Lebesgue Measure vs. Inaccessible Cardinals
Jonathan W. Hoyle
is made on p. 69:
"...R. M. Solovay showed that the axiom 'there exists an inaccessible
cardinal' implies that every set of reals is Lebesgue measurable...and
subsequently Saharon Shelah proved the converse."
Is this correct? Am I missing something?
Since the Axiom of Choice is all that is required to be able to
construct a non-Lebesgue measurable set, the above quote implies the
statement "the Axiom of Choice is true iff there are no inaccessible
cardinals." Yet in all my readings of inaccessible cardinals, I have
never heard this before.
In fact, in "Infinity and the Mind", Rudy Rucker spends 45 pages in
wonderful detail on the chapter entitled "The Transfinite Cardinals",
and never once mentions this result, despite describing in detail both
the Axiom of Choice and inaccessible cardinals.
Is the term "inaccessible cardinal" being used to describe something
different in Stewart's book than is traditionally meant?
Richard Carr
:Date: Fri, 07 May 1999 21:52:31 -0400
:From: Jonathan W. Hoyle <jho...@rochester.rr.com>
:Newsgroups: sci.math, sci.logic
:Subject: Lebesgue Measure vs. Inaccessible Cardinals
:
:In Ian Stewart's book "From Here to Infinity", the following statement
:is made on p. 69:
:
:"...R. M. Solovay showed that the axiom 'there exists an inaccessible
:cardinal' implies that every set of reals is Lebesgue measurable...and
:subsequently Saharon Shelah proved the converse."
If this is the exact quote then Stewart is incorrect. Solovay's result is
if ZFC+there exists an inaccessible cardinal is consistent then so is
ZF+a whole load of other things :).
One of these things was that every set of reals is Lebesgue measurable.
Although obviously one does not get AC in the model, one does get
countable choice and indeed a bit more.
Details of this can be found, for example, in A. Kanamori (The Higher
Infinite). It was a very early forcing.
Shelah showed that Con(ZF+every set of reals is Lebesgue measurable)
implies Con(ZF(C?)+there exists an inaccessible cardinal).
The paper you'd need to look at is Sh176.
:
:Is this correct? Am I missing something?
As stated it wasn't. You are alert rather than missing something.
:
:Since the Axiom of Choice is all that is required to be able to
:construct a non-Lebesgue measurable set, the above quote implies the
:statement "the Axiom of Choice is true iff there are no inaccessible
:cardinals." Yet in all my readings of inaccessible cardinals, I have
:never heard this before.
Since it is not possible to prove (in ZF(C)) the consistency of the
existence of inaccessible cardinals but it is possible to prove (in ZF)
the consistency of both ZFC and ZF+not AC, then it is no surprise you
haven't heard it. I think that the consistency of an inaccessible cardinal
also implies the consistency of V=L+there is an inaccessible cardinal or
at least of GCH+there is an inaccessible cardinal. GCH implies AC, V=L
implies a very strong version of AC, so that would totally destroy the
supposed equivalence. (The supposed equivalence also imply that it would
not be possible in ZF to prove ZF+not AC consistent, which certainly is
not the case.)
:
:In fact, in "Infinity and the Mind", Rudy Rucker spends 45 pages in
:wonderful detail on the chapter entitled "The Transfinite Cardinals",
:and never once mentions this result, despite describing in detail both
:the Axiom of Choice and inaccessible cardinals.
This is understandable.
:
:Is the term "inaccessible cardinal" being used to describe something
:different in Stewart's book than is traditionally meant?
I'm not sure, having not read Stewart's book. Generally inaccessible is
either going to mean weakly or strongly inaccessible and usually
inaccessible is reserved just for strongly inaccessible (so that if you
want to talk about weakly inaccessible you add the qualifying "weakly".
The nice thing about kappa being strongly inaccessible is that V_kappa
models ZFC (if V does).
(Under GCH, both notions of inaccessibilty are the same, of course.)
Richard Carr.
Pierre Asselin
>In Ian Stewart's book "From Here to Infinity", the following statement
>is made on p. 69:
>"...R. M. Solovay showed that the axiom 'there exists an inaccessible
>cardinal' implies that every set of reals is Lebesgue measurable...and
>subsequently Saharon Shelah proved the converse."
>Is this correct? Am I missing something?
If the quote is accurate, Stewart goofed. Solovay's result is
that the *consistency* of "there exists an inaccessible cardinal"
implies the *consistency* of "every set of reals is Lebesgue
measurable".
More precisely, let
ZF = Zermelo-Fraenkel
DC = the axiom of dependent choice
IC = there exists an uncountable inaccessible cardinal
LM = every set of reals is Lebesgue measurable
Solovay: Consis(ZF+IC) --> Consis(ZF+DC+LM)
Shelah: Consis(ZF+IC) <-- Consis(ZF+DC+LM)
Reference: Stan Wagon, "The Banach-Tarski Paradox", Cambridge
University Press 1985, ISBN 0-521-45704-1.
--
--Pierre Asselin, Westminster, Colorado
l...@netcom.com
Richard Carr
:Date: Sat, 8 May 1999 03:51:05 GMT
:From: Pierre Asselin <l...@netcom.com>
:Newsgroups: sci.math, sci.logic
:Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
:
:"Jonathan W. Hoyle" <jho...@rochester.rr.com> writes:
:
:>In Ian Stewart's book "From Here to Infinity", the following statement
:>is made on p. 69:
:
:>"...R. M. Solovay showed that the axiom 'there exists an inaccessible
:>cardinal' implies that every set of reals is Lebesgue measurable...and
:>subsequently Saharon Shelah proved the converse."
:
:>Is this correct? Am I missing something?
:
:If the quote is accurate, Stewart goofed. Solovay's result is
:that the *consistency* of "there exists an inaccessible cardinal"
:implies the *consistency* of "every set of reals is Lebesgue
:measurable".
:
:More precisely, let
:
: ZF = Zermelo-Fraenkel
: DC = the axiom of dependent choice
: IC = there exists an uncountable inaccessible cardinal
: LM = every set of reals is Lebesgue measurable
:
:Solovay: Consis(ZF+IC) --> Consis(ZF+DC+LM)
Indeed it was even more. If B=every set of reals has the Baire property
and PS=every set of reals has the perfect set property then Solovay got
Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).
I think he had ZFC+IC on the left, not ZF+IC.
:Shelah: Consis(ZF+IC) <-- Consis(ZF+DC+LM)
:
:
Jonathan W. Hoyle
sense. What Stewart apparently meant to say was that inaccessible
cardinals imply *the consistency of* every set of reals being Lebesgue
measurable. This sounds like an editing error in the text, although it
is conceivably possible that the author was just mistaken on this point.
I think I'll email the author about it, so that future printings will
have this fixed. Thanks again to you all.
Mike Oliver
Richard Carr wrote:
> Indeed it was even more. If B=every set of reals has the Baire property
> and PS=every set of reals has the perfect set property then Solovay got
> Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).
> I think he had ZFC+IC on the left, not ZF+IC.
Well, superficially at least there's some tiny question about what
it means to be an inaccessible cardinal in the absence of Choice.
That is, an inaccessible cardinal is one that's regular and strongly
limit, where kappa is strongly limit iff for every lambda < kappa,
2^lambda < kappa.
If we don't have Choice, what does this "<" mean? If A<B means there's
an injection from A to B and none from B to A (the usual meaning without
choice), then if kappa is inaccessible then Choice must hold a fortiori
up to rank kappa. In that case L(V_kappa) is an inner model of
ZFC+"there exists an inaccessible", so we would have Con(ZF+IC) --> Con(ZFC+IC).
Richard Carr
:Date: Sat, 08 May 1999 17:43:10 -0700
:From: Mike Oliver <oli...@math.ucla.edu>
:Newsgroups: sci.math, sci.logic
:Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
:
:
:
:
:
OK.
Mike Oliver
Richard Carr wrote:
> OK.
Mm, maybe not. I thought about it some more and I'm pretty sure I was wrong
about one part.
Certainly if kappa is inaccessible in the sense given then V_kappa |= AC,
but that doesn't in any obvious way mean that there's a wellordering
of V_kappa itself, particularly one that will show up in L(V_kappa).
In fact I now conjecture the following: If kappa is inaccessible in the
ground model, then there is a forcing notion adding a subset of kappa
such that in the extension, kappa is still inaccessible, but L(V_kappa) |= (not AC).
But I haven't checked this.
However we can still save the conclusion that Con(ZF+IC) --> Con(ZFC+IC).
For, suppose kappa is inaccessible in the sense given. Then surely kappa
is weakly inaccessible: Given lambda<kappa, lambda^+ can be injected into
the set of equivalence classes of wellorderings of lambda, the equivalence
relation being having the same length. So lambda^+ <= 2^2^lambda < kappa.
But now kappa is weakly inaccessible in L, but in L weak inaccessibility
and strong inacessibilty are the same. So L is an inner model of ZFC+IC.
Now there's still one point I haven't touched: I've been implicitly assuming
that an inaccessible cardinal must at least be an *ordinal*. What if we
relax this definition, and say that a set A is inaccessible if
1) it's strongly limit in the sense that for B \subset A there's
an injection from P(B) into A and no injection from A into P(B),
and
2) it's regular in the sense that for any index set I \subset A and any
family {B_i |i \in I} of subsets of A, there's an injection from
(Sigma_{i \in I} B_i) into A, and no injection in the opposite direction
(here Sigma represents the disjoint union).
It's not obvious to me that such an A must be wellorderable (a wellorderable
A would have as its cardinality an inaccessible in the sense that I've been
treating). Does anyone know?
Mike Oliver
I, Mike Oliver, wrote:
> Now there's still one point I haven't touched: I've been implicitly assuming
> that an inaccessible cardinal must at least be an *ordinal*. What if we
> relax this definition, and say that a set A is inaccessible if
>
> 1) it's strongly limit in the sense that for B \subset A there's
> an injection from P(B) into A and no injection from A into P(B),
>
> and
>
> 2) it's regular in the sense that for any index set I \subset A and any
> family {B_i |i \in I} of subsets of A, there's an injection from
> (Sigma_{i \in I} B_i) into A, and no injection in the opposite direction
> (here Sigma represents the disjoint union).
