> B, where B is a proper subset
of A.
Define a function g: N -> A as follows:
Select any element z of (non-empty) A/B and
define g(0) = z. Define g(n+1) = f^-1(g(n)).
If g is not 1-to-1, then f is not a bijection.
Therefore, g(N) is a denumerable subset of A.
(end proof)
-------------------
-- Every infinite set has a denumerable subset. --
Let A be an infinite set. There does not exist a
bijection between A and any finite ordinal
n = {0, 1, 2, ..., n-1}.
Lemma. There exists a sequence of functions
( g_0, g_1, g_2, ... ) such that g_n: n -> A,
is 1-to-1, and for m <= n and k <= m ,
g_n(k) = g_m(k).
g_0: {} -> A exists; it is the empty set.
Assume g_n, satisfying the above conditions, exists.
It cannot be onto A, or else it would be a bijection,
and A would be finite. Thus, A/g_n(n) is not empty.
Select any element of A/g_n(n) and name it a_n.
Define g_{n+1}(n) = a_n, otherwise
g_{n+1}(k) = g_n(k). Therefore g_{n+1} exists,
satisfying the above conditions.
By induction, g_n exists, for all n.
(end lemma)
Take a sequence of functions ( g_0, g_1, g_2, ... )
such that g_n: n -> A, is 1-to-1,
and for m <= n and k <= m, g_n(k) = g_m(k).
Define g \subset NxA as the union of all the g_n.
The relation g is single-valued, well-defined on N
and 1-to-1. Therefore g: N -> A is an injection
into A, and g(N) is a denumerable subset of A.
(end proof)
---------------------
Ross A. Finlayson
> herbzet wrote [to Dave Seaman]:
>
>> To save typing, I'll call the sentence "There exists
>> an infinite set with no denumerable subset" W (for weird).
>> Now if we substitute the axiom of choice for W we get ZFC
>> in which W is refuted. This is what suggested to me that
>> ZF + ~C proves W. But you say that ZF + ~C does not
>> decide W, if I understand you correctly. So we need
>> a yet stronger axiom to add to ZF in order to decide W
>> (in the affirmative).
>
That a set doesn't have a denumerable subset is basically calling it a
limit ordinal.
Cantor counts backwards, but always stops at limit ordinals. In his
original theory of infinity, Cantor counts backwards from infinity, and
forwards, ordinally.
In that way, to say that W is w, the statement of the existence by the
normal ordinal recognition of an infinite limit ordinal, there are just
all the finite ordinals. Where this is the only set that exists that is
not finite where all the finite sets have the property that they have
denumerable subsets, (except the empty set), is this limit ordinal
omega, the statement of its existence is the theorem that it exists,
except it's actually an axiom called "the axiom of infinity". So, ZF is
consistent with that a least infinite limit ordinal exists, because it
defines one as the only symbol in the language besides zero. Those are
the two constants in ZF(C), empty and infinity.
> Everyone in this sub-thread seems to agree that ZF alone
> will not prove ~W. I expect that means someone has
> proven it, but I can't seem to find it or it's reference.
> I would appreciate it if someone were to point this
> out to me.
>
> To be honest, I am somewhat skeptical that ZF cannot
> prove ~W without Choice, but I am also skeptical of my
> ability to detect my use of the Axiom of Choice --
> it just seems so obviously true. Am I using Choice
> in the proofs sketched below?
>
> There are two common definitions of "infinite set",
> which are not equivalent without AC. (1) A Dedekind
> infinite set has a bijection with a proper subset
> of itself. (2) An infinite set does not have a
> bijection with any of the finite ordinals.
>
I think there are four or perhaps five definitions that I have considered.
> It seems to me that I can prove "Every
> [infinite/Dedekind infinite] set has a denumerable
> subset." either way without using Choice.
> If I'm wrong, where did I use AC?
>
> Thanks,
> Jim Burns
>
>
Sometimes it varies on whether it would apply. Consider statements
involving the transfer principle. That is about how if a property
transcends the object then if that property applies to the objects
separately then it does to them together, the transfer principle, where
what is true for each object is true for all of them. So, in cases
involving quantification over the objects separately and together, their
collection is substructured in the original, about where the
transcendant property references are in the products of the algebraic
monads of objects' properties among objects. So, where the transfer
principle might apply, it doesn't for example to that the limit ordinal
of a collection of finite ordinals is a finite ordinal.
Regards,
Ross F.
Jim Burns
Ross, I am straining to respond positively to
your response, especially since you seem to be the
only response that my post has received.
However, even though I have carefully read your
entire post twice (and I would read it again if I
thought it would matter), my reactions cover the
gamut from "This makes no sense at all" all the
way to "He must be using these terms idiosyncratically,
since the usual meanings make this obviously wrong."
Jim Burns
Virgil
"Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> That a set doesn't have a denumerable subset is basically calling it a
> limit ordinal.
That would make the set of reals, and the set of points in a plane, and
all sorts of other non-ordinal sets into ordinal sets.
What sort of ordinal set of points is the surface of a sphere?
herbzet
Jim Burns wrote:
>
> Ross, I am straining to respond positively to
> your response, especially since you seem to be the
> only response that my post has received.
>
> However, even though I have carefully read your
> entire post twice (and I would read it again if I
> thought it would matter), my reactions cover the
> gamut from "This makes no sense at all" all the
> way to "He must be using these terms idiosyncratically,
> since the usual meanings make this obviously wrong."
>
> Jim Burns
Hi Jim,
Your original post was a response to a post of mine,
but I'm not really competent to answer so I didn't.
But your post, in intent, was really addressed to
anyone, and I assume someone will get around to
giving an informed response. It's the weekend,
have patience.
Ross is generally received as something of a
poet in sci.logic. I'm not really competent
to judge on that either.
Hi, Ross.
--
hz
Ross A. Finlayson
> Ross, I am straining to respond positively to
> your response, especially since you seem to be the
> only response that my post has received.
>
> However, even though I have carefully read your
> entire post twice (and I would read it again if I
> thought it would matter), my reactions cover the
> gamut from "This makes no sense at all" all the
> way to "He must be using these terms idiosyncratically,
> since the usual meanings make this obviously wrong."
>
> Jim Burns
>
Hi Jim,
It should be easier. Really I use standard terminology.
>> Sometimes it varies on whether it would apply. Consider statements
>> involving the transfer principle. That is about how if a property
>> transcends the object then if that property applies to the objects
>> separately then it does to them together, the transfer principle, where
>> what is true for each object is true for all of them. So, in cases
>> involving quantification over the objects separately and together, their
>> collection is substructured in the original, about where the
>> transcendant property references are in the products of the algebraic
>> monads of objects' properties among objects. So, where the transfer
>> principle might apply, it doesn't for example to that the limit ordinal
>> of a collection of finite ordinals is a finite ordinal.
That's clear and even relevant. Where the transfer principle applies:
what is true for each object is true for all of them _together_, as an
object it maintains that property. An example is given of the ordinals,
where in quantifying over all the ordinals less than a given ordinal,
the result (set) is that ordinal, the collection has the same property
as its elements. So, there's no quantifying upwards over ordinals
because the set collected from those elements wouldn't be a set.
Then the other part: that also applies.
Regards,
Ross F.
Ross A. Finlayson
Any well ordering's ordinal domain of its space-filling function (simply
in terms of space-filling curves) is said surface. How do you
well-order the line?
See, for each point in the line it's defined by all the points less than
it including itself.
There's a spiral space filling curve to fill the disc from the origin, a
ray to fill the line segment, then those other suitable primitive
functions can be composed in transformation relations, in tuples of
ordinals maintaining composite algebraic alignment, into coordinate
spaces, is the lines' structure and quite more than the two points that
define it. That is about the structure of the coordinate pair
supporting the space, to define them.
It's kind of like the proper class, for the finitary case, in terms of
illustrating that a limit ordinal doesn't have a greatest subelement
from among which to select some non-denumerable set has no
non-denumerable subset, that quantification doesn't allow the selection
of some non-denumerable (presumably non-empty) subset. There is a
similar non-resolution because of transfer principle concerns. In ZFC
well-order the contents and show via transfinite induction some feature
would have that a subset of those selected that is denumerable. This is
where the set has a particular structural feature to have that its
elements are only predicatizably addressible in the largest groups, thus
that without choice it's not possible to describe a large enough structure.
Consider for example continua in real analysis with no integration over
non-measurable regions of discontinuities. In the analysis of the
measurable areas, there are no areas that aren't denumerable (in
standard real analysis, in ZF with reals). To analyze the
non-measurable areas, in terms say that given other properties the
non-measurable areas can be recomposed into measurable areas, in terms
of analyzing the entire area, leads to completeness over the product
space, but in leaving them out, in the analysis there are denumerably
many structures, each a non-denumerable subset (here in standard measure
theory, where there can be more utility in a nonstandard theory in the
methods of exhaustion, in non-general terms). Similarly, in the
recomposition of the non-measurable areas, where there are
non-measurable areas in standard reals else Vitali and so on would apply
in the standard reals in terms of counterexamples of standard real
analysis that least positive infinitesimals don't exist, in the
recomposition of areas in the synthesis of integration across boundaries
of integration there might be, because they're real numbers, still no
way to analyze the non-measurable function.
Consider for example many have that given the definition of the elements
of a domain and their sum and then defining that in the terms of
infinite sums as an exact value, cancelling in algebraic systems as that
value, that it can be so maintained in that framework.
Regards,
Ross F.
Dave Seaman
> herbzet wrote [to Dave Seaman]:
>>
>> To save typing, I'll call the sentence "There exists
>> an infinite set with no denumerable subset" W (for weird).
>> Now if we substitute the axiom of choice for W we get ZFC
>> in which W is refuted. This is what suggested to me that
>> ZF + ~C proves W. But you say that ZF + ~C does not
>> decide W, if I understand you correctly. So we need
>> a yet stronger axiom to add to ZF in order to decide W
>> (in the affirmative).
> Everyone in this sub-thread seems to agree that ZF alone
> will not prove ~W. I expect that means someone has
> proven it, but I can't seem to find it or it's reference.
> I would appreciate it if someone were to point this
> out to me.
See below.
> To be honest, I am somewhat skeptical that ZF cannot
> prove ~W without Choice, but I am also skeptical of my
> ability to detect my use of the Axiom of Choice --
> it just seems so obviously true. Am I using Choice
> in the proofs sketched below?
> There are two common definitions of "infinite set",
> which are not equivalent without AC. (1) A Dedekind
> infinite set has a bijection with a proper subset
> of itself. (2) An infinite set does not have a
> bijection with any of the finite ordinals.
> It seems to me that I can prove "Every
> [infinite/Dedekind infinite] set has a denumerable
> subset." either way without using Choice.
> If I'm wrong, where did I use AC?
That proof is easy, but the converse does not hold in ZF. According to
Wikepedia, the equivalence of (1) and (2) is strictly weaker than the axiom of
countable choice (CC). References are given in the article.
<http://en.wikipedia.org/wiki/Dedekind-finite>
> Thanks,
> Jim Burns
It's true that for any n you can choose a suitable g_n, but in asserting
that you can thus define a sequence ( g_0, g_1, g_2, ... ), you have
invoked AC.
Remember that one equivalent form of AC is the multiplicative axiom: the
Cartesian product of a nonempty collection of nonempty sets is nonempty.
You have shown that the family of candidates for g_n is nonempty for each
n, but you cannot then conclude without AC that the Cartesian product,
the family of sequences ( g_0, g_1, g_2, ... ), is nonempty.
> Take a sequence of functions ( g_0, g_1, g_2, ... )
> such that g_n: n -> A, is 1-to-1,
> and for m <= n and k <= m, g_n(k) = g_m(k).
> Define g \subset NxA as the union of all the g_n.
> The relation g is single-valued, well-defined on N
> and 1-to-1. Therefore g: N -> A is an injection
> into A, and g(N) is a denumerable subset of A.
> (end proof)
> ---------------------
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>
Virgil
"Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> Jim Burns wrote:
> > Ross, I am straining to respond positively to
> > your response, especially since you seem to be the
> > only response that my post has received.
> >
> > However, even though I have carefully read your
> > entire post twice (and I would read it again if I
> > thought it would matter), my reactions cover the
> > gamut from "This makes no sense at all" all the
> > way to "He must be using these terms idiosyncratically,
> > since the usual meanings make this obviously wrong."
> >
> > Jim Burns
> >
>
> Hi Jim,
>
> It should be easier. Really I use standard terminology.
>
> >> Sometimes it varies on whether it would apply. Consider statements
> >> involving the transfer principle. That is about how if a property
> >> transcends the object then if that property applies to the objects
> >> separately then it does to them together, the transfer principle, where
> >> what is true for each object is true for all of them.
Each natural, being the set of prior naturals, is a finite set, so that
by that "transfer principle, the st of all naturals would also be
afinite set.
Virgil
"Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> Virgil wrote:
> > In article <gc82pk$ds4$1...@aioe.org>,
> > "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> >
> >
> >> That a set doesn't have a denumerable subset is basically calling it a
> >> limit ordinal.
> >
> > That would make the set of reals, and the set of points in a plane, and
> > all sorts of other non-ordinal sets into ordinal sets.
> >
> > What sort of ordinal set of points is the surface of a sphere?
>
> How do you well-order the line?
I don't. Lets see you do it. Even with the AoC, there is no known
explicit well ordering of the reals, and without it, such a well
ordering need not even be possible.
And a limit ordinal must be a well-ordered set, so (repeating my
original doubt) how do you make out that the set of reals is a limit
ordinal?
Ross A. Finlayson
> In article <gc9v73$qib$1...@aioe.org>,
> "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
>
>> Virgil wrote:
>>> In article <gc82pk$ds4$1...@aioe.org>,
>>> "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
>>>
>>>
>>>> That a set doesn't have a denumerable subset is basically calling it a
>>>> limit ordinal.