Oh oh, messed this up too. Of course in clause (1) I want to say |B| < |A| ;
as it stands we could have |B| = |A| which won't do. Similarly in clause
(2) I need |I| < |A| and |B_i| < |A| for each i \in I.
Dr Sinister
<Pine.LNX.4.10.990508...@cpw.math.columbia.edu>:
I'm not sure if any of you care to hear this, but so far no one has
addressed this naive issue I posted. Perhaps it will do better in this
thread. Here is what I posted earlier:
--
Dear Webster Kehr,
I have read some of your long paper, and I find it both interesting and
entertaining.
No doubt the following may be a naive question. But, hey, if we don't
ask, we don't learn.
If all the subsets of R(0,1) are countable, then what does that imply for
their measure? Specifically, how do you define a measure such that the
following function f(x) is Lebesgue integrable?
f(x) = 0 if x in R1 = {r1, r2, ...}, rn in R(0,1), R1 countable.
f(x) = 1 if x in complement(R1) wrt R(0,1).
Perhaps Nathan can comment on this as well.
--
No true Scotsman can be an atheist.
ilias kastanas 08-14-90
Mike Oliver <oli...@math.ucla.edu> wrote:
@
@Richard Carr wrote:
@> Indeed it was even more. If B=every set of reals has the Baire property
@> and PS=every set of reals has the perfect set property then Solovay got
@> Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).
@> I think he had ZFC+IC on the left, not ZF+IC.
@
@Well, superficially at least there's some tiny question about what
@it means to be an inaccessible cardinal in the absence of Choice.
@That is, an inaccessible cardinal is one that's regular and strongly
@limit, where kappa is strongly limit iff for every lambda < kappa,
@2^lambda < kappa.
@
@If we don't have Choice, what does this "<" mean? If A<B means there's
@an injection from A to B and none from B to A (the usual meaning without
@choice), then if kappa is inaccessible then Choice must hold a fortiori
@up to rank kappa. In that case L(V_kappa) is an inner model of
@ZFC+"there exists an inaccessible", so we would have Con(ZF+IC) --> Con(ZFC+IC).
If kappa is weakly inaccessible, then kappa is weakly inaccessible in
L... and hence strongly, too. So L is a model as desired.
Ilias
Richard Carr
:Date: 9 May 1999 03:50:53 GMT
:From: Dr Sinister <kickass.d...@hotmail.com>
:Newsgroups: sci.math, sci.logic
:Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
:
:Richard Carr <ca...@math.columbia.edu> wrote in
This is not true.
:<Pine.LNX.4.10.990508...@cpw.math.columbia.edu>:
:
:
Herman Rubin
>Richard Carr wrote:
>> Indeed it was even more. If B=every set of reals has the Baire property
>> and PS=every set of reals has the perfect set property then Solovay got
>> Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).
>> I think he had ZFC+IC on the left, not ZF+IC.
>Well, superficially at least there's some tiny question about what
>it means to be an inaccessible cardinal in the absence of Choice.
>That is, an inaccessible cardinal is one that's regular and strongly
>limit, where kappa is strongly limit iff for every lambda < kappa,
>2^lambda < kappa.
>If we don't have Choice, what does this "<" mean? If A<B means there's
>an injection from A to B and none from B to A (the usual meaning without
>choice), then if kappa is inaccessible then Choice must hold a fortiori
>up to rank kappa. In that case L(V_kappa) is an inner model of
>ZFC+"there exists an inaccessible", so we would have Con(ZF+IC) --> Con(ZFC+IC).
Weakly inaccessible does not involve AC at all; it is the statement
that and ordinal \kappa (larger than \omega) is not the sum of fewer
than \kappa smaller ordinals. Choice is not involved, as the ordinals
are well ordered.
Strong inaccessibility might be a little harder, but I think that
replacing 2^lambda < kappa by H(2^lambda) < kappa, where H is the
Hartogs function, will work adequately.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Mike Oliver
Herman Rubin wrote:
> Weakly inaccessible does not involve AC at all;
Yes, but in my usage (which I believe is standard) "inaccessible"
means "strongly inaccessible" unless the opposite is clear from context.
> Strong inaccessibility might be a little harder, but I think that
> replacing 2^lambda < kappa by H(2^lambda) < kappa, where H is the
> Hartogs function, will work adequately.
At least formally there seem to be a number of possibilities; I haven't
worked out all the implications between them. Let's say a set A
is "injectively strongly limit" if for any set B such that there's
an injection from B to A and none from A to B, there's also an injection
from P(B) to A and none from A to P(B).
OTOH A will be "surjectively strongly limit" if for
any B such that there's a surjection from A to B and none from B to A,
there's also a surjection from A to P(B) and none from P(B) to A.
A will be "injectively regular" if given an index set I with |I| < |A|
injectively and sets {B_i | i \in I}, each |B_i| < |A| injectively,
we have that | Sigma_{i \in I} B_i | < |A|, where the cardinalities
are interpreted injectively and Sigma is the disjoint union. I don't
care to define "surjectively regular" at the moment because I'm not
sure whether the cardinality of the index set should be interpreted
injectively or surjectively.
Say A is "injectively inaccessible" if A is injectively regular and
injectively strongly limit. If kappa is an initial ordinal and is
injectively inaccessible, then kappa is inaccessible in the sense
of my recent posts. In that case V_kappa |= AC; this takes a small
argument which I'll post if desired. I still don't know whether
necessarily L(V_kappa) |= AC -- anyone?
If kappa is an initial ordinal then there is no worry about the
definition of regularity, so then we can say kappa is "surjectively
inaccessible" if it's regular and surjectively strongly limit.
I don't know whether this is equivalent to its being injectively
inaccessible, or whether there's an implication in either direction.
I also don't know whether an injectively inaccessible set can
fail to be wellorderable.
The last three paragraphs contain at least three questions; does
anyone know the answers?
--
Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love")
Finger for PGP public key, or visit http://www.math.ucla.edu/~oliver.
1500 bits, fingerprint AE AE 4F F8 EA EA A6 FB E9 36 5F 9E EA D0 F8 B9
Mike Oliver
> If kappa is an initial ordinal then there is no worry about the
> definition of regularity, so then we can say kappa is "surjectively
> inaccessible" if it's regular and surjectively strongly limit.
> I don't know whether this is equivalent to its being injectively
> inaccessible, or whether there's an implication in either direction.
OK, got this one. The key is that if alpha is an *ordinal* and A
is a set, then |A| < |alpha| injectively iff |A| < |alpha| surjectively.
That's because either inequality implies that A may be wellordered,
so we can replace A by an initial ordinal.
Therefore *for*initial*ordinals* there's only one notion of inaccessibility.
For non-wellorderable cardinals I still don't know.
(Note that it's *not* true that |alpha| < |A| injectively iff |alpha| < |A|
surjectively. Counterexample: assume AD and let A=R, alpha=omega_1.)
Mike Oliver
I, Mike Oliver wrote:
>
> Therefore *for*initial*ordinals* there's only one notion of inaccessibility.
> For non-wellorderable cardinals I still don't know.
>
Spoke too soon once again. Herman's notion (where "kappa strongly limit"
is replaced by "For all lambda<kappa, the Hartogs number of P(lambda) is also
less than kappa") does appear to be different from the notions I was considering.
I was vaguely thinking that it was subsumed in "surjective inaccessibility" but
it's not; to say that the Hartogs number of A is less than kappa
is far weaker than to say there's a surjection from kappa to A and
none in the other direction. (It's equivalent to saying there's a beta<kappa
with no surjection A -->> beta .)
This definition is attractive in a couple of ways: 1) it seems to be
consistent with failures of AC below kappa and 2) I think it relativizes
down to transitive models of ZF.
There's one problem that occurs to me; maybe someone knows a way around
it. The definition as stated says that if lamda<kappa then Theta(P(lambda))<kappa
but it doesn't seem to say anything at all about P(P(lambda)). We could
change the definition to say "If Theta(A)<kappa then Theta(P(A))<kappa", but
now it's no longer obvious that it relativizes down; when you go to the
smaller model there could be more A's such that Theta(A)<kappa .
Randall Dougherty
Mike Oliver <oli...@math.ucla.edu> wrote:
>At least formally there seem to be a number of possibilities; I haven't
>worked out all the implications between them. Let's say a set A
>is "injectively strongly limit" if for any set B such that there's
>an injection from B to A and none from A to B, there's also an injection
>from P(B) to A and none from A to P(B).
>
>injectively and sets {B_i | i \in I}, each |B_i| < |A| injectively,
>we have that | Sigma_{i \in I} B_i | < |A|, where the cardinalities
>are interpreted injectively and Sigma is the disjoint union. I don't
>care to define "surjectively regular" at the moment because I'm not
>sure whether the cardinality of the index set should be interpreted
>injectively or surjectively.
>
>Say A is "injectively inaccessible" if A is injectively regular and
>injectively strongly limit. If kappa is an initial ordinal and is
>injectively inaccessible, then kappa is inaccessible in the sense
>of my recent posts. In that case V_kappa |= AC; this takes a small
>argument which I'll post if desired. I still don't know whether
>necessarily L(V_kappa) |= AC -- anyone?
Not necessarily -- one can use symmetric forcing methods (as described
in Chapter 21 of Jech's Set Theory) to construct a counterexample.
Start with the constructible universe L and an inaccessible cardinal
kappa. Consider the Easton forcing construction which adds beta^++
subsets of beta for each regular cardinal beta < kappa; let M be
the resulting generic extension. In M, kappa is still inaccessible.
Let G be the group of automorphisms of the forcing notion which, for
each beta, allows one to permute the beta^++ new subsets of beta
(for all beta simultaneously). For each alpha < kappa, one can consider the
subgroup of G consisting of those permutations which leave fixed the
new subsets of beta for beta < alpha. These subgroups generate a
normal filter; let N be the symmetric submodel of M obtained by using
only hereditarily symmetric names under this filter.
For any alpha < kappa, the part of the M-generic filter giving
just the new subsets of beta for beta < alpha is in N. One can then
show that V_kappa is the same in N as in M. On the other hand,
any subset of kappa in N has a heretitarily symmetric name, and
hence lies in L[S_alpha] for some alpha, where S_alpha is the sequence
of new subsets of beta for beta < alpha, and alpha < kappa.
But any well-ordering of V_kappa can be rearranged to have order
type kappa, and can then be used to get a subset of kappa
encoding _all_ of the new subsets of beta for all beta < kappa;
therefore, N (and hence L(V_kappa)) does not contain a well-ordering
of V_kappa.
>I also don't know whether an injectively inaccessible set can
>fail to be wellorderable.
Symmetric forcing methods work here as well. Start with L and kappa
as above, but this time force to add kappa new subsets of kappa,
with the standard <kappa-closed forcing notion. Let A
be the set of these kappa new sets. Let G be the group
of permutations of kappa (which acts on A in the obvious way),
and for each alpha < kappa, consider the subgroup which fixes the
first alpha elements of A. These again generate a normal filter,
and one gets a symmetric submodel N. Now A is in N, but the
subsets of A in N are precisely those that are bounded
(under the kappa-ordering of A; this ordering is _not_ in N)
and their complements. The complements all have the same
size as A (even in N), and the bounded sets have sizes precisely
all cardinals less than kappa. Hence, A is not well-orderable in N,
but is injectively inaccessible.
Randall Dougherty r...@math.ohio-state.edu
Department of Mathematics, Ohio State University, Columbus, OH 43210 USA
"I have yet to see any problem, however complicated, that when looked at in the
right way didn't become still more complicated." Poul Anderson, "Call Me Joe"
Randall Dougherty
Randall Dougherty <r...@math.ohio-state.edu> wrote:
>In article <3737726B...@math.ucla.edu>,
>Mike Oliver <oli...@math.ucla.edu> wrote:
>>I also don't know whether an injectively inaccessible set can
>>fail to be wellorderable.
>
>Symmetric forcing methods work here as well.
However, not the way I just described them. One should be able
to get a model with a set A having the given properties
(all subsets either have the same size as A or have size < kappa;
all sizes less than kappa occur; A is not well-orderable),
but using subsets of kappa in the way described does not work.
Probably the simplest way to do it is to build a permutation model
with the desired properties and apply the Jech-Sochor embedding theorem.
I'll check the details when I get to the office tomorrow.
Nathan the Great
> :Dear Webster Kehr,
> :
> :I have read some of your long paper, and I find it both interesting and
> :entertaining.
> :
> :No doubt the following may be a naive question. But, hey, if we don't
> :ask, we don't learn.
> :
> :If all the subsets of R(0,1) are countable, then what does that imply for
> :their measure?
I'm not familiar with the mathematical meaning of "measure". If R[0,1] is cut
into ten equal pieces: [0, .1], [.1, .2], [.2, .3], ... [.9, 1] each piece
will measure 1/10 unit. If R[0,1] is cut into n equal pieces, each piece will
measure 1/n. The sum of these pieces is n*(1/n) = 1. R[0,.1] measures the
same as R(0,.1) because points don't have extension and numbers, as Webster
would say, don't have thickness.
> Specifically, how do you define a measure such that the
> :following function f(x) is Lebesgue integrable?
Can you explain in English?
> :f(x) = 0 if x in R1 = {r1, r2, ...}, rn in R(0,1), R1 countable.
> :f(x) = 1 if x in complement(R1) wrt R(0,1).
How do *you* answer this question?
> :
> :Perhaps Nathan can comment on this as well.
> :
I will be happy to comment, once I figure out what you're talking about.
Nathan
John Savard
>Since the Axiom of Choice is all that is required to be able to
>construct a non-Lebesgue measurable set, the above quote implies the
>statement "the Axiom of Choice is true iff there are no inaccessible
>cardinals." Yet in all my readings of inaccessible cardinals, I have
>never heard this before.
From reading the replies to your post, I think your mistake is that
every set of reals could be Lebesgue measurable even if only something
slightly weaker than the Axiom of Choice is true.
Not being an expert on these matters, I made a nontechnical post
("Beyond the Merely Infinite") recently suggesting that perhaps there
are no sets (to which the axiom of the power set applies) whose
cardinality is an uncountable inaccessible cardinal - that while these
numbers could "exist", they would have to be banished from the realm
of set theory to allow the consequences of their existence to be,
somehow, comfortable for me.
Probably just a lucky coincidence, but this seems to indicate my
intuition is working well these days...
John Savard ( teneerf<- )
http://members.xoom.com/quadibloc/index.html
Randall Dougherty
Randall Dougherty <r...@math.ohio-state.edu> wrote:
>In article <7hambh$muv$1...@mathserv.mps.ohio-state.edu>,
>Randall Dougherty <r...@math.ohio-state.edu> wrote:
>>In article <3737726B...@math.ucla.edu>,
>>Mike Oliver <oli...@math.ucla.edu> wrote:
>>>I also don't know whether an injectively inaccessible set can
>>>fail to be wellorderable.
>>
>>Symmetric forcing methods work here as well.
>
>However, not the way I just described them. One should be able
>to get a model with a set A having the given properties
>(all subsets either have the same size as A or have size < kappa;
>all sizes less than kappa occur; A is not well-orderable),
>but using subsets of kappa in the way described does not work.
>Probably the simplest way to do it is to build a permutation model
>with the desired properties and apply the Jech-Sochor embedding theorem.
>I'll check the details when I get to the office tomorrow.
Yes, this approach via Jech-Sochor does work to give a non-well-orderable
but injectively inaccessible set A. A direct construction
along the lines I gave before also works, but one has to use
sets of new subsets of kappa, not just subsets of kappa, as
elements of the set A.
Mike Oliver
Randall Dougherty wrote:
> Start with the constructible universe L and an inaccessible cardinal
> kappa. Consider the Easton forcing construction which adds beta^++
> subsets of beta for each regular cardinal beta < kappa; let M be
> the resulting generic extension. In M, kappa is still inaccessible.
Where do you use the fact that you're increasing the cardinality
of 2^beta? I.e. would everything go through without extra argument
if you simply added *one* subset of beta using <beta-closed forcing,
for each beta<kappa?
> Let G be the group of automorphisms of the forcing notion which, for
> each beta, allows one to permute the beta^++ new subsets of beta
> (for all beta simultaneously).
I could read this as "G = Aut(P), and each element of G leads to permutations
of the new subsets of each beta" or "G is the group of all automorphisms
leading to such permutations". I think you mean the first, right? What
about the *old* subsets of beta? Presumably you could have a name for
a ground-model subset of beta such that *which* subset it names depends
on the generic object. Are such names not hereditarily symmetric?
> For each alpha < kappa, one can consider the
> subgroup of G consisting of those permutations which leave fixed the
> new subsets of beta for beta < alpha. These subgroups generate a
> normal filter; let N be the symmetric submodel of M obtained by using
> only hereditarily symmetric names under this filter.
So such a name is one for which for sufficiently large alpha, if
you fix all new subsets of beta for beta<alpha, then you also fix
the name? I suppose you have to deal with the transitive closure;
does there need to be a name for the transitive closure for which
a *single* alpha works, or can the alpha's be increasing as you
go down the epsilon tree?
> For any alpha < kappa, the part of the M-generic filter giving
> just the new subsets of beta for beta < alpha is in N. One can then
> show that V_kappa is the same in N as in M. On the other hand,
> any subset of kappa in N has a heretitarily symmetric name, and
> hence lies in L[S_alpha] for some alpha, where S_alpha is the sequence
> of new subsets of beta for beta < alpha, and alpha < kappa.
You mean the sequence of sets { {all new subsets of beta | beta<alpha }
or do we also know an ordering for the new subsets of beta for a *particular*
beta?
Bennett Standeven
On Wed, 12 May 1999, Nathan the Great wrote:
> Dr. Sinister, are you an evil Cantorian?
>
> > :Dear Webster Kehr,
> > :
> > :I have read some of your long paper, and I find it both interesting and
> > :entertaining.
> > :
> > :No doubt the following may be a naive question. But, hey, if we don't
> > :ask, we don't learn.
> > :
> > :If all the subsets of R(0,1) are countable, then what does that imply for
> > :their measure?
>
> I'm not familiar with the mathematical meaning of "measure". If R[0,1] is cut
> into ten equal pieces: [0, .1], [.1, .2], [.2, .3], ... [.9, 1] each piece
> will measure 1/10 unit. If R[0,1] is cut into n equal pieces, each piece will
> measure 1/n. The sum of these pieces is n*(1/n) = 1. R[0,.1] measures the
> same as R(0,.1) because points don't have extension and numbers, as Webster
> would say, don't have thickness.
Right, so:
1) Points have measure 0.
2) The unit segment has measure 1.
Now to clinch the proof, we need just one more assumption:
3) Measure is countably additive.
IE: If we take the segment (0, 1), the segment (1, 1.5), the segment (1.5,
1.75), etc. and put them all together, we get a set of measure 2.
(The significance of Lebesque measure here is that it has these three
properties)
Then:
Suppose the reals are countable. Then we can express (0,1) as the union of
a countable collection of points. But each point has measure zero, by (1).
Therefore, by (3) the unit interval must have measure 0. But by (2), it
must have measure 1. Therefore, if the reals are countable, then 1=0, and
so the reals are singleton.
Randall Dougherty
Mike Oliver <oli...@math.ucla.edu> wrote:
>Thanks. I see the general outline but I'm missing some details.
>
>Randall Dougherty wrote:
>> Start with the constructible universe L and an inaccessible cardinal
>> kappa. Consider the Easton forcing construction which adds beta^++
>> subsets of beta for each regular cardinal beta < kappa; let M be
>> the resulting generic extension. In M, kappa is still inaccessible.
>
>Where do you use the fact that you're increasing the cardinality
>of 2^beta? I.e. would everything go through without extra argument
>if you simply added *one* subset of beta using <beta-closed forcing,
>for each beta<kappa?
Increasing the cardinality of 2^beta isn't important (I just said that
because that version of the forcing is a bit more well-known). I do
need enough new subsets of beta to be able to permute them usefully,
but beta new subsets of beta will suffice (and adding beta new subsets
of beta is equivalent to adding one new subset of beta).
>> Let G be the group of automorphisms of the forcing notion which, for
>> each beta, allows one to permute the beta^++ new subsets of beta
>> (for all beta simultaneously).
>
>I could read this as "G = Aut(P), and each element of G leads to permutations
>of the new subsets of each beta" or "G is the group of all automorphisms
>leading to such permutations". I think you mean the first, right? What
>about the *old* subsets of beta? Presumably you could have a name for
>a ground-model subset of beta such that *which* subset it names depends
>on the generic object. Are such names not hereditarily symmetric?