>>> That would make the set of reals, and the set of points in a plane, and
>>> all sorts of other non-ordinal sets into ordinal sets.
>>>
>>> What sort of ordinal set of points is the surface of a sphere?
>> How do you well-order the line?
>
> I don't. Lets see you do it.
EF, the natural/unit equivalency function, modeled by real functions,
well-orders the unit line segment.
Even with the AoC, there is no known
> explicit well ordering of the reals, and without it, such a well
> ordering need not even be possible.
>
> And a limit ordinal must be a well-ordered set, so (repeating my
> original doubt) how do you make out that the set of reals is a limit
> ordinal?
The initial ordinal of a cardinal is an ordinal. The initial ordinal of
the reals is a limit ordinal (on average about twice as many as are in
any set less than it).
That these sets have natural well-orderings or are well-ordered is
considered in the standard and nonstandard. I don't care because I use
whatever standard definitions of the reals there are to satisfy their
properties, or show via that they don't exist that said theory is
incomplete.
ZF is inconsistent (quite).
Regards,
Ross F.
Virgil
"Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> Virgil wrote:
> > In article <gc9v73$qib$1...@aioe.org>,
> > "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> >
> >> Virgil wrote:
> >>> In article <gc82pk$ds4$1...@aioe.org>,
> >>> "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> >>>
> >>>
> >>>> That a set doesn't have a denumerable subset is basically calling it a
> >>>> limit ordinal.
> >>> That would make the set of reals, and the set of points in a plane, and
> >>> all sorts of other non-ordinal sets into ordinal sets.
> >>>
> >>> What sort of ordinal set of points is the surface of a sphere?
> >> How do you well-order the line?
> >
> > I don't. Lets see you do it.
>
> EF, the natural/unit equivalency function, modeled by real functions,
> well-orders the unit line segment.
References? There are over 9000 Google references, most of which are
irrelevant, so which one(s) should I look at?
MoeBlee
wrote:
> That a set doesn't have a denumerable subset is basically calling it a
> limit ordinal.
No, it's not.
MoeBlee
Ross A. Finlayson
Why don't you go write a Wiki page on it. How about a sci.math Wiki.
Some years a search on Google for "equivalency function" returned five
results.
Look at any of them. They are generally ordered among posts of similar
material. In these discussions about the equivalency function, there
are many here who could tell you exactly what it was claimed to be, in
its three or five minimal properties that define it.
EF, is a function. Its input is the domain of natural integers. Its
output is a real number from [0,1). The output ranges from zero to one
over the integers, i.e., it is strictly monotone increasing, but only
infinitesimally increasing, all the way from zero to one.
For any input of a finite number, one through infinity, EF(n) = 0. In a
notation, EF(oo) = 1, EF(infinity) equals one.
It ranges from zero to one in its natural total order in the natural
total order of the real numbers.
Then, it is not a real function, because standardly it's everywhere zero
so the fact that its integral evaluates to equal one instead of 1/2. If
there were a CDF of a uniform distribution of the natural integers, EF
would be its CDF.
It has properties outside the system of standard real numbers, in real
numbers, and modeled by real numbers. Those are often used.
EF(0) = 0
EF(n) < EF(n+1)
EF(n) <= 1
lim_n->oo EF(n) = 1
int_n=0..oo EF(n) = 1
EF is a cumulative distribution function, CDF.
Regards,
Ross F.
Virgil
"Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> Virgil wrote:
> > In article <gcb83t$82k$1...@aioe.org>,
> > "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> >> EF, the natural/unit equivalency function, modeled by real functions,
> >> well-orders the unit line segment.
> >
> > References? There are over 9000 Google references, most of which are
> > irrelevant, so which one(s) should I look at?
>
> Why don't you go write a Wiki page on it. How about a sci.math Wiki.
Writing about what one does know anything abut is your bag, not mine.
I have no idea what you mean by a "natural/unit equivalency function"
and there are too many too different definitions available on Google to
assume that I know which one of them you mean by it.
>
> Some years a search on Google for "equivalency function" returned five
> results.
Thirty years ago Google searches did not exist. So what?
>
> Look at any of them.
Pick for me one of them that means what you mean, provided, of course,
you have any idea what you mean.
>
> EF, is a function. Its input is the domain of natural integers. Its
> output is a real number from [0,1). The output ranges from zero to one
> over the integers,
As there is only one integer in the range of a function from N to [0,1),
and only the upper half of them in its domain, what you are saying is
nonsense.
> i.e., it is strictly monotone increasing, but only
> infinitesimally increasing, all the way from zero to one.
You must be working in non-standard analysis them, since for a function,
EF:N -> [0,1), which is strictly increasing, each EF(n+1) - EF(n), is a
non-infinitesimal, strictly positive real number, and the infinite
series with such terms, s_n = EF(n+1) - EF(n) necessarily converges to
a number no greater than 1.
> For any input of a finite number, one through infinity, EF(n) = 0.
Then it is not a real function, as no real function can have those
properties.
> In a
> notation, EF(oo) = 1, EF(infinity) equals one.
If EF is a real function and EF(n) = 0 for ALL n in N, then
lim_{n -> oo} EF(N)= 0, and EF(oo) is not even defined.
Further garbage snipped.
Ross A. Finlayson
No.
Regards,
Ross F.
Ross A. Finlayson
> In article <gcbffh$ae9$1...@aioe.org>,
> "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
>
>> Virgil wrote:
>>> In article <gcb83t$82k$1...@aioe.org>,
>>> "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
>
>>>> EF, the natural/unit equivalency function, modeled by real functions,
>>>> well-orders the unit line segment.
>>> References? There are over 9000 Google references, most of which are
>>> irrelevant, so which one(s) should I look at?
>> Why don't you go write a Wiki page on it. How about a sci.math Wiki.
>
> Writing about what one does know anything abut is your bag, not mine.
>
Sic.
> I have no idea what you mean by a "natural/unit equivalency function"
> and there are too many too different definitions available on Google to
> assume that I know which one of them you mean by it.
>
The N/U EF has too few properties, although they are far-reaching, that
they are difficult to understand, given that generally they are minimal
default properties in small numeric systems, for example with the domain
naturals and range reals. Because they are minimal default properties
(eg, "1", the multiplicative identity, or "0", the additive identity)
where the transform is reversible it's usable. It doesn't matter so
much that it's not a real function, although some of its ideal
properties aren't so modeled (eg area) by all the functions that model
the standard real properties.
>> Some years a search on Google for "equivalency function" returned five
>> results.
>
> Thirty years ago Google searches did not exist. So what?
>> Look at any of them.
>
> Pick for me one of them that means what you mean, provided, of course,
> you have any idea what you mean.
I wrote one write here in this previous message, apparently you ignored
that. I'll get to writing more LaTeX documents with the typeset layout
of a definition of EF, as I already have one.
>> EF, is a function. Its input is the domain of natural integers. Its
>> output is a real number from [0,1). The output ranges from zero to one
>> over the integers,
>
> As there is only one integer in the range of a function from N to [0,1),
> and only the upper half of them in its domain, what you are saying is
> nonsense.
>
>
No, it's carefully considered notation, because in some considerations
EF's range doesn't include the endpoint, it's half-open. Generally the
range of EF is [0,1], because the limit as its domain diverges, which it
does, equals one.
So, it's convergent everywhere standardly (with allowances for
definitions with the domain being discrete), and converges in the limit
to one.
>> i.e., it is strictly monotone increasing, but only
>> infinitesimally increasing, all the way from zero to one.
>
> You must be working in non-standard analysis them, since for a function,
> EF:N -> [0,1), which is strictly increasing, each EF(n+1) - EF(n), is a
> non-infinitesimal, strictly positive real number, and the infinite
> series with such terms, s_n = EF(n+1) - EF(n) necessarily converges to
> a number no greater than 1.
>
Sic.
>> For any input of a finite number, one through infinity, EF(n) = 0.
>
> Then it is not a real function, as no real function can have those
> properties.
>
Right. EF is not a standard real function. The unit impulse function
or Dirac delta, very useful in real analysis and modeled by real
functions, is not a real function. Infinitesimals aren't standard, in
the infinitesimal analysis.
>> In a
>> notation, EF(oo) = 1, EF(infinity) equals one.
>
> If EF is a real function and EF(n) = 0 for ALL n in N, then
> lim_{n -> oo} EF(N)= 0, and EF(oo) is not even defined.
>
You just said it wasn't.
> Further garbage snipped.
Please don't do that, it's annoying.
Regards,
Ross F.
James Burns
>
> Jim Burns wrote:
>
>>Ross, I am straining to respond positively to
>>your response, especially since you seem to be the
>>only response that my post has received.
>>
>>However, even though I have carefully read your
>>entire post twice (and I would read it again if I
>>thought it would matter), my reactions cover the
>>gamut from "This makes no sense at all" all the
>>way to "He must be using these terms idiosyncratically,
>>since the usual meanings make this obviously wrong."
>>
>>Jim Burns
>
>
> Hi Jim,
>
> Your original post was a response to a post of mine,
> but I'm not really competent to answer so I didn't.
> But your post, in intent, was really addressed to
> anyone, and I assume someone will get around to
> giving an informed response. It's the weekend,
> have patience.
Thank you for the advice. Sorry if my last post came
out a whine.
> Ross is generally received as something of a
> poet in sci.logic. I'm not really competent
> to judge on that either.
Alas, I'm not competent to judge poetry either.
Jim Burns
James Burns
> On Sat, 04 Oct 2008 10:15:53 -0400, Jim Burns wrote:
>
>>It seems to me that I can prove "Every
>>[infinite/Dedekind infinite] set has a denumerable
>>subset." either way without using Choice.
>>If I'm wrong, where did I use AC?
>
> That proof is easy, but the converse does not hold in ZF.
> According to Wikepedia, the equivalence of (1) and (2)
> is strictly weaker than the axiom of countable choice (CC).
> References are given in the article.
>
> <http://en.wikipedia.org/wiki/Dedekind-finite>
Thank you. I think I will need some time to work
through things.
In particular, I'm looking forward to the proof
that there is a model of ZF with an infinite set
without a denumerable subset.
[...]
>>By induction, g_n exists, for all n.
>>(end lemma)
>
> It's true that for any n you can choose a suitable g_n,
> but in asserting that you can thus define a sequence
> ( g_0, g_1, g_2, ... ), you have invoked AC.
>
> Remember that one equivalent form of AC is the
> multiplicative axiom: the Cartesian product of a
> nonempty collection of nonempty sets is nonempty. You have
> shown that the family of candidates for g_n is nonempty
> for each n, but you cannot then conclude without AC
> that the Cartesian product, the family of sequences
> ( g_0, g_1, g_2, ... ), is nonempty.
>
>>Take a sequence of functions ( g_0, g_1, g_2, ... )
>>such that g_n: n -> A, is 1-to-1,
>>and for m <= n and k <= m, g_n(k) = g_m(k).
I really expect that I'm wrong, but I just don't see
how assuming the absence of such a sequence does not
produce a contradiction.
(One way that /will not/ work, but took me some thought
to realize that: "Assume there is a longest sequence ...".\
BUT there doesn't have to be an infinite sequence
just because there is no maximum finite length,
just as there doesn't have to be an infinite natural.)
Again, thanks.
Jim Burns
James Burns
> Jim Burns wrote:
>
>> Ross, I am straining to respond positively to
>> your response, especially since you seem to be the
>> only response that my post has received.
>>
>> However, even though I have carefully read your
>> entire post twice (and I would read it again if I
>> thought it would matter), my reactions cover the
>> gamut from "This makes no sense at all" all the
>> way to "He must be using these terms idiosyncratically,
>> since the usual meanings make this obviously wrong."
>>
>> Jim Burns
>>
>
> Hi Jim,
>
> It should be easier. Really I use standard terminology.
I'll stick to the first, short example:
You wrote
: That a set doesn't have a denumerable subset
: is basically calling it a limit ordinal.
But not having a denumerable subset is (basically)
tha same as being finite, and no finite set is
a limit ordinal. (Anyway, even if some definitions
might include {} as a limit ordinal, one possible
exception would not a rule make.) So, that looks
exactly backwards.
The most charitable interpretation I see would
assume that you made a typo, and you meant
; That a set DOES have a denumerable subset
; is basically calling it a limit ordinal.
However, that's not true either. Sets, even
infinite sets, are generally NOT ordinals, for
example, {2,3,5,7,11,13,...}.
Perhaps you really meant
| That an ORDINAL DOES have a denumerable subset
| is basically calling it a limit ordinal.
But omega+1, omega+2, etc are a not limit ordinals
-- unless you mean something else by the term.
Or perhaps you meant
; That a set DOES have a denumerable subset is
; basically calling it BIJECTABLE WITH a limit ordinal.
Calling a unspecified set bijectable with some
ordinal tacitly assumes the axiom of choice, I
believe (since, then all sets would well-orderable).
But since the discussion you're responding to
is about what must be used in place of the axiom
of choice, throwing that into the discussion without
noting the use of choice would be some combination of
useless, confusing and irrelevant.
And that was your first sentence. Sigh.
Sorry, it's just too much work figuring out
how to respond to you.
Jim Burns
Dave Seaman
> Dave Seaman wrote:
> [...]
>>>By induction, g_n exists, for all n.
>>>(end lemma)
>>
>> It's true that for any n you can choose a suitable g_n,
>> but in asserting that you can thus define a sequence
>> ( g_0, g_1, g_2, ... ), you have invoked AC.
>>
>> Remember that one equivalent form of AC is the
>> multiplicative axiom: the Cartesian product of a
>> nonempty collection of nonempty sets is nonempty. You have
>> shown that the family of candidates for g_n is nonempty
>> for each n, but you cannot then conclude without AC
>> that the Cartesian product, the family of sequences
>> ( g_0, g_1, g_2, ... ), is nonempty.