I think I need to use more notation to clarify this. For each regular
beta < kappa, let P_beta be the forcing notion for adding
beta new subsets of beta: a condition is a partial function from
beta x beta to {0,1} whose domain has size less than beta.
Let P be the Easton product of P_beta for beta < kappa: a condition
is a sequence <p_beta : beta < kappa> where p_beta is in P_beta
and, for each regular gamma <= kappa, the set of beta < gamma
such that p_beta is nonempty has size less than gamma.
For each regular beta, any permutation pi of beta induces an
automorphism pi' of P_beta via the formula
pi'(p)(pi(eta),rho) = p(eta,rho)
for all eta < beta and rho < beta [here "=" means "both are defined with the
same value or both are undefined]. Let G_beta be the set of
all of these automorphisms pi' (so G_beta is a group of automorphisms
of P_beta). This is what I meant by "permuting the new subsets of beta."
(Ground model subsets of beta are not affected. In fact, this isn't
an action on sets at all; its an action on the forcing poset, which
yields an action on _names_. Yes, a name could be a name for
a ground model subset of beta such that which subset it names
depends on the generic object. Such a name could even be hereditarily
symmetric, if it only depends on a small part of the generic object.)
Any sequence <pi'_beta : beta < kappa> with pi'_beta in G_beta
gives an automorphism of the restricted product P; let G be
the group of all such automorphisms.
>> For each alpha < kappa, one can consider the
>> subgroup of G consisting of those permutations which leave fixed the
>> new subsets of beta for beta < alpha. These subgroups generate a
>> normal filter; let N be the symmetric submodel of M obtained by using
>> only hereditarily symmetric names under this filter.
>
>So such a name is one for which for sufficiently large alpha, if
>you fix all new subsets of beta for beta<alpha, then you also fix
>the name? I suppose you have to deal with the transitive closure;
>does there need to be a name for the transitive closure for which
>a *single* alpha works, or can the alpha's be increasing as you
>go down the epsilon tree?
Yes, that's right, and the alpha's can increase as you go down the
epsilon tree. For instance, each of the beta new subsets of beta
has a canonical name which is hereditarily symmetric (just take alpha
to be greater than beta). Then one can form a set whose members are all of
these subsets, for all beta; the canonical name for this is hereditarily
symmetric, because it is itself symmetric with alpha=0 and its
members are symmetric with varying alpha's. (The members of members and so
on are all canonically named ground model sets, hence completely symmetric.)
>> For any alpha < kappa, the part of the M-generic filter giving
>> just the new subsets of beta for beta < alpha is in N. One can then
>> show that V_kappa is the same in N as in M. On the other hand,
>> any subset of kappa in N has a heretitarily symmetric name, and
>> hence lies in L[S_alpha] for some alpha, where S_alpha is the sequence
>> of new subsets of beta for beta < alpha, and alpha < kappa.
>
>You mean the sequence of sets { {all new subsets of beta | beta<alpha }
>or do we also know an ordering for the new subsets of beta for a *particular*
>beta?
Again more notation will make things more precise. For each alpha,
one can naturally decompose P as a product of Q_alpha and Q^alpha,
where Q_alpha is the Easton product of P_beta for beta < alpha
and Q^alpha is the Easton product of P_beta for beta >= alpha.
Then any generic object H on P is a product H_alpha x H^alpha
where H_alpha is generic on Q_alpha and H^alpha is generic
on Q^alpha (even in L[H_alpha]). The canonical name for H_alpha
is hereditarily symmetric. If a subset of kappa in M has a name
which is symmetric at level alpha, then if a forcing condition
p decides whether a certain gamma is in kappa, then the part of p
in Q_alpha also decides whether gamma is in kappa; hence,
the named subset of kappa is in L[H_alpha]. This is what I
meant above when talking about S_alpha (which is supposed to contain
the information needed to reconstruct H_alpha: the list of
beta new subsets of beta for each beta < alpha, in the order specified
by the generic object).
Dr Sinister
<37398E64...@ashland.baysat.net>:
>Dr. Sinister, are you an evil Cantorian?
No, I'm not an evil Cantorian.
I'm not a mathematician either, so I don't really care what the outcome
of this debate. I don't have any dogma to follow, nor any papers to
defend. And I don't sponge my living off the government like 99% of
mathematicians seem to do.
>> :Dear Webster Kehr,
>> :
>> :I have read some of your long paper, and I find it both interesting
>> :and entertaining.
>> :
>> :No doubt the following may be a naive question. But, hey, if we don't
>> :ask, we don't learn.
>> :
>> :If all the subsets of R(0,1) are countable, then what does that imply
>> :for their measure?
>
>I'm not familiar with the mathematical meaning of "measure". If R[0,1]
>is cut into ten equal pieces: [0, .1], [.1, .2], [.2, .3], ... [.9, 1]
>each piece will measure 1/10 unit. If R[0,1] is cut into n equal
>pieces, each piece will measure 1/n. The sum of these pieces is n*(1/n)
>= 1. R[0,.1] measures the same as R(0,.1) because points don't have
>extension and numbers, as Webster would say, don't have thickness.
Fine. I will use this idea in a moment.
>> Specifically, how do you define a measure such that the
>> :following function f(x) is Lebesgue integrable?
>
>Can you explain in English?
I am not here to write dissertations. Measurable sets are defined in many
books.
>> :f(x) = 0 if x in R1 = {r1, r2, ...}, rn in R(0,1), R1 countable.
>> :f(x) = 1 if x in complement(R1) wrt R(0,1).
>
>How do *you* answer this question?
As far as I know, this integral = 1.
Now, if R1<R is countable, then it has measure zero. Perhaps one can
redefine a measure for R1<R where R is countable and R1 does not
necessarily have measure zero, as you pointed out above.
Fine, if R1=[0,0.1] then you can assign it some 'measure' 1/10 and then
integrate.
Let R1={0<p/q<1: p<<N,q<<N}. This set is countable. The integral of f(x)
over this set is 1 (f as defined above). How do you construct a measure
for this which agrees with the result obtained from Lebesgue integration?
[snip]
>I will be happy to comment, once I figure out what you're talking about.
I am talking about the measurability of subsets of R. This is something
non-cantorians have to define, unless they don't mind dispensing with
integration altogether.
Andrew Boucher
----------
In article <8DC625CEEsin...@news.globalserve.net>,
Dr Sinister <drsin...@my-dejanews.com> wrote:
>I am talking about the measurability of subsets of R. This
is something
>non-cantorians have to define, unless they don't mind
dispensing with
>integration altogether.
>
I thought Riemann integration worked just fine.
Measurability (and Lebesgue integration) is needed for the
funny sets.
Mike McCarty
Andrew Boucher <Helene....@wanadoo.fr> wrote:
)
)----------
)In article <8DC625CEEsin...@news.globalserve.net>,
)Dr Sinister <drsin...@my-dejanews.com> wrote:
)
)>I am talking about the measurability of subsets of R. This
)is something
)>non-cantorians have to define, unless they don't mind
)dispensing with
)>integration altogether.
)>
)I thought Riemann integration worked just fine.
)Measurability (and Lebesgue integration) is needed for the
)funny sets.
Not really true. The purpose of using Lebesgue integration (or Daniell,
or whatever) is not that it works for "funny sets", because that is not
the common case. The reason is that the "nice" properties one wants it
to have aren't there for the R. Integral.
It's similar to the insistence on actual infinities when dealing with
the integers. We don't *really* need an infinite number of integers.
But if we don't have them, then we can't always form the sum A+B, and
proofs get needlessly complicated with all kinds of provisos and so on
to state the precise conditions under which the conclusion is true.
In the real world, we *don't* need to integrate singular continuous
densities. But if we don't have the tools necessary to handle them,
then our proofs need all kinds of weird add-on assumptions which get in
the way of making progress.
Likewise, when we state some property holds p.p, in the backs of our
minds we know we are really saying it holds *always*, well, except for
some weird cases which don't matter, and we can lump them together that
way. Measure theory lets us do this simplification. Not using it makes
things complicated.
Mike
--
----
char *p="char *p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
This message made from 100% recycled bits.
I don't speak for Alcatel <- They make me say that.
Herman Rubin
Mike McCarty <jmcc...@sun1307.ssd.usa.alcatel.com> wrote:
.................
>Not really true. The purpose of using Lebesgue integration (or Daniell,
>or whatever) is not that it works for "funny sets", because that is not
>the common case. The reason is that the "nice" properties one wants it
>to have aren't there for the R. Integral.
There are "reasonable" functions taking on only the values 0 and 1
which are Lebesgue integrable, but not Riemann integrable, and which
cannot be changed on a set of measure 0 to be so integrable. One way
to get such is to express numbers between 0 and 1 to some base b, and
consider sets A_i defined by the digits a_j for j between (i-1)^2 and
i^2 are all 0. Then consider the indicator function of the set of
points belonging to an even number of A_i. Since the measure of A_i
is b^{-(2i-1)}, almost all points belong to only a finite number of
the A_i, so this is defined except for a set of measure 0. But this
set and its complement both have positive measure in any non-trivial
interval.
This is such a function, and when normalized provides an example of a
probability density which is not Riemann integrable.
Andrew Boucher
----------
In article <7hhft1$hb0$1...@relay1.dsccc.com>,
jmcc...@sun1307.ssd.usa.alcatel.com (Mike McCarty) wrote:
>In article <7hhchi$mrr$1...@wanadoo.fr>,
>Andrew Boucher <Helene....@wanadoo.fr> wrote:
>)
>)----------
>)In article
<8DC625CEEsin...@news.globalserve.net>,
>)Dr Sinister <drsin...@my-dejanews.com> wrote:
>)
>)>I am talking about the measurability of subsets of R.
This
>)is something
>)>non-cantorians have to define, unless they don't mind
>)dispensing with
>)>integration altogether.
>)>
>)I thought Riemann integration worked just fine.
>)Measurability (and Lebesgue integration) is needed for the
>)funny sets.
>
>Not really true. The purpose of using Lebesgue integration
(or Daniell,
>or whatever) is not that it works for "funny sets", because
that is not
>the common case. The reason is that the "nice" properties
one wants it
>to have aren't there for the R. Integral.