>>
>>>Take a sequence of functions ( g_0, g_1, g_2, ... )
>>>such that g_n: n -> A, is 1-to-1,
>>>and for m <= n and k <= m, g_n(k) = g_m(k).
> I really expect that I'm wrong, but I just don't see
> how assuming the absence of such a sequence does not
> produce a contradiction.
Do you see why the multiplicative axiom is equivalent to AC? A member of
the Cartesian product is essentially a choice function.
> (One way that /will not/ work, but took me some thought
> to realize that: "Assume there is a longest sequence ...".\
> BUT there doesn't have to be an infinite sequence
> just because there is no maximum finite length,
> just as there doesn't have to be an infinite natural.)
Precisely. Putting it another way, you don't need AC if only finitely
many choices are needed. AC becomes important when you need to make an
infinite number of choices.
Consider an infinite number of pairs of shoes. You don't need AC to
choose one shoe from each pair. Just take the left shoe from each pair.
Now consider an infinite number of pairs of socks. In this case, you
can't choose one sock from each pair except by invoking AC.
James Burns
> On Mon, 06 Oct 2008 12:02:49 -0400, James Burns wrote:
>>Dave Seaman wrote:
>>>>Take a sequence of functions ( g_0, g_1, g_2, ... )
>>>>such that g_n: n -> A, is 1-to-1,
>>>>and for m <= n and k <= m, g_n(k) = g_m(k).
>
>>I really expect that I'm wrong, but I just don't see
>>how assuming the absence of such a sequence does not
>>produce a contradiction.
>
> Do you see why the multiplicative axiom is equivalent
> to AC? A member of the Cartesian product is
> essentially a choice function.
Yes. I don't have a problem with the various equivalent
formulations of AC -- I probably could even show some
of them are equivalent off the top of my head.
I also do not have a problem believing that
there are propositions that require us to make
an infinite number of choices "at the same time".
It's that this particular proposition is so straight-
forward. I don't yet see why I can't elaborate
the choice procedure enough -- in this particular
case -- that the "Make countable choices" step becomes
"Make one choice" showing up countably many times.
Please understand that I am not asking for more
explanation here. I am responding now basically out of
gratitude for your earlier help, but I think I will
be more effective on my own for a little while,
reading, scribbling half-glimpsed ideas, and so on.
Thank you again for your help.
>>(One way that /will not/ work, but took me some thought
>>to realize that: "Assume there is a longest sequence ...".\
>>BUT there doesn't have to be an infinite sequence
>>just because there is no maximum finite length,
>>just as there doesn't have to be an infinite natural.)
>
> Precisely. Putting it another way, you don't need
> AC if only finitely many choices are needed. AC
> becomes important when you need to make an infinite
> number of choices.
>
> Consider an infinite number of pairs of shoes. You
> don't need AC to choose one shoe from each pair.
> Just take the left shoe from each pair.
>
> Now consider an infinite number of pairs of socks.
> In this case, you can't choose one sock from each
> pair except by invoking AC.
And yet, Finite Choice is enough to leave no pair of socks
without a chosen sock. This is where I start to boggle.
Jim Burns
Ross A. Finlayson
Hi Jim,
Thanks for the attention. Perhaps we were talking past each other,
because I was thinking that the statement was "it can not be shown in ZF
that there are no non-denumerable sets without denumerable subsets",
that there were particular sets that offered no separation into
denumerably many subsets.
Not having any subsets means the set is empty, not having any
denumerable subsets but being non-empty would have the set being
non-denumerable (uncountable) yet with no separation of its elements
except into non-denumerably many.
The first uncountable ordinal only has countable proper subsets. (Here
there might be a general reference to the notion that uncountable sets
are countable unions of countable sets, which has come up several times
in eg Hausdorff, if I recall correctly Smullyan).
Each set is equivalent to a particular ordinal, the initial ordinal of
cardinals of equivalent sets, although in ZF there is not the trichotomy
of cardinals, because without choice it is not feasible to shown
functions between the two constructs. Yet, each set defined by its
elements does have a cardinal in ZF, and, cardinals are trichotomous,
because each cardinal has an initial ordinal, and ordinals are trichotomous.
So, if each sets has a cardinal then that cardinal is an ordinal and
there is method of separation via transfinite induction over choice
sequences of the set. Yet, in ZF the set's cardinals might not be
comparable to other sets', for example initial ordinals of various
cardinals', cardinals, without a choice sequence confounding separation.
Regards,
Ross F.
Dave Seaman
No. Any finite choice leaves infinitely many pairs of socks with no
chosen sock.
Herbert Newman
>
> Not having any subsets means the set is empty ...
>
Except that the empty set, 0, has 0 as it's (only) subset.
0 c 0.
(I guess you meant _proper subset_ here.)
Herb
Ross A. Finlayson
Hi Herb,
Sure, that is so. Each element in 0 the subset is an element of 0 the set.
Thanks,
Ross F.
MoeBlee
wrote:
> Not having any subsets means the set is empty,
No, it doesn't. There is no set that doesn't have any subsets.
> The first uncountable ordinal only has countable proper subsets.
Where did you get that idea?
> (Here
> there might be a general reference to the notion that uncountable sets
> are countable unions of countable sets,
What kind of notion is that? It's not a theorem of set theory.
> which has come up several times
> in eg Hausdorff, if I recall correctly Smullyan).
What remarks by Hausdorff or Smullyan are you talking about? It's not
a theorem of set theory that uncountable sets are countable unions of
countable sets.
> Each set is equivalent to a particular ordinal,
I guess you mean that in ZFC, every set is 1-1 with some ordinal.
> the initial ordinal of
> cardinals of equivalent sets,
Yes, in ZFC, for every set there is a least ordinal 1-1 with the set.
> although in ZF there is not the trichotomy
> of cardinals, because without choice it is not feasible to shown
> functions between the two constructs.
In ZF, cardinal trichotomy is not provable. But also in ZF, it is not
provable that it is not the case that for any two sets, one of them
injects into the other.
> Yet, each set defined by its
> elements does have a cardinal in ZF,
Without choice (or the numeration theorem) we are NOT ensured that
every set is 1-1 with some ordinal.
> and, cardinals are trichotomous,
> because each cardinal has an initial ordinal, and ordinals are trichotomous.
That's in ZFC, not ZF.
> So, if each sets has a cardinal then that cardinal is an ordinal and
> there is method of separation via transfinite induction over choice
> sequences of the set.
It's not apparent what is meant by "there is a method of separation"
and "transfinite induction over choice sequences of a set".
> Yet, in ZF the set's cardinals might not be
> comparable to other sets', for example initial ordinals of various
> cardinals', cardinals, without a choice sequence confounding separation.
In ZF it's not ensured that every set has a cardinal, in the sense of
a least ordinal that is 1-1 with the set.
Anyway, your remark, "That a set doesn't have a denumerable subset is
basically calling it a limit ordinal" is foolish.
Don't you EVER get just a LITTLE bit tired of always being thoroughly
uninformed about set theory while you spout about it?
MoeBlee
Dave Seaman
> 0 c 0.
On the other hand, Ax(Ay(~(ycx))->x=0) is true.
Every set that that has no subsets is empty.
MoeBlee
> On Mon, 6 Oct 2008 22:46:37 +0200, Herbert Newman wrote:
> > On Mon, 06 Oct 2008 13:23:35 -0700 Ross A. Finlayson wrote:
>
> >> Not having any subsets means the set is empty ...
>
> > Except that the empty set, 0, has 0 as it's (only) subset.
> > 0 c 0.
> > (I guess you meant _proper subset_ here.)
>
> On the other hand, Ax(Ay(~(ycx))->x=0) is true.
> Every set that that has no subsets is empty.
Right, but if 'means' is taken in the sense of 'means the same as',
then
Ay ~y subset of x
doesn't mean
x=0.
MoeBlee
Chris Menzel
<r...@tiki-lounge.com.invalid> said:
> Not having any subsets means the set is empty,
No, the empty has itself as a subset.
> not having any denumerable subsets but being non-empty would have the
> set being non-denumerable (uncountable)
No finite set has a denumerable subset.
> The first uncountable ordinal only has countable proper subsets.
Nonsense. w1 - {0} is uncountable and a proper subset of w1.
Ross A. Finlayson
> On Mon, 06 Oct 2008 13:23:35 -0700, Ross A. Finlayson
> <r...@tiki-lounge.com.invalid> said:
>> Not having any subsets means the set is empty,
>
> No, the empty has itself as a subset.
>
Sure, it does.
>> not having any denumerable subsets but being non-empty would have the
>> set being non-denumerable (uncountable)
>
> No finite set has a denumerable subset.
>
OK, where denumerable means countably infinite (as opposed to countable)
that's so. Otherwise it was being used in the same sense as "countable".
>> The first uncountable ordinal only has countable proper subsets.
>
> Nonsense. w1 - {0} is uncountable and a proper subset of w1.
>
The first uncountable ordinal only has countable elements. Each of
those is a countable union of countable sets.
Thanks,
Ross F.
Ross A. Finlayson
> On Oct 6, 1:23 pm, "Ross A. Finlayson" <r...@tiki-lounge.com.invalid>
> wrote:
>
>> Not having any subsets means the set is empty,
>
> No, it doesn't. There is no set that doesn't have any subsets.
>
Sure, the empty set has no proper subsets, and is its own subset.
>> The first uncountable ordinal only has countable proper subsets.
>
> Where did you get that idea?
>
That's a good question. The ordinals that comprise the least
uncoountable ordinal are each countable, much as the ordinals that
comprise the "least" infinite ordinal are finite. Infinite sets are
equivalent.
>> (Here
>> there might be a general reference to the notion that uncountable sets
>> are countable unions of countable sets,
>
> What kind of notion is that? It's not a theorem of set theory.
>
>> which has come up several times
>> in eg Hausdorff, if I recall correctly Smullyan).
>
> What remarks by Hausdorff or Smullyan are you talking about? It's not
> a theorem of set theory that uncountable sets are countable unions of
> countable sets.
>
I'm talking about the general remarks that it is so.
>> Each set is equivalent to a particular ordinal,
>
> I guess you mean that in ZFC, every set is 1-1 with some ordinal.
>
Yeah, that is what is written.
That was where, sets generally are ordinals, so, it is not so relevant.
It's relevant outside of ZF, which is inconsistent, anywhere outside
ZF (else ZF would be complete, which could only be shown non-externally
to ZF were ZF inconsistent, which it is).
Not my problem.
Thanks,
Ross F.
MoeBlee
wrote:
> MoeBlee wrote:
> > On Oct 6, 1:23 pm, "Ross A. Finlayson" <r...@tiki-lounge.com.invalid>
> > wrote:
> >> The first uncountable ordinal only has countable proper subsets.
>
> > Where did you get that idea?
>
> That's a good question. The ordinals that comprise the least
> uncoountable ordinal are each countable,
The ordinals that are members of the least uncountable ordinal are all
countable.
That doesn't entail that every proper subset of the least uncountable
ordinal is countable.
> much as the ordinals that
> comprise the "least" infinite ordinal are finite. Infinite sets are
> equivalent.
Infinite sets are equivalent (equivalent in cardinality, I suppose you
mean) to WHAT?
> >> (Here
> >> there might be a general reference to the notion that uncountable sets
> >> are countable unions of countable sets,
>
> > What kind of notion is that? It's not a theorem of set theory.
>
> >> which has come up several times
> >> in eg Hausdorff, if I recall correctly Smullyan).
>
> > What remarks by Hausdorff or Smullyan are you talking about? It's not
> > a theorem of set theory that uncountable sets are countable unions of
> > countable sets.
>
> I'm talking about the general remarks that it is so.
WHAT "general remarks"?
It is NOT a theorem of set theory that uncountable sets are countable
unions of countable sets.
> >> Each set is equivalent to a particular ordinal,
>
> > I guess you mean that in ZFC, every set is 1-1 with some ordinal.
>
> Yeah, that is what is written.
It's what is proven.
> That was where, sets generally are ordinals,
What is that supposed to MEAN? "generally ordinals". No, sets are NOT
generally ordinals. Rather, in ZFC, for every set there is a well
ordering of that set that is isomorphic to some ordinal. But that
doesn't make sets "generally ordinals".
> so, it is not so relevant.
> It's relevant outside of ZF, which is inconsistent, anywhere outside
> ZF (else ZF would be complete, which could only be shown non-externally
> to ZF were ZF inconsistent, which it is).
I have no idea what that jumble of words is supposed to mean.
> Not my problem.
Of course it's not your problem when you spew misinformation about a
subject you know nothing about; just like it's not a baby's problem
when, with a smile on its face, it pees on the kitchen floor.
MoeBlee
Ross A. Finlayson
> On Oct 6, 6:37 pm, "Ross A. Finlayson" <r...@tiki-lounge.com.invalid>
> wrote:
>> MoeBlee wrote:
>>> On Oct 6, 1:23 pm, "Ross A. Finlayson" <r...@tiki-lounge.com.invalid>
>>> wrote:
>
>>>> The first uncountable ordinal only has countable proper subsets.
>>> Where did you get that idea?
>> That's a good question. The ordinals that comprise the least
>> uncoountable ordinal are each countable,
>
> The ordinals that are members of the least uncountable ordinal are all
> countable.
>
> That doesn't entail that every proper subset of the least uncountable
> ordinal is countable.
>
>> much as the ordinals that
>> comprise the "least" infinite ordinal are finite. Infinite sets are
>> equivalent.
>
> Infinite sets are equivalent (equivalent in cardinality, I suppose you
> mean) to WHAT?
>
Infinite sets are irregular. Infinite sets are equivalent (to each other).
>>>> (Here
>>>> there might be a general reference to the notion that uncountable sets
>>>> are countable unions of countable sets,
>>> What kind of notion is that? It's not a theorem of set theory.
>>>> which has come up several times
>>>> in eg Hausdorff, if I recall correctly Smullyan).
>>> What remarks by Hausdorff or Smullyan are you talking about? It's not
>>> a theorem of set theory that uncountable sets are countable unions of
>>> countable sets.
>> I'm talking about the general remarks that it is so.
>
> WHAT "general remarks"?
>
> It is NOT a theorem of set theory that uncountable sets are countable
> unions of countable sets.
>
I mean the general attribution that "uncountable sets are countable
unions of countable sets".
>>>> Each set is equivalent to a particular ordinal,
>>> I guess you mean that in ZFC, every set is 1-1 with some ordinal.
>> Yeah, that is what is written.
>
> It's what is proven.
>
So it is written.
>> That was where, sets generally are ordinals,
>
> What is that supposed to MEAN? "generally ordinals". No, sets are NOT
> generally ordinals. Rather, in ZFC, for every set there is a well
> ordering of that set that is isomorphic to some ordinal. But that
> doesn't make sets "generally ordinals".
>
>> so, it is not so relevant.
>> It's relevant outside of ZF, which is inconsistent, anywhere outside
>> ZF (else ZF would be complete, which could only be shown non-externally
>> to ZF were ZF inconsistent, which it is).
>
> I have no idea what that jumble of words is supposed to mean.
>
Here, I'll summarize: ZF's universe contains itself.
>> Not my problem.
>
> Of course it's not your problem when you spew misinformation about a
> subject you know nothing about; just like it's not a baby's problem
> when, with a smile on its face, it pees on the kitchen floor.
>
> MoeBlee
>
Not my problem. Cute, not my problem.
I would say that actually I'm quite familiar with the usage in context
of these terms, with a general sense of their established meaning and
import within the framework of discussion. Furthermore, on the other
side, I have written several, perhaps even more, posts with meaningful
content, i.e., actually advancing the discussion, the mathematical
discussion. In establishing a framework and vocabulary for these few
constructs established alongside the standard, and against it, I'm
looking forward to culling them for material, as I do generally, when
I'm not inventing something.
Thanks,
Ross F.
Virgil
"Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> I mean the general attribution that "uncountable sets are countable
> unions of countable sets".
Even a countable union of pairwise disjoint countable sets, which will
be as "big" as such a union can get, is countable.
How wrong can Ross get?
Ross A. Finlayson
That's attributed to Hausdorff, and perhaps Smullyan. Also I say it.
Virgil, you're damn annoying, your rhetorical question is not even a
veiled insult. That's outright rude, and you're a buffoon trapped in
the past. Next time you come crawling back from licking your wounds,
expect another. We should play an online video game against each other.
There are no applications of the trans-finite where measure theory makes
use of only countable additivity.
That's not to say set theory doesn't have a large collection of results
about the trans-finite cardinals of regular (i.e. finite) set
arithmetic. Reframed correctly, where I see the transfinite cardinals
is in the large finite combonatorics, on their way to the infinite.
(That is where, there is no universe in ZF, no, "Absolute" infinite, no
big infinity in ZF, no greatest.) In the cumulative hierarchy and
cumulative limit hierarchy, and that is basically it instead of there
being the cumulative limit limit hierarchy, in the cumulative n-limit
hierarchy, in ordinals, mapping scalar infinities to limit ordinals in
the large and small along the lines of the Infinity Arithmetic of
Yaroslav Sergeyev, as he describes in his book that is little more than
a pamphlet compared to operations, describing in two paragraphs a
reasonable (for his purposes) axiomatization. In the scalar, fluents
and fluxions, second derivatives, limit ordinals, these other systems of
arithmetic of basically the large (infinites) in partitioning the
infinites into regular, uniform boundaries, to regular, uniform
partitions, (with the division into the boundaries and the partitions,
except where the boundary is the point partition, just like real
numbers) in the methods of exhaustion, are perfectly analytic. They
don't use transfinite cardinals.
So, when Fraenkel tells you countable unions of countable sets can be
uncountable, I forget where he said that, but I seem to recall he did,
there are more sources than two.
Transfinite cardinals are just a safety net for standard real analysis.
Also they might have applications in finite combinatorics. Also, they
help explain the ordinal process. ZF is obviously not a complete system
of infinity, it specifically proscribes itself from so being.
That the properties can transcend the objects with the transfer
principle, for example that collections retain the traits of the
elements, is a convenient feature of those things. It's like if a
bucket of apples are from Mexico, each apple is from Mexico. Is the set
of objects from Mexico? No, the apple is an object, not a set. The
apples' properties apply, to each apple from that basket separately and
together, simply the membership operation, of the set-like object,
applies, the property is transcended, the transfer principle applies.
Then, in the proviso of induction, over the cumulative hierarchy of
ordinals, i-e as is done in the method of transfinite induction, because
that's a way induction operates in ordinals, and as well it's a
mechanism that works where the ordinals permeate the cardinals.
So, the transfer principle is and was actually relevant in these
descriptions of the relations between the finitely many and infinitely
many objects, or sets, of that post.
Thanks,
Ross F.
MoeBlee
wrote:
> MoeBlee wrote:
> > On Oct 6, 6:37 pm, "Ross A. Finlayson" <r...@tiki-lounge.com.invalid>
> > wrote:
> >> MoeBlee wrote:
> >>> On Oct 6, 1:23 pm, "Ross A. Finlayson" <r...@tiki-lounge.com.invalid>
> >>> wrote:
>
> >>>> The first uncountable ordinal only has countable proper subsets.
> >>> Where did you get that idea?
> >> That's a good question. The ordinals that comprise the least
> >> uncoountable ordinal are each countable,
>
> > The ordinals that are members of the least uncountable ordinal are all
> > countable.
>
> > That doesn't entail that every proper subset of the least uncountable
> > ordinal is countable.
>
> >> much as the ordinals that
> >> comprise the "least" infinite ordinal are finite. Infinite sets are
> >> equivalent.
>
> > Infinite sets are equivalent (equivalent in cardinality, I suppose you
> > mean) to WHAT?
>
> Infinite sets are irregular.
'irregular' means what?
> Infinite sets are equivalent (to each other).
It is not the case that any two infinite sets have a bijection between
them.
> >>>> (Here
> >>>> there might be a general reference to the notion that uncountable sets
> >>>> are countable unions of countable sets,
> >>> What kind of notion is that? It's not a theorem of set theory.
> >>>> which has come up several times
> >>>> in eg Hausdorff, if I recall correctly Smullyan).
> >>> What remarks by Hausdorff or Smullyan are you talking about? It's not
> >>> a theorem of set theory that uncountable sets are countable unions of
> >>> countable sets.
> >> I'm talking about the general remarks that it is so.
>
> > WHAT "general remarks"?
>
> > It is NOT a theorem of set theory that uncountable sets are countable
> > unions of countable sets.
>
> I mean the general attribution that "uncountable sets are countable
> unions of countable sets".
There is no such. It is not a theorem of set theory that uncountable
sets are countable unions of contable sets.
> >>>> Each set is equivalent to a particular ordinal,
> >>> I guess you mean that in ZFC, every set is 1-1 with some ordinal.
> >> Yeah, that is what is written.
>
> > It's what is proven.
>
> So it is written.
>
> >> That was where, sets generally are ordinals,
>
> > What is that supposed to MEAN? "generally ordinals". No, sets are NOT
> > generally ordinals. Rather, in ZFC, for every set there is a well
> > ordering of that set that is isomorphic to some ordinal. But that
> > doesn't make sets "generally ordinals".
>
> >> so, it is not so relevant.
> >> It's relevant outside of ZF, which is inconsistent, anywhere outside
> >> ZF (else ZF would be complete, which could only be shown non-externally
> >> to ZF were ZF inconsistent, which it is).
>
> > I have no idea what that jumble of words is supposed to mean.
>
> Here, I'll summarize: ZF's universe contains itself.
What do you THINK you mean by "ZF's universe"? If ZF has a model (I'll
omit this 'if' clause from now on), then it has many different models
and with models that don't have the same universe as certain other
models. There is no single model and universe of ZF to takes as "ZF's
universe" as if there is but one universe for a model of ZF. And I
don't think you'll find a universe for a model of ZF such that that
universe is a member of itself. Indeed if ZF itself is our meta-
theory, then no set, thus no universe, is a member of itself.
> I would say that actually I'm quite familiar with the usage in context
> of these terms, with a general sense of their established meaning and
> import within the framework of discussion.
You would say, thus you would add just one more falsehood. You know
JACK about set theory.
> Furthermore, on the other
> side, I have written several, perhaps even more, posts with meaningful
> content, i.e., actually advancing the discussion, the mathematical
> discussion. In establishing a framework and vocabulary for these few
> constructs established alongside the standard, and against it, I'm
> looking forward to culling them for material, as I do generally, when
> I'm not inventing something.
You add misinformation, misunderstanding, confustion, and gibberish.
Your "invention" is as meaningful and inventive as the puke of a prom
girl at the end of her first night out drinking.
MoeBlee
MoeBlee
wrote:
> Virgil wrote:
> > In article <gcei2j$gu...@aioe.org>,
> > "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
>
> >> I mean the general attribution that "uncountable sets are countable
> >> unions of countable sets".
> That's attributed to Hausdorff, and perhaps Smullyan.
No, it's not.
> Also I say it.
You say all kinds of idiotic things.
> Virgil, you're damn annoying, your rhetorical question is not even a
> veiled insult. That's outright rude,
You deserve insult.
MoeBlee
Virgil
"Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> Virgil wrote:
> > In article <gcei2j$guj$1...@aioe.org>,
> > "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> >
> >> I mean the general attribution that "uncountable sets are countable
> >> unions of countable sets".
> >
> > Even a countable union of pairwise disjoint countable sets, which will
> > be as "big" as such a union can get, is countable.
> >
> > How wrong can Ross get?
>
> That's attributed to Hausdorff, and perhaps Smullyan. Also I say it.
A countable union of countable sets is provably countable as follows:
Index each sets by a natural and within each set, each member by a
natural. This is possible because of those assumed countablities.
The each element in the union has at least one ordered pair index of
form (set-index, member-index) in NxN.
One can then single index these in sequence by ordering (m,n) indices so
that:
(1) If m1 = m2 and n1=n2 then (m1,n1) = (m2,n2)
(2) If m1 + n1 < m2 + n2 then (m1,n1) < (m2,n2)
(3) If m1 + n1 = m2 + n2 and m1 < n2 then (m1,n1) < (m2,n2)
>
> Virgil, you're damn annoying
I am also quite often right, which is probably what is most annoying.
The countability of a countable union of countable sets is often a
homework problem in a set theory course, so is hardly esoteric.
Ross A. Finlayson
That's attributed to Hausdorff, and perhaps Smullyan. Also I say it.
That's outright rude, expect another.
MoeBlee
wrote:
> MoeBlee wrote:
> > On Oct 7, 12:59 am, "Ross A. Finlayson" <r...@tiki-lounge.com.invalid>
> > wrote:
> >> Virgil wrote:
> >>> In article <gcei2j$gu...@aioe.org>,
> >>> "Ross A. Finlayson" <r...@tiki-lounge.com.invalid> wrote:
> >>>> I mean the general attribution that "uncountable sets are countable
> >>>> unions of countable sets".
>
> >> That's attributed to Hausdorff, and perhaps Smullyan.
>
> > No, it's not.
>
> >> Also I say it.
>
> > You say all kinds of idiotic things.
>
> >> Virgil, you're damn annoying, your rhetorical question is not even a
> >> veiled insult. That's outright rude,
>
> > You deserve insult.
> That's attributed to Hausdorff, and perhaps Smullyan. Also I say it.
There is no reason to believe that Hausdorff or Smullyan ever made the
incorrect claim that uncountable sets are the countable union of
countable sets.
> That's outright rude, expect another.
Yes, that you continue to spew misinformation is a form of rudeness.
> So, when Fraenkel tells you countable unions of countable sets can be
> uncountable,
What he may have said is that without a choice axiom, we can't prove
that all countable unions of countable sets are countable. But that is
not a claim that all uncountable sets are countable unions of
countable sets.
I REALLY REALLY do not understand why you lack the most basic
motivation to pick up a textbook on set theory so that you could learn
about it aside from the mountain of confusions you've created for
yourself from your ignorance.
MoeBlee
Ross A. Finlayson
...
> Yes, that you continue to spew misinformation is a form of rudeness.
>
>> So, when Fraenkel tells you countable unions of countable sets can be
>> uncountable,
>
> What he may have said is that without a choice axiom, we can't prove
> that all countable unions of countable sets are countable. But that is
> not a claim that all uncountable sets are countable unions of
> countable sets.
>
> I REALLY REALLY do not understand why you lack the most basic
> motivation to pick up a textbook on set theory so that you could learn
> about it aside from the mountain of confusions you've created for
> yourself from your ignorance.
>
> MoeBlee
Hi,
Perhaps the situation would be easier to understand if you might
acknowledge that I do read, and that I do use standard terminology,
generally standardly, when it exists. Your overgeneralization and wrong
assumption is leading you to confuse yourself.
Also, that there are these echoes of earlier students of the field might
lead you to reconsider that these are unfounded statements.
Above it was so claimed that a countable union of countable sets is not
uncountable, yet, as you claim it is not possible to so prove that in
ZF. So, the reasonable observer might ascertain that where those
conditions are mutually exclusive, that it is decideable and
undecideable, that's a ground for a claim of contradiction.
Not all who wander are lost.