>
>It's similar to the insistence on actual infinities when
dealing with
>the integers. We don't *really* need an infinite number of
integers.
>But if we don't have them, then we can't always form the
sum A+B, and
>proofs get needlessly complicated with all kinds of
provisos and so on
>to state the precise conditions under which the conclusion
is true.
No agree. First, we do really need an infinite number of
natural numbers, because otherwise I would have to assume
the world is necessarily finite, and I see no reason to
suppose this is the case. Finiteness even strikes me as
incoherent. If n is the largest natural number, isn't {me} U
{1..n} of size n+1? (You *do* have to suppose that I am not
a natural number!) Secondly, even if we
*didn't need* an infinite number of natural numbers, I would
still say my intuition believes in them, so I would assume
it in any case. (Which do *you* prefer? And by that, not
which do you prefer to use, but which do you think of as
actually existing? An unbounded or a bounded set of natural
numbers?)
Which is not to say that your point doesn't show insight,
only that I wouldn't use little itty-bitty trustworthy N to
back it up. I think this type of argument has more force if
you would use the real numbers R. Do we really "need" them?
We do, but only because they make things much, much
simpler.
But just because we make the leap once, doesn't mean we
should always make the leap. Each time should be
scrutinized. And ultimately I think even the analogy with R
breaks down. R is used by the physicist and the engineer.
The guys that build bridges use R. I don't think they're
using non-Riemann integration.
>
>In the real world, we *don't* need to integrate singular
continuous
>densities.
We agree.
>But if we don't have the tools necessary to handle them,
>then our proofs need all kinds of weird add-on assumptions
which get in
>the way of making progress.
I guess we agree that, if you have all kinds of weird
add-ons, you can take that as an indication that you're on
the wrong track. We don't agree whether the conclusion
should be: accept measurability a la Lebesgue; or take
another look at what you're proving and why. Maybe the
add-ons aren't as weird as all that, or maybe they reveal
something about reality. Maybe we don't even need the
proposition to begin with. I don't know how to begin
analysis without the real numbers; but I can get a good way
without Lebesgue integration. After all, we didn't have it
until a hundred years ago...
Mike Oliver
Andrew Boucher wrote:
> No agree. First, we do really need an infinite number of
> natural numbers, because otherwise I would have to assume
> the world is necessarily finite, and I see no reason to
> suppose this is the case.
Do you have any reason to suppose it's *not* the case? Besides
the fact that it makes things easier?
But in any case, you don't need to *have* infinitely many naturals
in order to avoid *assuming* a necessarily finite world; you can
simply remain agnostic on whether the world is necessarily finite.
> Finiteness even strikes me as
> incoherent. If n is the largest natural number, isn't {me} U
> {1..n} of size n+1?
How do you know you can form this set?
> Secondly, even if we
> *didn't need* an infinite number of natural numbers, I would
> still say my intuition believes in them, so I would assume
> it in any case.
My intuition has no problems with V.
> I don't know how to begin
> analysis without the real numbers; but I can get a good way
> without Lebesgue integration. After all, we didn't have it
> until a hundred years ago...
Which would be a stronger point if not for the fact that most of
the best mathematics is less than a hundred years old.
Andrew Boucher
----------
In article <373E12CE...@math.ucla.edu>, Mike Oliver
<oli...@math.ucla.edu> wrote:
>
>
>Andrew Boucher wrote:
>> No agree. First, we do really need an infinite number of
>> natural numbers, because otherwise I would have to assume
>> the world is necessarily finite, and I see no reason to
>> suppose this is the case.
>
>Do you have any reason to suppose it's *not* the case?
Besides
>the fact that it makes things easier?
>
>But in any case, you don't need to *have* infinitely many
naturals
>in order to avoid *assuming* a necessarily finite world;
you can
>simply remain agnostic on whether the world is necessarily
finite.
No agree. If it's possible the world is infinite, I want an
infinite set of naturals. Just in case.
>
>> Finiteness even strikes me as
>> incoherent. If n is the largest natural number, isn't
{me} U
>> {1..n} of size n+1?
>
>How do you know you can form this set?
Well gee whiz. To be equally flippant, what about the union
axiom? (I look deep inside myself, and I know {me} exists.
The set of naturals exist, and since this set = {1..n}, then
{1..n} exists. Ok so I forgot 0...)
>
>> Secondly, even if we
>> *didn't need* an infinite number of natural numbers, I
would
>> still say my intuition believes in them, so I would
assume
>> it in any case.
>
>My intuition has no problems with V.
Last time I checked, that was another thread...
>
>> I don't know how to begin
>> analysis without the real numbers; but I can get a good
way
>> without Lebesgue integration. After all, we didn't have
it
>> until a hundred years ago...
>
>Which would be a stronger point if not for the fact that
most of
>the best mathematics is less than a hundred years old.
Which would be a stronger point if beauty wasn't in the eye
of the beholder.
Mike Oliver
"Andrew Boucher" <Helene....@wanadoo.fr> wrote:
>
> ----------
> In article <373E12CE...@math.ucla.edu>, Mike Oliver
> <oli...@math.ucla.edu> wrote:
>> But in any case, you don't need to *have* infinitely many naturals
>> in order to avoid *assuming* a necessarily finite world; you can
>> simply remain agnostic on whether the world is necessarily
>> finite.
>
> No agree. If it's possible the world is infinite, I want an
> infinite set of naturals. Just in case.
But you haven't demonstrated that it's possible that the world
is infinite. What if the assumption that the world is infinite
leads to a contradiction? In that case it's *impossible* that
the world is infinite. And your failure to find this contradiction
doesn't demonstrate that it's not there.
>>> Finiteness even strikes me as
>>> incoherent. If n is the largest natural number, isn't {me} U
>>> {1..n} of size n+1?
>>How do you know you can form this set?
> Well gee whiz. To be equally flippant, what about the union
> axiom? (I look deep inside myself, and I know {me} exists.
> The set of naturals exist, and since this set = {1..n}, then
> {1..n} exists. Ok so I forgot 0...)
I wasn't being flippant. What *about* the union axiom? I don't
claim there's any certainty there. You're the one (apparently)
making claims of certainty. I claim only knowledge in a pragmatic
sense, on which level the mental operations which justify arbitrarily-
large integers do not appear all that different from the ones
which justify, say, supercompact cardinals.
--== Sent via Deja.com http://www.deja.com/ ==--
---Share what you know. Learn what you don't.---
Andrew Boucher
In article <7hm0e5$cvd$1...@nnrp1.deja.com>, Mike Oliver
<oli...@math.ucla.edu> wrote:
You can't demonstrate everything, or you can only
demonstrate it by assuming axioms pretty close to what you
are trying to prove. Possibility (logical possibility) is
often one such thing. It's *possible* that the world began
existing just five minutes ago. But I don't think you can
prove it. It's not a proof that tells me that a square
circle is impossible, and a red circle is possible.
(Similarly, it's not a proof that's going to tell me that
Julius Caesar is not the number one, or that I'm not a
natural number.)
>
>>>> Finiteness even strikes me as
>>>> incoherent. If n is the largest natural number, isn't
{me} U
>>>> {1..n} of size n+1?
>
>>>How do you know you can form this set?
>
>> Well gee whiz. To be equally flippant, what about the
union
>> axiom? (I look deep inside myself, and I know {me}
exists.
>> The set of naturals exist, and since this set = {1..n},
then
>> {1..n} exists. Ok so I forgot 0...)
>
>I wasn't being flippant. What *about* the union axiom? I
don't
>claim there's any certainty there. You're the one
(apparently)
>making claims of certainty. I claim only knowledge in a
pragmatic
>sense, on which level the mental operations which justify
arbitrarily-
>large integers do not appear all that different from the
ones
>which justify, say, supercompact cardinals.
I wasn't making claims of certainty, because I don't
find this (psychological) notion very useful. Since you
weren't being flippant, then you can forget about my
flippant reply (and I apologize humbly), and I'll give you
my real one now.
I think it's possible that the external world is
infinite, and I think the world (in the larger sense, i.e.
including numbers) *is* infinite. Clearly, in a formal
setting, this is often difficult to prove (that's why ZF
assumes it!). But I wasn't presenting a formal argument. I
tried to make the informality clear by using me--not exactly
a formal concept--as an element and the word "incoherent",
rather than "inconsistent."
So, first, to ask me how I "know" I can form my set
appeared to me off-the-mark, because I thought it was
obvious I was using naive set theory. Of course you can
outlaw that, say by pointing out that naive set theory is
contradictory. But in any case the use of set theory is not
essential to my argument--it was just a quick way of getting
it across. I have three sons, regardless of whether or not
you allow me to form the set of my three sons. Similarly,
myself plus the numbers between 1 and n are n+1 objects,
regardless of whether these form an officially sanctioned
set. In brief, you can count (finite numbers of) objects,
even if you can't prove they form a set.
Secondly, I provided what I thought was a good--albeit a
short--proof. If there's a largest natural number n, then
the natural numbers 1 through n are n objects. Add in me
(who is not a natural number). Or even use 0 instead. It
looks like you should be able to conclude that there are n+1
objects. The reply "How do you know?" appeared
inappropriate. It really seems the ball is in your court.
Where do *you* think the argument falls down. Call me lazy,
but I don't want to do all the work here.
Your last comment is interesting, but I think you have
got just a bit carried away and made it too categorical (or
maybe I am just misinterpreting it). Not *all* knowledge is
pragmatic, unless you mean this is some trivial way. Sure,
you can dredge up pragmatic reasons to justify any
proposition, but that's not why you believe it. Do you
really believe that you exist or that 2+2=4 because of
pragmatic reasons?
On the other hand, I do agree that *some* mathematical
knowledge--including analysis--is basically pragmatic. E.g.
I accept the existence of the real numbers for pragmatic
reasons. But we apparently have a much different tolerance
of pragmatism. You have not said this, and forgive me if I
am maligning your viewpoint, but I would not believe in a
proposition simply because it makes my proofs shorter, or
because I can prove more with it (unless those additional
theorems had some pragmatic value).
Bennett Standeven
On Sun, 16 May 1999, Mike Oliver wrote:
> In article <7hlq41$p4$1...@wanadoo.fr>,
> "Andrew Boucher" <Helene....@wanadoo.fr> wrote:
> >
> > ----------
> > In article <373E12CE...@math.ucla.edu>, Mike Oliver
> > <oli...@math.ucla.edu> wrote:
> >> But in any case, you don't need to *have* infinitely many naturals
> >> in order to avoid *assuming* a necessarily finite world; you can
> >> simply remain agnostic on whether the world is necessarily
> >> finite.