These notions of alternative foundations, for foundations they are in
the mathematical logic and foundations of mathematics, arise from
perceived needs in their use, not in denial of existing work. When
there are identified reasonable and in some senses better alternatives,
they can stand on their own regardless of whether they offend your
sensibilities.
The reals are well-ordered? Well-order them.
Regards,
Ross F.
MoeBlee
wrote:
> Perhaps the situation would be easier to understand if you might
> acknowledge that I do read, and that I do use standard terminology,
It would be better for you to understand that whatever else you read,
you've not sat down to understand the material in an introductory
textbook on set theory, and that you do not even come close to having
a grasp of basic standard terminology in the subject.
> Above it was so claimed that a countable union of countable sets is not
> uncountable, yet, as you claim it is not possible to so prove that in
> ZF.
What you claimed is: an uncountable set is a countable union of
countable sets. And that is incorrect. There is no theorem of ordinary
set theory that an uncountable set is a countable union of countable
sets. Moreover, with the axiom of choice, we prove that it is NOT the
case that there exists an uncountable set that is a countable union of
countable sets.
MoeBlee
MoeBlee
> Moreover, with the axiom of choice, we prove that it is NOT the
> case that there exists an uncountable set that is a countable union of
> countable sets.
P.S. The axiom of countable choice is sufficient for that.
MoeBlee
David R Tribble
> The reals are well-ordered? Well-order them.
You say this quite a lot, as though the fact that since no one
has yet constructed a well-ordering of the reals, then there
must not in fact actually be a well-ordering of them.
> Infinite sets are equivalent.
> Infinite sets are irregular. Infinite sets are equivalent (to each other).
Irregular?
You also say that quite a lot, as though you've actually
given a disproof of the powerset axiom or Cantor's
diagonal theorem. But I must have missed it, even after
having read hundreds of your posts.
Ross A. Finlayson
Hi David,
In well-ordering the reals, you might recall the particular discussion
about the application of the nested intervals to ordinals instead of
integers and the proviso over ordinals of schemata for transfinite
induction to show how the results were parallel, where there was ready
use made of the consequences of the well-ordering principle a.k.a Axiom
of Choice a.k.a. Zorn's Lemma, via Ono, etcetera.
In well-ordering the reals, regardless of whether in the realm of study
of the axiom of determinacy or the Woodin UCLA camp there are claims
that the reals _are_ "projectively" well-ordered, bringing to mind the
extension of the reals to the two-point compactification with the
extended reals, there are as well considerations that the continuum of
real numbers finds Dedekind/Cauchy/Eudoxus insufficient to represent
them all, where as well the EF (evaluation in their linear order) is
insuficient for all purposes. Together, they form a more cohesive
description, better axioms, for the continuum of the real numbers, where
for example rationals and irrationals are each dense,
Infinite sets are non-well-founded, not regular (vis-a-vis the axioms of
regularity/foundation, the only axiom that restricts comprehensions in
ZF, except to the extent that the definition of the symbol omega
restricts comprehension over the true Spinozan continuum of naturals
with their implicit compactification, being natural), they're irregular.
Now, it is well known that for example the universe would be irregular
or that the order type of ordinals would be irregular and the Russell
set would be irregular, regardless of their specification, that those
bonus, sputnik, or implicit elements extra the specification are
generally references to the object itself. As well, there are various
schools on consideration of irregular sets, with the large sets and
various considerations of what they should be. That the sets are
irregular, that they don't have the properties satisfying regularity, is
of their properties.
In terms of some "disproof" of the powerset axiom, showing it false,
instead I argue that the regularity and infinity axioms are actually the
false ones. In terms of the diagonal theorem, I present about a dozen
arguments of its interpretation in terms of various and a particular
example of a (nonstandard, non-real) function, EF, the equivalency
function, and so discuss it with regards to the sets of natural integers
and real numbers. With regard to the powerset results accordingly, I
agree that, although infinite sets are equivalent, there are particular
addressing schemes that do not admit bijections among set and powerset.
However, the infinite sets are irregular and thus via some other
addressing, that the sets being infinite are equivalent, because they're
inexhaustible.
That isn't to say that I don't see tremendous utility in the asymptotics
of the number series and etcetera, or for that matter other methods of
the infinit(esim)al analysis, continuum analysis. Indeed, there are
extant in the continuous and discrete many quite general methods of
partition and summation over the indices which are not finite. Far from
it, I think only half of the integers are even.
Even something as simple as a claim that "of the integers, half of them
are even", is not uncontroversial. In the language of number theory,
where that is interpreted that the asymptotic density of the evens
within the integers is 1/2, that is certainly true. In the
probabilistic sense, the statement is modeled by the finite,
asymptotically, and I even observed that there was set-theoretic
machinery in ZFC to maintain a uniform distribution over them all.
Again, that leads right back to these particular results that shape the
consensus on their numbers and the realm of their usage, in this case
the argument of Vitali that there would thus exist among the real
numbers a constant the sum over the infinite indices of the naturals
equals between one and three, an infinitesimal constant which I believe
he called c. That such a thing would exist, these infinitesimals in the
real numbers or continuum, is argued since the two and three thousand
years ago, at least. So, half of the integers are even, because they're
numbers, and the most accessible statement agreeing with expectation
about the numbers is via the definition of the asymptotic density of the
one within the other. Where the putative foundations don't allow that
as a statement, it's not, and where it is a statement in the domain of
interest, in this case the universe of integers and all functions
between them etcetera, the structure, the context, it is and is not.
Where it is generally seen that it is the duty of foundations to support
and make rigorous the application of formalizations in derivation of
true statements, particularly for those elementary considerations that
are nearly axiomatic in the systems of definitions and rules that
comprise those higher level mathematical theories, in this case of integers.
Thanks,
Ross F.
Chris Menzel
<r...@tiki-lounge.com.invalid> said:
> Chris Menzel wrote:
>> On Mon, 06 Oct 2008 13:23:35 -0700, Ross A. Finlayson
>> <r...@tiki-lounge.com.invalid> said:
>>> Not having any subsets means the set is empty,
>>
>> No, the empty has itself as a subset.
>
> Sure, it does.
>
>>> not having any denumerable subsets but being non-empty would have
>>> the set being non-denumerable (uncountable)
>>
>> No finite set has a denumerable subset.
>
> OK, where denumerable means countably infinite (as opposed to
> countable) that's so. Otherwise it was being used in the same sense
> as "countable".
Well don't use it that way. "denumerable" means "countably infinite".
>>> The first uncountable ordinal only has countable proper subsets.
>>
>> Nonsense. w1 - {0} is uncountable and a proper subset of w1.
>
> The first uncountable ordinal only has countable elements.
Quite so.
> Each of those is a countable union of countable sets.
Pretty obviously.
So therefore...what? You think that, from this, it follows that every
proper subset of w1 (the first uncountable ordinal) is countable?
Including the set w1 - {0}? You might want to think that one through.
MoeBlee
wrote:
> Infinite sets are non-well-founded, not regular (vis-a-vis the axioms of
> regularity/foundation, the only axiom that restricts comprehensions in
> ZF, except to the extent that the definition of the symbol omega
> restricts comprehension over the true Spinozan continuum of naturals
> with their implicit compactification, being natural), they're irregular.
If we ever need someone to write copy for another Sokal hoax, we'll
definitely give you a call. In the meantime though, keep your day job.
MoeBlee
Mariano Suárez-Alvarez
Heh.
The level of absurdity is raising lately around here?
Or is it simply that my resistance to it is lowering?
-- m
Ross A. Finlayson
Hi, no, it is not so absurd.
Consider for example that via Skolemization, where there's a model of a
theory there's a standard countable model. If you look at model theory
you see that models are ordinals. So, for example ZF has a standard
countable model, but, ZF's universe is irregular, the model is that
universe, so, any least infinity is irregular, because it models an
irregular universe.
In terms of absurdity you might wonder why there aren't applications of
the trans-finite cardinals in analysis, yet there is regular and
widespread use among many fields of nonstandard, nonreal functions, with
applications.
Similarly, in terms of physical experiment and theories of natural
physics, in the consideration that physical objects are modeled by
mathematical objects, and that the functions between those objects as
well model physical objects, there are infinitely many of them, and
altogether they are the physical universe, and, also, the mathematical
model is its own powerset. So, in reality, the universe of physical
and/or mathematical things is irregular.
There is no set-theoretical universe in regular set theory. The regular
set theory is missing all of itself. Yet, in any supertheory of ZF that
has irregular items, ZF's universe is the Russell set and contains itself.
Similarly, age-old notions of the "point at infinity" and so on reflect
the notional Ouroboros of the infinity, that consumes itself, contains
itself. That is reflected in useful applications with the projectively
extended real number line, the consistency of meromorphological
transforms in modern analysis, and etcetera.
Thanks,
Ross F.
MoeBlee
The second one you mentioned. The daily rate of posted crank nonsense
is a universal constant.
MoeBlee
MoeBlee
wrote:
> Consider for example that via Skolemization, where there's a model of a
> theory there's a standard countable model.
What do you mean by 'standard model'?
> If you look at model theory
> you see that models are ordinals.
Nicely done! Just blurt out some arbitrary assertion as if is true.
> So, for example ZF has a standard
> countable model,
Since you're mentioning ordinals, yes, (if ZF has a model then) ZF has
a model whose universe is the set of natural numbers; but I don't
think that model is standard in the sense of 'e' mapping to the
membership relation on the universe.
> but, ZF's universe is irregular, the model is that
> universe, so, any least infinity is irregular, because it models an
> irregular universe.
Nicely done again.
Similarly, as to further of your post not included here.
Virtually anyone who understands set theory has read at least one
textbook on the subject. Would you please tell me what you have
against doing that? Why do you find it so onerous to get an
introductory textbook on set theory (or even better, first one on the
predicate calculus), settle back in a comfortable chair, turn to page
one, and start reading?
MoeBlee
James Burns
This brings up a question of correct terminology:
If the daily rate of posted crank nonsense were
constant over, not only this universe, but over
all alternate universes, it presumably would be
a more-than-universal constant. What would be
a good way of expressing this?
Jim Burns
Ross A. Finlayson
Hi Moeblee,
You gripe about that often. I recall one time you lamented that nobody
else studies (or rather that I hadn't enough) and I mentioned I had a
copy of Suppes' Axiomatic Set Theory sitting here. Weren't you happy?
I have a copy of it sitting here, and, I've read much of it. Other
books within reach include Tarski's Introduction to Logic, Halmos'
Finite-Dimensional Vector spaces, here's even a Hardy's Pure
Mathematics. I'm not unfamiliar with their contents. Just like many
here I have a math or logic degree.
Get a grip, you're not the only person with a library or penchant for study.
Look at model theory, you'll see models are ordinals, functionally or
mechanistically, for example in consideration of Cohen's independence
results with the ordinals and their disorder, in that example, as has
been discussed. (Model theory is an unnecessary abstraction.)
That aside I hope you can see that I attempt to address these issues
holistically, with a sense of their meaning in a bigger picture or the
big picture, these sweeping claims of universalism vis-a-vis regular set
theories' restriction of comprehension lead to one or the other, and not
both, holding forth. Yet, where ZF is incomplete (or it's
inconsistent), and ZF-R has a "greater" universe than ZF, with or
without antifoundational support, then via separation ZF's universe thre
is the Russell set, containing itself, irregular.
Thanks,
Ross F.
MoeBlee
wrote:
> You gripe about that often. I recall one time you lamented that nobody
> else studies (or rather that I hadn't enough)
Why write "nobody else" where you correct it immediately anyway? I
don't complain that "nobody else" studies; that would be ridiculous.
> and I mentioned I had a
> copy of Suppes' Axiomatic Set Theory sitting here. Weren't you happy?
> I have a copy of it sitting here, and, I've read much of it.
I can only imgagine. What's your technique? Read a word from page 82,
then a word from page 43, then three words from page 114 etc., until
eventually you've read enough words to say you've read much of it?
> Other
> books within reach include Tarski's Introduction to Logic, Halmos'
> Finite-Dimensional Vector spaces, here's even a Hardy's Pure
> Mathematics. I'm not unfamiliar with their contents. Just like many
> here I have a math or logic degree.
You have a degree in math or a degree in logic? What you have is an
especially high degree of confusion.
> Get a grip, you're not the only person with a library or penchant for study.
You, get a grip...I've never said ANYTHING LIKE that I am the only
person to have books and to study them.
> Look at model theory, you'll see models are ordinals, functionally or
> mechanistically, for example in consideration of Cohen's independence
> results with the ordinals and their disorder, in that example, as has
> been discussed. (Model theory is an unnecessary abstraction.)
That bizarre jumble you just said is something you learned on the road
to your degree in math or logic? Something you learned from your
reading? You're too prolific by far.
> That aside I hope you can see that I attempt to address these issues
> holistically, with a sense of their meaning in a bigger picture or the
> big picture, these sweeping claims of universalism vis-a-vis regular set
> theories' restriction of comprehension lead to one or the other, and not
> both, holding forth. Yet, where ZF is incomplete (or it's
> inconsistent), and ZF-R has a "greater" universe than ZF, with or
> without antifoundational support, then via separation ZF's universe thre
> is the Russell set, containing itself, irregular.
Brilliant. I wish I had a degree in mathematics or logic so that I
were at a level to converse with you.
MoeBlee
David R Tribble
>> The reals are well-ordered? Well-order them.
>
David R Tribble wrote:
>> You say this quite a lot, as though the fact that since no one
>> has yet constructed a well-ordering of the reals, then there
>> must not in fact actually be a well-ordering of them.
>
Ross A. Finlayson wrote:
> In well-ordering the reals, you might recall the particular discussion
> about the application of the nested intervals to ordinals instead of
> integers and the proviso over ordinals of schemata for transfinite
> induction to show how the results were parallel,
[snip]
> In terms of some "disproof" of the powerset axiom, showing it false,
> instead I argue that the regularity and infinity axioms are actually the
> false ones.
[snip]
Sorry, but I usually get lost after the first seven or eight
subordinate clauses in your single-sentence paragraphs.
Scanning quickly over your post, I see a lot of phrases that
look like other stuff you've posted in the past. Still can't say
that any of it actually makes sense to me, perhaps someone
else has more time and patience to form a coherent synopsis
of it and respond accordingly...
I will, though, make the rather obvious comment that only
those uneducated in the ways of mathematics and logic are
wont to say such incredibly silly things like "those axioms
are false".
Ross A. Finlayson
> Ross A. Finlayson wrote:
>>> The reals are well-ordered? Well-order them.
>
> David R Tribble wrote:
>>> You say this quite a lot, as though the fact that since no one
>>> has yet constructed a well-ordering of the reals, then there
>>> must not in fact actually be a well-ordering of them.
>
> Ross A. Finlayson wrote:
>> In well-ordering the reals, you might recall the particular discussion
>> about the application of the nested intervals to ordinals instead of
>> integers and the proviso over ordinals of schemata for transfinite
>> induction to show how the results were parallel,
> [snip]
>> In terms of some "disproof" of the powerset axiom, showing it false,
>> instead I argue that the regularity and infinity axioms are actually the
>> false ones.
> [snip]
>
> Sorry, but I usually get lost after the first seven or eight
> subordinate clauses in your single-sentence paragraphs.
>
Perhaps it would be bettered if I offered to replace the signal along
parallel phase adjustment tree.
Seriously, though, program channel phase adjustment basically reduces in
arithmetic those sentences.
> Scanning quickly over your post, I see a lot of phrases that
> look like other stuff you've posted in the past. Still can't say
> that any of it actually makes sense to me, perhaps someone
> else has more time and patience to form a coherent synopsis
> of it and respond accordingly...
>
> I will, though, make the rather obvious comment that only
> those uneducated in the ways of mathematics and logic are
> wont to say such incredibly silly things like "those axioms
> are false".
Well it's not necessarily true, they're axioms.
Thanks,
Ross F.
lwa...@lausd.net
wrote:
> MoeBlee wrote:
> > Virtually anyone who understands set theory has read at least one
> > textbook on the subject. Would you please tell me what you have
> > against doing that? Why do you find it so onerous to get an
> > introductory textbook on set theory (or even better, first one on the
> > predicate calculus), settle back in a comfortable chair, turn to page
> > one, and start reading?
> else studies (or rather that I hadn't enough) and I mentioned I had a
> copy of Suppes' Axiomatic Set Theory sitting here. Weren't you happy?
> I have a copy of it sitting here, and, I've read much of it. Other
> books within reach include Tarski's Introduction to Logic, Halmos'
> Finite-Dimensional Vector spaces, here's even a Hardy's Pure
> Mathematics. I'm not unfamiliar with their contents. Just like many
> here I have a math or logic degree.
And so I return to the set theory debates.
When I last left, I promised that I would not post in the set
theory threads until I found a set theory textbook. And so it
is only fitting that I make my first post right in the middle
of yet another debate about set theory textbooks.
I was opposed to buying a textbook because I don't believe
that one should have to spend money to post on sci.math. But
after making a series of bad posts in a certain thread --
where every time I tried to correct myself I instead kept
digging myself into a deeper hole -- I finally relented and
decided to buy a set theory textbook.
So I went to my local bookstore and purchased the least
expensive set theory text I could find. The text I chose is
not one of the pricier texts listed in this thread, or
mentioned anywhere else on sci.math. It is most likely a
text that no one else here on sci.math owns. I reiterate
that I chose the book solely because of its price, which
was around US$25.
Levy (1978) is the author. But who is Levy? The only
reference to the name Levy anywhere on sci.math, except as
part of the duo "Feferman and Levy," in the original 9000+
post WM thread about a fortnight ago, concerning the fact
that the countable union of countable sets is countable is
independent of ZF. I have no proof that this Levy is the
same as Feferman and Levy, and since this Levy actually
refers to Feferman and Levy in the text, it's safe to
assume that these are different Levys.
BTW, the second lowest price for a set theory text was the
Tarski text mentioned in RF's post above. Of course, I've
heard of Tarski. Perhaps I should've purchased Tarski
instead since the name Tarski probably carries more weight
here than Levy (unless Levy is really Feferman and Levy). If
posters start complaining that I don't have an authoritative
text, then maybe I'll buy Tarski too. If RF can afford more
than one text, then so can I.
Then again, Levy frequently cites more wellknown authors
in his theorems. For example, he writes on p. 77:
1.6 Theorem (Proof by induction on finite sets --
Zermelo 1909). If phi(0) and for every finite a and for
every z!ea if phi(a) then phi(au{z}), then we have phi(a)
for every finite set a.
If one doesn't consider Levy to be an authoritative source,
surely that poster must accept Zermelo as authoritative --
he is, after all, the Z in ZFC.
And I feel that any set theory text should be sufficient --
as long as it's not one of those popular texts that confuse
aleph_1 and c, for example.
So this is the plan. Whenever I want to state a result in a
standard theory, I try citing both an Internet source and a
theorem in the book. The former is for those who don't have
a copy of the same book I have and want to be able to just
click on the link and read. The latter is for those who only
trust book sources rather than web sources and prefer that I
back up everything I say with a book. If possible, I won't
cite the middleman Levy, but go straight to the original
prover of the theorem (such as Zermelo in my example).
Of course, if I'm posting something about a nonstandard
theory (such as infinitesimals), then this won't appear in
Levy or most other texts, so I provide only a web link.
Although I've conceded the point about buying a textboook, I
still hold my ground on posting to Google Groups rather than
a full-priced newsreader. I still don't believe that one
should have to pay to post, whether through purchasing a
book or subscribing for a newsreader. But I'll take the
one-time cost of a textbook over a monthly newsreader fee.
Chris Menzel
<r...@tiki-lounge.com.invalid> said:
> MoeBlee wrote:
>> ...
>> Virtually anyone who understands set theory has read at least one
>> textbook on the subject. Would you please tell me what you have
>> against doing that? Why do you find it so onerous to get an
>> introductory textbook on set theory (or even better, first one on the
>> predicate calculus), settle back in a comfortable chair, turn to page
>> one, and start reading?
>>
>> MoeBlee
>
> Hi Moeblee,
>
> You gripe about that often. I recall one time you lamented that
> nobody else studies (or rather that I hadn't enough) and I mentioned I
> had a copy of Suppes' Axiomatic Set Theory sitting here. Weren't you
> happy? I have a copy of it sitting here, and, I've read much of it.
> Other books within reach include Tarski's Introduction to Logic,
> Halmos' Finite-Dimensional Vector spaces, here's even a Hardy's Pure
> Mathematics. I'm not unfamiliar with their contents.
Perhaps you have read the words and can quote some of the theorems, but
it is quite clear that you do not understand many of the most basic
elements of those texts.
> Look at model theory, you'll see models are ordinals,
A fine illustration of the point. It is simply not possible to
understand basic model theory and simultaneously believe that "models
are ordinals". Models are simply not ordinals -- for one thing, a model
include a function that interprets the elements of a language -- nor
need they contain any ordinals.
Herbert Newman
>
> Levy (1978) is the author. But who is Levy?
>
I guess it's the one who co-authored
- Fraenkel, Bar-Hillel, Levy: Foundations of Set Theory.
(A rather well-known book.)
- Levy, Addison, Henkin, Tarski: The Fraenkel-Mostowski Method for
Independence Proofs in Set Theory.
Concerning the book you bought (if so):
- Levy: Basic Set Theory:
"Geared towards upper-level undergraduate and graduate students..."
Hmmm... Maybe not the best choice for an introduction to set theory. But
who knows?
Anyway, Levy certainly will be an "authoritative source" concerning set
theory.
Herb
Aatu Koskensilta
> Levy (1978) is the author. But who is Levy?
Azriel Levy is a renowned logician.
> The only reference to the name Levy anywhere on sci.math, except as
> part of the duo "Feferman and Levy," in the original 9000+ post WM
> thread about a fortnight ago, concerning the fact that the countable
> union of countable sets is countable is independent of ZF. I have no
> proof that this Levy is the same as Feferman and Levy, and since
> this Levy actually refers to Feferman and Levy in the text, it's
> safe to assume that these are different Levys.
No, they're the same.
> Although I've conceded the point about buying a textboook, I
> still hold my ground on posting to Google Groups rather than
> a full-priced newsreader.
How about a free newsreader such as Gnus?
> I still don't believe that one should have to pay to post, whether
> through purchasing a book or subscribing for a newsreader. But I'll
> take the one-time cost of a textbook over a monthly newsreader fee.
There is a newsreader that comes with a monthly fee? This idea must be
due to some confusion on your part.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
georgie
snipped...
> I will, though, make the rather obvious comment that only
> those uneducated in the ways of mathematics and logic are
> wont to say such incredibly silly things like "those axioms
> are false".- Hide quoted text -
I will add that only those that have been programmed in mathematics
like religious fanatics fail to realize that axioms aren't the "truth
and the
way". They're just axioms. People who object to things like the
diagonal argument simply disagree with the fanatical belief in certain
axioms. Those "disbelievers" aren't inferrior idiots just because
the
mathematical fanatics can't stand to see someone disagree with their
holy axioms.
Aatu Koskensilta
> People who object to things like the diagonal argument simply
> disagree with the fanatical belief in certain axioms. Those
> "disbelievers" aren't inferrior idiots just because the mathematical
> fanatics can't stand to see someone disagree with their holy axioms.
So which of the axioms used in the diagonal argument do you object to?
herbzet
lwa...@lausd.net wrote:
> And so I return to the set theory debates.
Nice to see you again.
[...]
> Although I've conceded the point about buying a textboook, I
> still hold my ground on posting to Google Groups rather than
> a full-priced newsreader. I still don't believe that one
> should have to pay to post, whether through purchasing a
> book or subscribing for a newsreader. But I'll take the
> one-time cost of a textbook over a monthly newsreader fee.
A newsreader is a piece of software, very like an e-mail program.
It appears that your browser is Mozilla 4.0, so (if I'm not
mistaken) you can download the /free/ Thunderbird newsreader
at the Mozilla site.
Many ISP's have their own Usenet servers, so their customers
already have /free/ access (that is, it comes with their
account). You should check with your ISP.
Other people subscribe to a news provider, but there are
/free/ news servers, such as at http://motzarella.org
which is what I use, being a cheapskate.
To register at motzarella (a trivial task), and to install
and configure a newsreader (not hard) might take a little
effort one afternoon, but it's worth it.
If you're using your work computer, you might have to
call the network administration techies to help out,
which, admittedly, can be a pain at some organizations.
--
hz
Daryl McCullough
>
>georgie <geo_...@yahoo.com> writes:
>
>> People who object to things like the diagonal argument simply
>> disagree with the fanatical belief in certain axioms. Those
>> "disbelievers" aren't inferrior idiots just because the mathematical
>> fanatics can't stand to see someone disagree with their holy axioms.
>
>So which of the axioms used in the diagonal argument do you object to?
The Axiom of Idiocy, which says that anyone who doesn't believe is
ZF is an iferrior idiot. The question of whether this axiom is independent
of the others has never been settled, although Cohen was able to show
that it follows from ZF + "there exists a ridiculously large cardinal".
--
Daryl McCullough
Ithaca, NY
Herbert Newman
>
> I will add that only those that have been programmed in mathematics
> like religious fanatics fail to realize that axioms aren't the "truth
> and the way".
>
Right. Though they are at least the "starting point" for formal proofs in
math.
>
> They're just axioms.
>
That's true. But more important: some sets of axioms are the basis of
certain mathematical theories (say set theory, group theory, etc.).
>
> People who object to things like the diagonal argument simply disagree
> with the fanatical belief in certain axioms.
>
Oh, if it only were that simple. Actually many of them are just CRANKs,
incapable of logical and/or mathematical reasoning.
>
> Those "disbelievers" aren't inferior idiots just because ...
>
The sad truth is: some (if not many or most) are.
Herb
Brian Chandler
> Ross A. Finlayson wrote:
...and wrote and wrote...
> Sorry, but I usually get lost after the first seven or eight
> subordinate clauses in your single-sentence paragraphs.
That might be the idea. I've often thought that possibly Ross is
really extremely clever -- that's assuming of course that he's
actually a program.
Brian Chandler
georgie
> georgie <geo_c...@yahoo.com> writes:
> > People who object to things like the diagonal argument simply
> > disagree with the fanatical belief in certain axioms. Those
> > "disbelievers" aren't inferrior idiots just because the mathematical
> > fanatics can't stand to see someone disagree with their holy axioms.
>
> So which of the axioms used in the diagonal argument do you object to?
>
> --
>
> "Wovon man nicht sprechen kann, darüber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
I didn't say I objected to any of them. I'm just not so fanatical
that it
makes me incapable of recognizing the fact that the objectors aren't
idiots. They just aren't working from the same set of axioms.
georgie
>
> > People who object to things like the diagonal argument simply disagree
> > with the fanatical belief in certain axioms.
>
> Oh, if it only were that simple. Actually many of them are just CRANKs,
> incapable of logical and/or mathematical reasoning.
Many of them or all of them?
Virgil
<63644047-a158-49b1...@i20g2000prf.googlegroups.com>,
georgie <geo_...@yahoo.com> wrote:
How about those anti-mathematical fanatics who argue that no one should
be allowed to assume axioms which conflict with their own oddball
beliefs?
Virgil
<d4da9744-8f4e-4d1e...@z6g2000pre.googlegroups.com>,
georgie <geo_...@yahoo.com> wrote:
Mathematicians should be capable of considering the consequences of
accepting any set of axioms. That does not mean that they regard such
axiom sets as gospel.