> >
> > No agree. If it's possible the world is infinite, I want an
> > infinite set of naturals. Just in case.
>
> But you haven't demonstrated that it's possible that the world
> is infinite. What if the assumption that the world is infinite
> leads to a contradiction? In that case it's *impossible* that
> the world is infinite. And your failure to find this contradiction
> doesn't demonstrate that it's not there.
>
> >>> Finiteness even strikes me as
> >>> incoherent. If n is the largest natural number, isn't {me} U
> >>> {1..n} of size n+1?
>
> >>How do you know you can form this set?
>
> > Well gee whiz. To be equally flippant, what about the union
> > axiom? (I look deep inside myself, and I know {me} exists.
> > The set of naturals exist, and since this set = {1..n}, then
> > {1..n} exists. Ok so I forgot 0...)
>
> I wasn't being flippant. What *about* the union axiom? I don't
> claim there's any certainty there. You're the one (apparently)
> making claims of certainty. I claim only knowledge in a pragmatic
> sense, on which level the mental operations which justify arbitrarily-
> large integers do not appear all that different from the ones
> which justify, say, supercompact cardinals.
>
What do you mean by "certainty"?
Jonathan W. Hoyle
not terms anyone uses outside this board). There are Mathematicians,
and then there is Nathan.
Sorry, couldn't resist.
Jonathan W. Hoyle
misstatement, and he thanked me and said that it will be updated if or
when there is another printing of the book. Yes, as many of the other
posters here mentioned, the statement should read that the existence of
inaccessible cardinals implies *the consistency of* all sets of reals
being Lebesgue measurable, not that it implies that they are.
Thanks all.
Richard Carr
:Date: Sun, 16 May 1999 19:25:50 -0400
:From: Jonathan W. Hoyle <jho...@rochester.rr.com>
:Newsgroups: sci.math, sci.logic
:Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
:
:There is no such thing as "Cantorians" and "non-Cantorians" (these are
:not terms anyone uses outside this board). There are Mathematicians,
:and then there is Nathan.
:
:Sorry, couldn't resist.
:
:
Unfortunately, there is a book called "Cantorian Set Theory and Limitation
of Size". Of course, Cantorian here is used as an adjective to describe
the theory, not as a word to describe people. If one qualified
subscribers to the theory as Cantorians also this still doesn't give any
weight to Nathan who appears to think that all Cantorians are evil, if
only because they can follow simple arguments that he can't.
Mike Oliver
"Andrew Boucher" <Helene....@wanadoo.fr> wrote:
> I wasn't making claims of certainty, because I don't
> find this (psychological) notion very useful.
"Certainty" maybe wasn't the exact appropriate word. You were
responding to what I thought was an excellent post by Mike
McCarty wherein he showed how the generalization to Lebesgue
integration--not needed to integrate actual functions that arise
in practice, but needed to simplify the overall structure of
the theory--was analogous to the use of arbitrarily large integers,
which are *also* not needed to describe the observable world but
which simplify the overall theory.
You claimed that there was a qualitative difference here,
and referred to "little itty-bitty trustworthy N". That was
what I was calling a "claim of certainty". Maybe that wasn't
the right word; my claim is that the two situations are in
fact qualitatively very similar.
> Secondly, I provided what I thought was a good--albeit a
> short--proof.
> [...]
> Where do *you* think the argument falls down. Call me lazy,
> but I don't want to do all the work here.
All right, bear with me a little; there's nothing in your post
to which my argument will be *directly* responsive but hopefully
it'll show where I'm coming from (I quoted the above for some
sort of context).
First, for the duration of this post, let's interpret "possible"
*not* to mean "true in some possible world," but rather "when
assumed, does not lead in any *known* way to anything that directly
contradicts observation of the physical world, nor make it impossible
to reason usefully about that world".
With this meaning of "possible" I agree that it's possible the universe
is infinite, but I also think it's possible that it's *impossible*
that the universe is infinite. Suspend disbelief for a moment
and consider my example.
Are you familiar with the theory Q, called Robinson arithmetic? It's
a very weak theory indeed. I don't have the precise axioms at my
fingertips, but I think that you take the first four Peano
axioms, and then you violently restrict induction, so that not
only does the predicate have to be Delta_0 (i.e. have only bounded
quantifiers), but you're only allowed to conclude that this predicate
holds *up*to*a*bound*. I.e. if P is Delta_0 with one free variable
and you can prove P(0) and (all n<N)(P(n) --> P(n+1)), then
you're allowed to conclude (all n<N)(P(n)). (Corrections welcomed
if I've messed this up.)
Now consider some simple definition of some outrageously large
number, say Graham's number or Ackermann(100,100). I think
it is a reasonable bet that Q does not prove the existence of
either of these numbers via a proof that's short enough actually
to be written down.
Now suppose that someone were to find a proof of 0=1 from
the theory "Q + Graham's number exists", but that the proof
did not go through (at least via a proof short enough to be
written down) to conclude Q |= 0=1.
Would not the most reasonable conclusion be that Graham's number
is simply too large to exist; that it is *logically*impossible*
for there to be numbers that large?
And yet, life would go on, wouldn't it? We'd still be able to
use our old familiar numbers to count the number of panini rustici
we need at the supermarket without worrying that there'll be
a different number of them when we get home. The mathematics
that designs our airplane wings would still work--not surprising,
since the wing and the airflow can be modeled using only integers
below a much smaller bound. And so on.
The difficulty would be in trying to conceptualize all that
mathematics in a uniform framework. That's where we *really*
use arbitrarily large integers, and also where we really use
Lebesgue integration, and elementary embeddings from V into
a transitive inner model, and so on.
Mike Oliver
Andrew Boucher
----------
In article <7hnmch$en5$1...@nnrp1.deja.com>, Mike Oliver
<oli...@math.ucla.edu> wrote:
>In article <7hndc7$9mo$1...@wanadoo.fr>,
> "Andrew Boucher" <Helene....@wanadoo.fr> wrote:
>> I wasn't making claims of certainty, because I don't
>> find this (psychological) notion very useful.
>
>"Certainty" maybe wasn't the exact appropriate word. You
were
>responding to what I thought was an excellent post by Mike
>McCarty wherein he showed how the generalization to
Lebesgue
>integration--not needed to integrate actual functions that
arise
>in practice, but needed to simplify the overall structure
of
>the theory--was analogous to the use of arbitrarily large
integers,
>which are *also* not needed to describe the observable
world but
>which simplify the overall theory.
His was a good post. Still, don't touch N!
I'll respond to specific things below, but again just to
summarize my position in a nutshell: The situation is not
qualitatively the same. The status of arithmetic is
different than the status of set theory or analysis. The
axioms of N (e.g. the Peano axioms) are logically necessary,
so arithmetic theorems are logically necessary. The axioms
of Cantorian set theory (and real analysis) are not
logically necessary, and one should accept or reject them
for, as you put it, pragmatic reasons.
So of course you and any other Cantorian would prefer modern
set theory to be "just like" arithmetic. Even though you
have a philosophy of pragmatism, by putting the two on the
same level, the set theory benefits from arithmetic's true
and special status. If Cantorian set theory is just like
arithmetic, then, because there's nothing wrong with the
second, there can't be anything wrong with the first.
>
>You claimed that there was a qualitative difference here,
>and referred to "little itty-bitty trustworthy N". That
was
>what I was calling a "claim of certainty". Maybe that
wasn't
>the right word; my claim is that the two situations are in
>fact qualitatively very similar.
>
>> Secondly, I provided what I thought was a
good--albeit a
>> short--proof.
>> [...]
>> Where do *you* think the argument falls down. Call me
lazy,
>> but I don't want to do all the work here.
>
>All right, bear with me a little; there's nothing in your
post
>to which my argument will be *directly* responsive but
hopefully
>it'll show where I'm coming from (I quoted the above for
some
>sort of context).
>
>First, for the duration of this post, let's interpret
"possible"
>*not* to mean "true in some possible world," but rather
"when
>assumed, does not lead in any *known* way to anything that
directly
>contradicts observation of the physical world, nor make it
impossible
>to reason usefully about that world".
Yikes! To be honest I'm not so comfortable with this
interpretation (e.g. you explain "possible" in terms of
"impossible"), but okay, I don't think it changes any of our
arguments one way or another.
>
>With this meaning of "possible" I agree that it's possible
the universe
>is infinite, but I also think it's possible that it's
*impossible*
I'd probably jump out a window myself--from surprise. If I
had to conclude something, I might just well say that indeed
Graham's number is too large to exist. But I don't know,
because I don't expect it to happen. If someone proved that
PA was inconsistent, I would have the same problem. It
would be very disconcerting, to say the least.
I don't expect it to happen, because it's logically
impossible for "Q + Graham's number exists" to be
inconsistent. (And I would use induction to *prove* it!)
So, although your example is obviously more sophisticated, I
would liken your scenario to someone supposing that 1 = 2,
and then asking for my reaction.
I would think it's more likely that every experience in
the world which tells us what 1+1 equals, suddenly says that
1+1 = 3 rather than 1+1 = 2. When I add a liter of water to
another liter of water, poof! there's now three liters of
water. That, at least, is logically possible.
>We'd still be able to
>use our old familiar numbers to count the number of panini
rustici
>we need at the supermarket without worrying that there'll
be
>a different number of them when we get home. The
mathematics
>that designs our airplane wings would still work--not
surprising,
>since the wing and the airflow can be modeled using only
integers
>below a much smaller bound. And so on.
What I like most about your example is that it happened!
(Of course this may be the reason why you're presenting
it...) Set theory was shown to be inconsistent, the
explanation was that some sets are too big, and people did
go quietly on their way. It still hasn't filtered down to
the mainstream, where people worry about naive formation of
sets. (Oh my gosh! If I loan you five dollars, maybe the
set of my five dollars and your five dollars won't exist!
Maybe I'll never get my money back!)
But on the other hand a bound to the natural numbers
would be different.
1/ While the mathematics that designs wings would still
work, I think we would begin to wonder whether there are
unknown, weird effects which we are not modelling.
Basically, before we began, we would check that the number
of things being modelled was well below the bound, or we
would have to worry about a "Graham" effect.
2/ Consider a program with an infinite loop. What
would happen once it has gone through more steps than the
Graham number? Not possible, you say, given the proposed
lifetime of the universe? But I don't have to start at step
1. I can start at Graham number - x, where x is small,
since I can reason what the state of the computer would be
if it were at step Graham number - x, and begin running it
there. What would happen??
>
>The difficulty would be in trying to conceptualize all that
>mathematics in a uniform framework. That's where we
*really*
>use arbitrarily large integers, and also where we really
use
>Lebesgue integration, and elementary embeddings from V into
>a transitive inner model, and so on.
But you agree, I hope, that arbitrarily large integers come
before the rest. If the natural numbers are finite, then
forget the real numbers, Lebesgue integration and V. On the
other hand, the natural numbers could be (!!) infinite,
without V. So, even if you don't agree with anything else I
said, I think you must still say the two are *not*
qualitatively the same, because the situation is not
symmetric. If you want V, you have to give me N first.
David Petry
Mike Oliver wrote in message <7hnmch$en5$1...@nnrp1.deja.com>...
>First, for the duration of this post, let's interpret "possible"
>*not* to mean "true in some possible world," but rather "when
>assumed, does not lead in any *known* way to anything that directly
>contradicts observation of the physical world, nor make it impossible
>to reason usefully about that world".
It boggles my mind that you see no need to incorporate the concept
of falsifiability into your definition of "possible". That is, you allow
that something is "possible" even if it has no observable implications
for the physical world.
However, I can see that arguing on the Internet wouldn't be nearly as
much fun if everyone were held to such impossibly high standards as
to require falsifiability.
Richard Carr
:Date: Mon, 17 May 1999 12:00:21 -0700
:From: David Petry <david...@mindspring.com>
:Newsgroups: sci.math, sci.logic
:Subject: Re: Not so nasty Lebesgue, but not Riemann, integrable functions
:
:
:Mike Oliver wrote in message <7hnmch$en5$1...@nnrp1.deja.com>...
:
:
:>First, for the duration of this post, let's interpret "possible"
:>*not* to mean "true in some possible world," but rather "when
:>assumed, does not lead in any *known* way to anything that directly
:>contradicts observation of the physical world, nor make it impossible
:>to reason usefully about that world".
:
:It boggles my mind that you see no need to incorporate the concept
:of falsifiability into your definition of "possible". That is, you allow
:that something is "possible" even if it has no observable implications
:for the physical world.
Have you no imagination?
:
:However, I can see that arguing on the Internet wouldn't be nearly as
:
:
:
:
:
Mike McCarty
Andrew Boucher <Helene....@wanadoo.fr> wrote:
)
)----------
)In article <7hhft1$hb0$1...@relay1.dsccc.com>,
)jmcc...@sun1307.ssd.usa.alcatel.com (Mike McCarty) wrote:
[snip]
)I guess we agree that, if you have all kinds of weird
)add-ons, you can take that as an indication that you're on
)the wrong track. We don't agree whether the conclusion
)should be: accept measurability a la Lebesgue; or take
)another look at what you're proving and why. Maybe the
)add-ons aren't as weird as all that, or maybe they reveal
)something about reality. Maybe we don't even need the
)proposition to begin with. I don't know how to begin
)analysis without the real numbers; but I can get a good way
)without Lebesgue integration. After all, we didn't have it
)until a hundred years ago...
I was responding to a question about why one would go to the effort of
developing/learning about measure theory, not trying to show that the
Legesgue integral was the *right* one. In fact, I specifically mentioned
the Danielle integral as an alternative to Legesgue.
)>
)>Likewise, when we state some property holds p.p, in the
)backs of our
)>minds we know we are really saying it holds *always*, well,
)except for
)>some weird cases which don't matter, and we can lump them
)together that
)>way. Measure theory lets us do this simplification. Not
)using it makes
)>things complicated.
)>
)>Mike
)>
)>--
)>----
)>char *p="char
)*p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,3
)4);}
)>This message made from 100% recycled bits.
)
)>I don't speak for Alcatel <- They make me say that.
)
)
Mike Oliver
David Petry wrote:
>
> Mike Oliver wrote in message <7hnmch$en5$1...@nnrp1.deja.com>...
>
>> First, for the duration of this post, let's interpret "possible"
>> *not* to mean "true in some possible world," but rather "when
>> assumed, does not lead in any *known* way to anything that directly
>> contradicts observation of the physical world, nor make it impossible
>> to reason usefully about that world".
> It boggles my mind that you see no need to incorporate the concept
> of falsifiability into your definition of "possible". That is, you allow
> that something is "possible" even if it has no observable implications
> for the physical world.
The high paladin of falsifiability, Karl Popper, to whom my own foundational
ideas are much indebted, claimed in "Conjectures and Refutations" that
falsifiability was the "demarcation" between science and non-science;
that is, that an unfalsifiable claim was not a scientific claim (which
is not the same as calling it meaningless). In his earlier days on
the fringes of the Vienna Circle he *may* have made the more fundamentalist
assertion that such propositions were actually meaningless.
But never ever *ever* have I heard anyone say that if a proposition
is unfalsifiable then it is for that reason *impossible*. This is
either a novel Petrian formulation of the falsifiability criterion,
or a stunning irrelevancy. I vote for the latter.
--
Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love")
Finger for PGP public key, or visit http://www.math.ucla.edu/~oliver.
1500 bits, fingerprint AE AE 4F F8 EA EA A6 FB E9 36 5F 9E EA D0 F8 B9
David Petry
Mike Oliver wrote in message <37409C6F...@math.ucla.edu>...
>
>
>David Petry wrote:
>> It boggles my mind that you see no need to incorporate the concept
>> of falsifiability into your definition of "possible". That is, you allow
>> that something is "possible" even if it has no observable implications
>> for the physical world.
>But never ever *ever* have I heard anyone say that if a proposition
>is unfalsifiable then it is for that reason *impossible*.
My claim is that it is neither meaningful to assert that an unfalsifiable
proposition is possible, nor that it is impossible.
Mike Oliver
David Petry wrote:
>
> Mike Oliver wrote in message <37409C6F...@math.ucla.edu>...
>> But never ever *ever* have I heard anyone say that if a proposition
>> is unfalsifiable then it is for that reason *impossible*.
> My claim is that it is neither meaningful to assert that an unfalsifiable
> proposition is possible, nor that it is impossible.
Ah. Well in that case you should have no quarrel with my argument, in
context. Against someone arguing "A is logically necessary" I was
arguing " ~A is possible ", but all I need is " ~A is not impossible ".
Then if ~A is not falsifiable, you apparently agree that ~A is not
impossible, which is all that is needed to refute "A is necessary".
Mike Oliver
"Andrew Boucher" <Helene....@wanadoo.fr> wrote:
> The axioms of N (e.g. the Peano axioms) are logically necessary,
> so arithmetic theorems are logically necessary.
I say you have not justified this claim. See one of the other
articles in this set of followups--I'm splitting these up because
the posts are getting too long to exist :-).
> So of course you and any other Cantorian would prefer modern
> set theory to be "just like" arithmetic. [....] If Cantorian set
> theory is just like arithmetic, then, because there's nothing wrong
> with the second, there can't be anything wrong with the first.
If this is my plan I must be rather inept at it, since I've
been arguing that in fact there *can* (as far as we know) be
something wrong with arithmetic. Now it's true that I would like
people to go on from that to see that the risks we take in accepting
set theory are not different *in*kind* from those we take in accepting
arithmetic. But that doesn't say there's no further risk in accepting
set theory, nor that one who uses arithmetic is *compelled* to accept
this extra risk. It does say that it doesn't make sense to establish
the two fields as separate philosophical categories.
Mike Oliver
"Andrew Boucher" <Helene....@wanadoo.fr> wrote:
>> [Mike Oliver wrote:]
>> Now suppose that someone were to find a proof of 0=1 from
>> the theory "Q + Graham's number exists", [....]
> I'd probably jump out a window myself--from surprise.
Is that how you respond whenever something happens that you thought
was impossible? Remind me not to go for a stroll on your street :-)
> If I had to conclude something, I might just well say that indeed
> Graham's number is too large to exist. But I don't know,
> because I don't expect it to happen.
Nor do I. Obviously. Still, the situation is *imaginable*,
and if it were true, then I don't see how the world would appear
different, except that we would have this proof to look at.
None of our experiences would be any different, and if you
believe that intuition is shaped by experience and evolution
(the latter being racial experience), then our intuitions
would also be the same. Therefore neither experience nor
intuition can be the basis of an argument that this can't happen.
> I would think it's more likely that every experience in
> the world which tells us what 1+1 equals, suddenly says that
> 1+1 = 3 rather than 1+1 = 2. When I add a liter of water to
> another liter of water, poof! there's now three liters of
> water. That, at least, is logically possible.
Really? If every time you found two different things there
were magically a third such thing, this would surprise you *less*
than the purely abstract strangeness of finding a string of
symbols that represented a derivation of 0=1 from the formal theory
Q+"Graham's number exists"? You must have much more faith
in abstract reasoning than I have.
Mike Oliver
"Andrew Boucher" <Helene....@wanadoo.fr> wrote:
> What I like most about your example is that it happened!
> explanation was that some sets are too big, and people did
> go quietly on their way.
If you're talking about the Russell paradox, I believe that's a
historical misconception. Russell's paradox was a problem for Russell
and his Logicist buddies; it was never a problem AFAIK for Zermelo, who
was starting from quite different intuitive motivations (the iterative
hierarchy). If you're talking about the earlier Cantor's paradox,
you might have a point, but by that time set theory had not become
established.
> But on the other hand a bound to the natural numbers
> would be different.
> 1/ While the mathematics that designs wings would still
> work, I think we would begin to wonder whether there are
> unknown, weird effects which we are not modelling.
Well, sure, but there are *always* weird effects that we
aren't modeling. My confidence in a newly-designed wing doesn't
come from any claim that it's _logically_necessary_ that
the wing will work; it comes from the overall track record
of the methodology used to design it. None of that would change.
> Basically, before we began, we would check that the number
> of things being modelled was well below the bound, or we
> would have to worry about a "Graham" effect.