For example ZFC and NF are incompatible, but mathematicians are, or at
least should be, capable of considering the consequences of either axiom
set.
Virgil
<95da4608-f8ac-4f80...@r38g2000prr.googlegroups.com>,
georgie <geo_...@yahoo.com> wrote:
Many of them. A few, too few, actually learn how not to be cranks.
georgie
> In article
> <63644047-a158-49b1-b034-e8f7ffa6e...@i20g2000prf.googlegroups.com>,
>
> - Show quoted text -
Yeah! How about them!
Also - How about them Dodgers, Hey?
georgie
>
> georgie <geo_c...@yahoo.com> wrote:
> > On Oct 9, 9:41 am, Herbert Newman <nomail@invalid> wrote:
>
> > > > People who object to things like the diagonal argument simply disagree
> > > > with the fanatical belief in certain axioms.
>
> > > Oh, if it only were that simple. Actually many of them are just CRANKs,
> > > incapable of logical and/or mathematical reasoning.
>
> > Many of them or all of them?
>
> Many of them. A few, too few, actually learn how not to be cranks.
But before they conform to YOUR beliefs, you believe they are cranks,
right?
MoeBlee
> I was opposed to buying a textbook because I don't believe
> that one should have to spend money to post on sci.math.
The point is not whether one needs to spend money to post. The point
is that a person who repeatedly posts incorrect claims about a
technical subject (I'm talking about cranks, not you) should instead
first familiarize himself with the basics of that technical subject,
and that is best done by studying a good textbook.
> It is most likely a
> text that no one else here on sci.math owns.
Why do you think that?
> Levy (1978) is the author.
Do you mean Azriel Levy's 'Basic Set Theory'. Other posters on
sci.math do own it.
> But who is Levy?
Azriel Levy is among the the most prolific, famous, and referenced set
theorists.
> BTW, the second lowest price for a set theory text was the
> Tarski text mentioned in RF's post above.
That book just briefly touches on the subject of set theory itself.
Anyway, since you're concerned with price, if you can purchase by
Internet, then a used copy of the Dover edition of Suppes's 'Axiomatic
Set Theory' can often be found for just a few bucks plus shippping.
And that book is particularly recommended as a good start.
The Levy book is great, but misnamed in my opinion. It's somewhat more
advanced than an introductory (basic) text. When reading Levy I'd keep
in mind that he "straddles" between set theory and class theory, as
his treatment of set abstraction notation lets him refer to proper
classes even though he's officially in ZFC. That's okay as long as one
understands that in that context mention of proper classes can be (and
officially is) reduced to only sets. As to the level of the text, you
will notice that even the appendix that Levy includes to justify his
treatment of set abstraction notation is quite complicated and
requires some mathematical (even set theoretical) sophistication.
MoeBlee
MoeBlee
> People who object to things like the
> diagonal argument simply disagree with the fanatical belief in certain
> axioms.
NOT TRUE.
It is offered over and over and over to cranks that they are welcome
not to accept certain axioms. The problem with cranks is not over what
axioms to accept, but rather the refusal of cranks to recognize that,
whether we accept the axioms or not, the diagonal argument does
provide a proof from its axioms. More specifically, given the exact
formalized rules of even just intuitionisitc logic, there is a
sequence of formulas, each of which is either an axiom of Z set theory
or derived by the rules from previous entries in the sequence ,and
such that the last formula is that there is no function from a set
onto its power set, or, in the case of the question of the continuum,
no function from the set of natural numbers onto the set of real
numbers. And that is a FINITE matter in the sense that it is a matter
of pure inspection of a finite sequence of formulas.
MoeBlee
Ross A. Finlayson
Hello,
The Tarski book is good, I look at it now. I use it as a mouse pad so
it is generally within a foot of me. It has 239 pages. On page 75 are
Relations among Classes. Those are normal parts of classical logic.
Classical logic and modern set-theoretical logic, with fuzzy logic and
so on and also the type theory with the calculi and so on, generally
agree on the classical logic.
I bought it from the library.
I hope you don't mind if I call you Larry, Larry. As regards to your
brief hiatus from posting on sci.math, generally I would describe it as
Muckenheim trying to move on from his discussions to a continuance of
his discussions.
Tarski is beneficial from the regular completeness. Also so, is
universalist, via classical logic with the regular completeness. It's
less to prove it so.
Regards,
Ross F.
Ross A. Finlayson
Here, Tarski says on Cantor:
"CANTOR's set theory is one of those mathematical disciplines which are
in a state of especially intensive development. Its ideas and lines of
thought have penetrated into almost all branches of mathematics and have
exerted everywhere a most stimulating and fertilizing influence."
-- Alfred Tarski, "Introduction to Logic", Eighth Printing, 1959, page 69
Tarski is the same as of the Banach-Tarski results, with the regular
decomposition results. From his quote he generally had a positive
opinion on Cantor. Cantor agrees with much or most of the classical.
Regards,
Ross F.
MoeBlee
> The Tarski book is good,
The Tarski book is sweet old classic. But it's more in the nature of a
summary. It would provide a nice introductory read, but I don't think
it's extensive enough to provide one with a solid mastery of working
in the first order calculus. For that kind of thing, a book such as
Kalish/Montague/Mar is preferable. And of course, the Tarski book is
not a textbook in set theory, though it does touch on the subject.
MoeBlee
lwa...@lausd.net
Thanks.
> > Although I've conceded the point about buying a textboook, I
> > still hold my ground on posting to Google Groups rather than
> > a full-priced newsreader. I still don't believe that one
> > should have to pay to post, whether through purchasing a
> > book or subscribing for a newsreader. But I'll take the
> > one-time cost of a textbook over a monthly newsreader fee.
> A newsreader is a piece of software, very like an e-mail program.
> It appears that your browser is Mozilla 4.0, so (if I'm not
> mistaken) you can download the /free/ Thunderbird newsreader
> at the Mozilla site.
Nope. I just use the default IE browser that comes with my
Windows machine. I know -- I'm cheap.
> Many ISP's have their own Usenet servers, so their customers
> already have /free/ access (that is, it comes with their
> account). You should check with your ISP.
My ISP is AT&T, which was specifically mentioned in one of the
Google debate threads as having no Usenet. I could've had a
better ISP, but once again, I know -- I'm cheap.
> Other people subscribe to a news provider, but there are
> /free/ news servers, such as athttp://motzarella.org
> which is what I use, being a cheapskate.
Thanks for the info. For now, I'll stick to Google, but if
enough sci.math posters join the "killfile everything from
Google" bandwagon, I'd be forced to reconsider one of the
suggestions mention in herbzet's post.
lwa...@lausd.net
> On Oct 9, 12:42 am, lwal...@lausd.net wrote:
> > It is most likely a
> > text that no one else here on sci.math owns.
> Why do you think that?
It's because no one else has referred to the book
anywhere on sci.math. But then later on, MoeBlee
gives the real reasons why sci.math posters don't
recommend the book:
> The Levy book is great, but misnamed in my opinion. It's somewhat more
> advanced than an introductory (basic) text. When reading Levy I'd keep
> in mind that he "straddles" between set theory and class theory, as
> his treatment of set abstraction notation lets him refer to proper
> classes even though he's officially in ZFC.
OK, so I see that Levy's book really isn't "basic,"
and it deals with a class theory as well as pure
set theory.
> Anyway, since you're concerned with price, if you can purchase by
> Internet, then a used copy of the Dover edition of Suppes's 'Axiomatic
> Set Theory' can often be found for just a few bucks plus shippping.
> And that book is particularly recommended as a good start.
Thanks for the info. I'll stick with Levy for now, but I
may take up MoeBlee's suggestion and go with Dover/Suppes
if I struggle with Levy, maybe after the new year.
lwa...@lausd.net
> On Oct 9, 8:00 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> > So which of the axioms used in the diagonal argument do you object to?
> that it
> makes me incapable of recognizing the fact that the objectors aren't
> idiots. They just aren't working from the same set of axioms.
I agree with georgie wholeheartedly. There's nothing wrong
with using theories other than ZFC.
Maybe the following is a way to bridge the gap between the
rigor of ZFC and the intuition of the so-called "cranks."
We all know that we can find a model for many theories by
considering the cumulative hierarchy of sets. So for the
theory ZF-Infinity, V_omega is a model. If we include
Infinity but not the Replacement Schema, then we have
V_(omega+omega) as a model. Allowing for proper classes
adds one more level to the cumulative hierarchy, so that
if V_inaccessible is a model of ZFC, then we have that
V_(inaccessible+1) is a model of NBG.
So we may now ask, given a set of axioms, which is the
smallest ordinal alpha such that V_alpha is a model for
that set of axioms (and membership "e" actually maps
to itself, avoid Lowenheim-Skolem)? Notice that we have
trivial examples -- for example, what about the theory
whose only axioms are Extensionality and Empty Set? Then
V_1 is the simplest model. For V_1 = {0}, and 0 is the
only set whose existence can be proved in the theory.
So here are some models and theories:
V_1 -- Extensionality+Empty Set
V_2 -- Extensionality+Empty Set+Proper Classes
V_omega -- ZF-Infinity
V_(omega+1) -- ZF-Infinity+Proper Classes
V_(omega+2) -- Pocket Set Theory (Holmes)
V_(omega+omega) -- Z
V_(omega+omega+1) -- Z+Proper Classes
V_inaccessible -- ZFC
V_(inaccessble+1) -- NBG
What about Extensionality+Empty Set+Pairing? We
see that this theory proves the existence of 0 as
well as {n} for every _Zermelo_ (_not_ von Neumann)
natural n. Thus this theory proves the existence of
every Zermelo natural n -- and the smallest member
of the cumulative hierarchy containing every
Zermelo natural is V_omega. Of course V_omega is a
model of this theory, but there are simpler models
where no set has cardinality more than two. This
simpler model isn't in the cumulative hierarchy.
But what if we wanted to go in reverse? Instead of
starting with a theory and finding a model, what if
we began with an ordinal alpha, and asked for a
natural set of axioms such that V_alpha is a model
of that theory?
The purpose of this is that if we can determine
which member of the cumulative hierarchy most
represents the universe of sets that a "crank"
believes exists, then we can find a set of axioms
corresponding to the "crank's" theory. And so we
would have found a way to make the "crank" level
of rigor.
Consider tommy1729, for example. Originally, he
believed in the existence of aleph_0, aleph_1,
and aleph_2, but not aleph_3. This seems to
point towards V_(omega+3) as a model. Lately, he
has accepted aleph_omega as a cardinal, so maybe
we can use V_(omega+omega+1). Then again, he has
disapproved of ordinals, so maybe we shouldn't
actually use omega+3 or omega+omega+1 to describe
his theory.
(And before anyone protests, I already know that
card(V_omega+3) = beth_3, not aleph_3. But we
are discussing models here. Thus, in Z we can
prove the existence of aleph_n for every natural
n, but not aleph_omega. This is _not_ the same as
stating that Z proves beth_n = aleph_n for every
natural n.)
As for WM, he seems to have proposed at least three
different theories. (I'm almost glad that Google
has split WM's 9000+ thread into a dozen or more
subthreads, since then we can discuss each of his
theories in a separate thread.)
Sometimes WM supports the existence of infinite sets
but not uncountable sets. Then V_(omega+1) is a
model of this theory.
Sometimes WM rejects infinite sets, but accepts
arbitrary finite sets. We already know that finitists
work in the universe V_omega.
And sometimes WM even rejects the existence of
arbitrary finite sets, especially when he keeps
arguing about floor(pi*googleplex). I'll save that
for its own thread, since I'm not sure whether this
can fit into the cumulative hierarchy.
georgie
> On Oct 9, 5:42 am, georgie <geo_c...@yahoo.com> wrote:
>
> > People who object to things like the
> > diagonal argument simply disagree with the fanatical belief in certain
> > axioms.
>
> NOT TRUE.
YES TRUE. (This is the logical counter argument to NOT TRUE
that you so elegantly used, except YES TRUE arguments always
beat out NOT TRUE arguments due to the axiom of YES TRUE
is better than NOT TRUE. But hey, we're not fanatics, right?)
> It is offered over and over and over to cranks that they are welcome
> not to accept certain axioms. The problem with cranks is not over what
> axioms to accept, but rather the refusal of cranks to recognize that,
> whether we accept the axioms or not, the diagonal argument does
> provide a proof from its axioms. More specifically, given the exact
> formalized rules of even just intuitionisitc logic, there is a
> sequence of formulas, each of which is either an axiom of Z set theory
> or derived by the rules from previous entries in the sequence ,and
> such that the last formula is that there is no function from a set
> onto its power set, or, in the case of the question of the continuum,
> no function from the set of natural numbers onto the set of real
> numbers. And that is a FINITE matter in the sense that it is a matter
> of pure inspection of a finite sequence of formulas.
This sounds like you're using the axiom of intuition that states that
the guy who is more fanatical gets to state that his intuition beats
out people who are less fanatical.
When someone argues that there is no such thing as infinite numbers
and you insist that there are, either you are using an axiom that
states
that there are infinite numbers or you're insisting that your
intuition about
the existence of infinite numbers is the only correct intuition.
Only
fanatics would insist that their axioms are the "truth" or only their
intuition is the "truth". Either way, it's very similar to religious
extremism.
David R Tribble
> This brings up a question of correct terminology:
> If the daily rate of posted crank nonsense were
> constant over, not only this universe, but over
> all alternate universes, it presumably would be
> a more-than-universal constant. What would be
> a good way of expressing this?
Perhaps that would apply only to Ross's irregular universes.
Of course, since those universes are their own powersets,
presumably the constant would be a more-than-power(ful)
constant.
herbzet
lwa...@lausd.net wrote:
> herbzet wrote:
> > It appears that your browser is Mozilla 4.0, so (if I'm not
> > mistaken) you can download the /free/ Thunderbird newsreader
> > at the Mozilla site.
>
> Nope. I just use the default IE browser that comes with my
> Windows machine. I know -- I'm cheap.
A bit of confusion here on my part:
1) If you're using Outlook Express for e-mail, it already has
a news reader built in. At the top of the first page of the
help file for Outlook Express (in my system) it says
"Introducing Outlook Express
Microsoft Outlook Express puts the world of online communication
on your desktop. Whether you want to exchange e-mail with colleagues
and friends or join newsgroups to trade ideas and information, the
tools you need are here."
Do you use Outlook Express for e-mail? It comes with Windows
(well, it came with my version).
2) In the full header to your post it has these two lines:
User-Agent: G2/1.0
X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 6.0; SLCC1;
which I /guessed/ meant you were using "Mozilla 4.0". But it turns out
that /my/ post header says I'm using "Mozilla 4.79", when what I'm
/actually/ using is Netscape Communicator 4.79.
(The current Mozilla "Firefox" browser is only in version 3 according
to their website, which is what I guessed, wrongly, that you were using.)
I'm using the news reader that comes built-in with Netscape
Communicator. Are you using Netscape Navigator/Communicator 4.0?
At work maybe?
3) I don't really need to hear the answers to these questions;
my point is, whether you are using a Netscape browser or
IE, you probably /already have/ a newsreader resident on
your machine, although you might need to set it up.
Then all you need to do is get a /free/ account at
http://motzarella.org (a trivial exercise - you can
do it right now) and you're all set.
You might like it better. It doesn't cost anything to try.
OK, I'm done. :-)
--
hz
Virgil
<cbb43af4-8e60-4f2b...@v56g2000hsf.googlegroups.com>,
georgie <geo_...@yahoo.com> wrote:
Not necessarily. Those who are constructionists need not be cranks, but
many of those who are cranks seem to need to be, at least occasionally,
constructionists.
Non-crank constructionists are usually a good deal less compulsive about
proselytizing.
Aatu Koskensilta
> I didn't say I objected to any of them. I'm just not so fanatical
> that it makes me incapable of recognizing the fact that the
> objectors aren't idiots.
What objections do you have in mind?
--
Aatu Koskensilta (aatu.kos...@uta.fi)
David R Tribble
>> People who object to things like the diagonal argument simply
>> disagree with the fanatical belief in certain axioms. Those
>> "disbelievers" aren't inferrior idiots just because the mathematical
>> fanatics can't stand to see someone disagree with their holy axioms.
>
Aatu Koskensilta wrote:
>> So which of the axioms used in the diagonal argument do you object to?
>
georgie wrote:
> I didn't say I objected to any of them. I'm just not so fanatical that it
> makes me incapable of recognizing the fact that the objectors aren't
> idiots. They just aren't working from the same set of axioms.
More to the point, they just aren't working from the same rules
of logic.
At some point you realize that their failure to acknowledge or
even comprehend the contradictions derived from their own
arguments is more than just simply refusing to accept the
standard axioms. The problem goes much deeper.
Add to that the narcissistic and paranoid tendencies to insult
and threaten anyone who disagrees with them in the slightest
detail, and that just increases the evidence of crankhood.
MoeBlee
> On Oct 9, 12:11 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Oct 9, 12:42 am, lwal...@lausd.net wrote:
> > > It is most likely a
> > > text that no one else here on sci.math owns.
> > Why do you think that?
>
> It's because no one else has referred to the book
> anywhere on sci.math. But then later on, MoeBlee
> gives the real reasons why sci.math posters don't
> recommend the book:
I didn't say anything about why other people have or have not
recommended the book. (By the way, Chris Menzel does say it is his
favorite book set theory textbook, though I have no idea whether he
thinks it is the best in the role of an introduction.)
Anyway, again you show your penchant for odd inferences. I bet there
are lots of books that haven't been mentioned on sci.math but that are
owned by various posters. I wouldn't conclude that no poster owns a
certain book just because it hadn't been mentioned. And, though I
don't recall whether on sci.math or on sci.logic, on at least one of
those two news groups, the Levy book HAS been mentioned, discussed,
and referenced at least a few times.
> I'll stick with Levy for now, but I
> may take up MoeBlee's suggestion and go with Dover/Suppes
> if I struggle with Levy, maybe after the new year.
The new year just passed. ; )
MoeBlee
MoeBlee
> On Oct 9, 2:25 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Oct 9, 5:42 am, georgie <geo_c...@yahoo.com> wrote:
>
> > > People who object to things like the
> > > diagonal argument simply disagree with the fanatical belief in certain
> > > axioms.
>
> > NOT TRUE.
>
> YES TRUE. (This is the logical counter argument to NOT TRUE
> that you so elegantly used, except YES TRUE arguments always
> beat out NOT TRUE arguments due to the axiom of YES TRUE
> is better than NOT TRUE. But hey, we're not fanatics, right?)
But "NOT TRUE" wasn't my ARGUMENT. The argument is:
> > It is offered over and over and over to cranks that they are welcome
> > not to accept certain axioms. The problem with cranks is not over what
> > axioms to accept, but rather the refusal of cranks to recognize that,
> > whether we accept the axioms or not, the diagonal argument does
> > provide a proof from its axioms. More specifically, given the exact
> > formalized rules of even just intuitionisitc logic, there is a
> > sequence of formulas, each of which is either an axiom of Z set theory
> > or derived by the rules from previous entries in the sequence ,and
> > such that the last formula is that there is no function from a set
> > onto its power set, or, in the case of the question of the continuum,
> > no function from the set of natural numbers onto the set of real
> > numbers. And that is a FINITE matter in the sense that it is a matter
> > of pure inspection of a finite sequence of formulas.
>
> This sounds like you're using the axiom of intuition that states that
> the guy who is more fanatical gets to state that his intuition beats
> out people who are less fanatical.
What in the world are you talking about? I talked about the question
of formal derivation of theorems using rules and axioms. I said
NOTHING about intuition. You COMPLETELY missed the point: WHATEVER
one's intuitions, the matter of whether a certain finite sequence of a
certain unequivocally defined kind exists is not a matter of intuition
but rather of pure finite inspection. I don't object to people having
different intuitions that lead to rejecting certain axioms.
> When someone argues that there is no such thing as infinite numbers
> and you insist that there are, either you are using an axiom that
> states
> that there are infinite numbers or you're insisting that your
> intuition about
> the existence of infinite numbers is the only correct intuition.
PLEASE, READ what I wrote. I am NOT insisting that one must accept an
axiom that entails that there exist infinite sets and infinite
cardinal and ordinal numbers. And I am NOT insisting that the
intuition that there are NOT such infinite objects is incorrect.
Rather, just go back to what I actually wrote, as you just quoted it,
please.
> Only
> fanatics would insist that their axioms are the "truth" or only their
> intuition is the "truth". Either way, it's very similar to religious
> extremism.
And that is just the kind of thing we find from cranks. The fallback
crank position is that his principles are the ones that are faithful
to "reality" and that certain other intuitions about the matter are
therefore wrong.
Meanwhile, take, for example, me. You will not find me ever to have
posted that the axioms of ZFC are true in some unqualified sense or
that other intuitions about the matter are not true.
MoeBlee
Michael Press
stevend...@yahoo.com (Daryl McCullough) wrote:
> In article <87r66pz...@alatheia.dsl.inet.fi>, Aatu Koskensilta says...
> >
> >georgie <geo_...@yahoo.com> writes:
> >
> >> People who object to things like the diagonal argument simply
> >> disagree with the fanatical belief in certain axioms. Those
> >> "disbelievers" aren't inferrior idiots just because the mathematical
> >> fanatics can't stand to see someone disagree with their holy axioms.
> >
> >So which of the axioms used in the diagonal argument do you object to?
>
> The Axiom of Idiocy, which says that anyone who doesn't believe is
> ZF is an iferrior idiot. The question of whether this axiom is independent
> of the others has never been settled, although Cohen was able to show
> that it follows from ZF + "there exists a ridiculously large cardinal".
The argument has come to be known as farcing.
--
Michael Press
Jesse F. Hughes
Evidently, you can't comprehend plain English.
Moe argued that it is undeniable that Z proves |R| > |N|. He did not
argue that the Z is the truth.
You're too damned stupid to know the difference between these two
claims.
--
Jesse F. Hughes
Did you lay down in heaven? Did you wake up in hell?
I bet you never guessed that it would be so hard to tell.
-- The Flatlanders, /Judgment Day/
** Posted from http://www.teranews.com **
georgie
> georgie <geo_c...@yahoo.com> writes:
> > I didn't say I objected to any of them. I'm just not so fanatical
> > that it makes me incapable of recognizing the fact that the
> > objectors aren't idiots.
>
> What objections do you have in mind?
The ones that they have.
georgie
> What in the world are you talking about?
By convention, people who can't follow the discussion are
called CRANKS around here.
> I talked about the question
> of formal derivation of theorems using rules and axioms. I said
> NOTHING about intuition. You COMPLETELY missed the point: WHATEVER
> one's intuitions, the matter of whether a certain finite sequence of a
> certain unequivocally defined kind exists is not a matter of intuition
> but rather of pure finite inspection. I don't object to people having
> different intuitions that lead to rejecting certain axioms.
Your a CRANK. You can't even understand the simplest statements.
> > When someone argues that there is no such thing as infinite numbers
> > and you insist that there are, either you are using an axiom that
> > states
> > that there are infinite numbers or you're insisting that your
> > intuition about
> > the existence of infinite numbers is the only correct intuition.
>
> PLEASE, READ what I wrote. I am NOT insisting that one must accept an
> axiom that entails that there exist infinite sets and infinite
> cardinal and ordinal numbers. And I am NOT insisting that the
> intuition that there are NOT such infinite objects is incorrect.
> Rather, just go back to what I actually wrote, as you just quoted it,
> please.
Begging will not get you out of CRANKDOM.
> > Only
> > fanatics would insist that their axioms are the "truth" or only their
> > intuition is the "truth". Either way, it's very similar to religious
> > extremism.
>
> And that is just the kind of thing we find from cranks. The fallback
> crank position is that his principles are the ones that are faithful
> to "reality" and that certain other intuitions about the matter are
> therefore wrong.
>
> Meanwhile, take, for example, me. You will not find me ever to have
> posted that the axioms of ZFC are true in some unqualified sense or
> that other intuitions about the matter are not true.
You just don't get it, do you?
georgie
You sir, are an ass if you think that was the ONLY argument moe ever
put forward. Also you must be a CRANK if you don't understand the
issue.
Aatu Koskensilta
> The Axiom of Idiocy, which says that anyone who doesn't believe is
> ZF is an iferrior idiot. The question of whether this axiom is independent
> of the others has never been settled, although Cohen was able to show
> that it follows from ZF + "there exists a ridiculously large cardinal".
I think you have rather in mind Cantor's 1987 paper in the _Journal of
Pointless Twaddle_ (p. 56674 - 423). Curiously, in that paper Cantor
notes the diagonal argument depends on nothing more than minimal logic
and predicative comprehension.
Aatu Koskensilta
> OK, so I see that Levy's book really isn't "basic,"
> and it deals with a class theory as well as pure
> set theory.
No. It is just that Levy is more concerned with the formal details of
translating class-speak to the official framework of ZFC. This is
nothing but a formal technicality of no mathematical substance.
Aatu Koskensilta
> We all know that we can find a model for many theories by
> considering the cumulative hierarchy of sets. So for the
> theory ZF-Infinity, V_omega is a model. If we include
> Infinity but not the Replacement Schema, then we have
> V_(omega+omega) as a model. Allowing for proper classes
> adds one more level to the cumulative hierarchy, so that
> if V_inaccessible is a model of ZFC, then we have that
> V_(inaccessible+1) is a model of NBG.
V_kappa+1 with kappa an inaccessible is a huge overkill for
NBG. Perhaps you're thinking of Morse-Kelley set theory here?
> So we may now ask, given a set of axioms, which is the
> smallest ordinal alpha such that V_alpha is a model for
> that set of axioms (and membership "e" actually maps
> to itself, avoid Lowenheim-Skolem)?
There are models of ZFC of form V_kappa with kappa way below the first
inaccessible.
georgie
> stevendaryl3...@yahoo.com (Daryl McCullough) writes:
> > The Axiom of Idiocy, which says that anyone who doesn't believe is
> > ZF is an iferrior idiot. The question of whether this axiom is independent
> > of the others has never been settled, although Cohen was able to show
> > that it follows from ZF + "there exists a ridiculously large cardinal".
>
> I think you have rather in mind Cantor's 1987 paper in the _Journal of
> Pointless Twaddle_ (p. 56674 - 423). Curiously, in that paper Cantor
> notes the diagonal argument depends on nothing more than minimal logic
> and predicative comprehension.
You must not have read the revised paper from 1989. In that paper,
Cantor
proves that fanaticism is all that is required to hijack a newsgroup.
Aatu Koskensilta
> You must not have read the revised paper from 1989. In that paper,
> Cantor proves that fanaticism is all that is required to hijack a
> newsgroup.
How about a broccoli? As to your earlier comments, it's a pertinent
observation that those in the news who valiantly battle the lie that
is the diagonal argument virtually never object to any axioms, but
rather blather in an incoherent fashion about supposed logical flaws
in the proof in a way that is an indication of nothing more than their
failure to understand the argument. It is of course entirely possible
to find infinitary set theory intellectually repugnant for all sorts
of intelligible reasons, but such dislike, when reasonable, is not
combined with inane refutations of the diagonal argument, which, as
mathematics goes, is simplicity itself.