Well, if the bound is anywhere near Graham's number, then the
number of things we're modeling is always always ALWAYS going
to be *much* less than that, no matter how long the human race
survives or how much technological progress we make.
> 2/ Consider a program with an infinite loop. What
> would happen once it has gone through more steps than the
> Graham number? Not possible, you say, given the proposed
> lifetime of the universe? But I don't have to start at step
> 1. I can start at Graham number - x, where x is small,
> since I can reason what the state of the computer would be
> if it were at step Graham number - x, and begin running it
> there. What would happen??
Nothing special, I imagine. What is this supposed to prove?
> But you agree, I hope, that arbitrarily large integers come
> before the rest. If the natural numbers are finite, then
> forget the real numbers, Lebesgue integration and V. On the
> other hand, the natural numbers could be (!!) infinite,
> without V. So, even if you don't agree with anything else I
> said, I think you must still say the two are *not*
> qualitatively the same, because the situation is not
> symmetric. If you want V, you have to give me N first.
Well, there's two issues here. One is that if the existence
of Graham's number is inconsistent, then yes, certainly all
stronger theories ( I Delta_0 + "exponentiation is total", PA,
PA+"PA is 1-consistent", analysis, ZF, ZF+"exists a measurable",
etc.) are all also inconsistent. Whereas there could be one
theory in this list that is inconsistent but all weaker theories
could be consistent.
But this is a difference of degree, not of kind. I see nothing
very special about PA's position in this spectrum--there are theories
a little bit weaker than PA and theories a little bit stronger.
Why should PA be the dividing line?
David Petry
Mike Oliver wrote in message <374101A5...@math.ucla.edu>...
> Against someone arguing "A is logically necessary" I was
>arguing " ~A is possible ", but all I need is " ~A is not impossible ".
>Then if ~A is not falsifiable, you apparently agree that ~A is not
>impossible, which is all that is needed to refute "A is necessary".
You certainly can't conclude from "~A is not falsifiable" that
"A is logically necessary" is false. In fact, it's easy to provide
lots of counterexamples.
Mike Oliver
That wasn't the structure of my argument. Still, name one such
counterexample. It seems to me that if you know that A is logically
necessary, then that in itself falsifies ~A, thereby showing directly
that ~A is in fact falsifiable.
Andrew Boucher
In article <7hr2go$o5j$1...@nnrp1.deja.com>, Mike Oliver
<oli...@math.ucla.edu> wrote:
>In article <7hof76$buc$1...@wanadoo.fr>,
> "Andrew Boucher" <Helene....@wanadoo.fr> wrote:
>
>>> Now suppose that someone were to find a proof of 0=1
from
>>> the theory "Q + Graham's number exists", [....]
>
>> I'd probably jump out a window myself--from surprise.
>
>Is that how you respond whenever something happens that you
thought
>was impossible?
for arithmetic.
> Remind me not to go for a stroll on your street :-)
>> If I had to conclude something, I might just well say
that indeed
>> Graham's number is too large to exist. But I don't know,
>> because I don't expect it to happen.
>
>Nor do I. Obviously. Still, the situation is
*imaginable*,
>and if it were true, then I don't see how the world would
appear
>different, except that we would have this proof to look at.
psychological meanings, so I would prefer to avoid it. I'd
like to stick to logical possibility and logical necessity.
To make you happy ;-) I'll use "imagine", but you should
just translate "can imagine P" as "P is logically possible".
If this seems unfair, see my comments a bit further below.
What you can imagine is someone you trust informing you that
there's an inconsistency, or you can imagine reading it in a
journal. But you can't imagine it's *the case*. You can't
imagine that Q + Graham number exists is inconsistent.
(Repeat the same argument with 1 = 2. Again, you can
imagine someone claiming in a journal to have proved that 1
= 2. But you can't actually imagine 1 = 2, or that you can
prove that 1 = 2 in PA.)
I'll put it again, but in a slightly different way. You can
imagine a computer, programmed to provide proofs in PA, to
finish up one of its outputs with 1 = 2 (or PA is
inconsistent). That is because you can imagine the computer
making a mistake. But you can't imagine it capable of
actually providing a proof--what counts as and is a proof.
You can't imagine an inconsistency.
Now I guess you may think I am being unfair. After all you
gave a definition of "possible" which is different from
"logically possible". Although I'm not clear about the
definition in detail, by your meaning you want to say, "I
don't know today whether PA will one day be proven
inconsistent, so it's possible." For the advancement of our
discussion, I accept the intent of your definition, and
agree that given this meaning, you can say a lot of the
things you're saying.
The problem is, no matter how you use "possible", it's still
*not* logically possible that PA will be or is inconsistent.
And that's my assertion. To repeat myself: arithmetic is
logically necessary, analysis and Cantorian set theory are
not. So your different use of "possible" is not here nor
there, or at least, it's not disagreeing with what I'm
saying. It's interesting, but it's off to the side.
But you shouldn't introduce another meaning, because I think
where you disagree with me is on the question of whether
arithmetic is logically necessary. If you've asked me how I
"know," I've almost certainly replied, "It just is." Or
rather: arithmetic theorems follow validly from logically
necessary axioms and so are logically necessary. And then,
when you ask me how I "know" about the logical necessity of
the axioms, that's when I reply, "It just is." 0 is a
natural number: it just necessarily is so. Induction: it
just necessarily is so. And so on, down the list. I can't
give any deeper explanation, just as I can't explain in more
detail why "Red is a color" is logically necessary. It just
*is*.
So clearly on *that* point we can disagree. I say, "It just
is" and you say, "It just isn't." Maybe I'm mistaken, but I
think we disagree (honestly) on that point. I may question
your sincerity or your sanity under my breath, but
basically, I realize you can stay in your castle and repeat,
"It just isn't," and there's not much I can do about it. (I
can try to draw you out and fight, like asking how you
justify "Red is a color", but then you can always refuse.)
But that's the only point of contention, because we agree on
the rest. I agree with you that analysis and Cantorian set
theory must be judged on their pragmatic value. I probably
disagree with you that analysis *is* pragmatic while
Cantorian set theory is *not*, but now that's another thread
(which I'm not starting, because after this discussion, I need a rest!).
>None of our experiences would be any different, and if you
>believe that intuition is shaped by experience and
evolution
>(the latter being racial experience), then our intuitions
>would also be the same. Therefore neither experience nor
>intuition can be the basis of an argument that this can't
happen.
>
>> I would think it's more likely that every experience
in
>> the world which tells us what 1+1 equals, suddenly says
that
>> 1+1 = 3 rather than 1+1 = 2. When I add a liter of water
to
>> another liter of water, poof! there's now three liters of
>> water. That, at least, is logically possible.
>
>Really? If every time you found two different things there
>were magically a third such thing, this would surprise you
*less*
>than the purely abstract strangeness of finding a string of
>symbols that represented a derivation of 0=1 from the
formal theory
>Q+"Graham's number exists"? You must have much more faith
>in abstract reasoning than I have.
using psychological terms like "surprise" and "expect" in my
two previous paragraphs, and then "likely", which can have a
more logical meaning, in the paragraph you cite. The
inconsistency can't happen: it's logically impossible. The
magic can happen: it's logically possible. In possible
world jargon, there's a possible world in which the second
happens, but not the first. So (trivially) the first is
less likely. Sorry for that!
Andrew Boucher
----------
In article <7hr2iq$o75$1...@nnrp1.deja.com>, Mike Oliver
<oli...@math.ucla.edu> wrote:
>In article <7hof76$buc$1...@wanadoo.fr>,
> "Andrew Boucher" <Helene....@wanadoo.fr> wrote:
>> 2/ Consider a program with an infinite loop. What
>> would happen once it has gone through more steps than the
>> Graham number? Not possible, you say, given the proposed
>> lifetime of the universe? But I don't have to start at
step
>> 1. I can start at Graham number - x, where x is small,
>> since I can reason what the state of the computer would
be
>> if it were at step Graham number - x, and begin running
it
>> there. What would happen??
>
>Nothing special, I imagine. What is this supposed to
prove?
There's a largest natural number, G. A computer program
presumably can't run for more steps than G. I'm saying you
can cheat, and duplicate what the program would do by
starting the computer not at step 1, but at G - x. What
does this computer do after it runs x+1 steps? Presumably
the same thing that it would do after running one plus G
steps, this time beginning at step 1. But it can't...
But ok fair enough. Again, you're asking me to consider a
world which is impossible, where 1 = 2. So basically coming
up with practical difficulties is not the right response,
although I'd like to play the game. The right response is:
your world is not possible.
>
>> But you agree, I hope, that arbitrarily large integers
come
>> before the rest. If the natural numbers are finite, then
>> forget the real numbers, Lebesgue integration and V. On
the
>> other hand, the natural numbers could be (!!) infinite,
>> without V. So, even if you don't agree with anything
else I
>> said, I think you must still say the two are *not*
>> qualitatively the same, because the situation is not
>> symmetric. If you want V, you have to give me N first.
>
>Well, there's two issues here. One is that if the
existence
>of Graham's number is inconsistent, then yes, certainly all
>stronger theories ( I Delta_0 + "exponentiation is total",
PA,
>PA+"PA is 1-consistent", analysis, ZF, ZF+"exists a
measurable",
>etc.) are all also inconsistent. Whereas there could be
one
>theory in this list that is inconsistent but all weaker
theories
>could be consistent.
>
>But this is a difference of degree, not of kind. I see
nothing
>very special about PA's position in this spectrum--there
are theories
>a little bit weaker than PA and theories a little bit
stronger.
>Why should PA be the dividing line?
I never said (and I don't believe) PA is the dividing line.
But okay, I'll respond.
1/ You need PA just to be able to reason about what you can
prove in any first-order system.
2/ It's not because you have found a spectrum that there is
not a difference in kind. For instance, suppose PA is
consistent, but that ZF + "exists a measurable" is
inconsistent. Then there's a big difference in kind--one is
consistent, the other not--regardless of your spectrum.
You ask what is special about PA's position in the
spectrum. Of course, I'm not saying there is something
privileged about its position, only that there is a
difference between arithmetic and analysis. PA and anything
to the left are logically necessary. Maybe there are a few
things to the right which are also logically necessary. But
systems too far to the right (such as analysis) aren't
logically necessary.
Similarly, you can have: (1) arithmetic consistent and
analysis inconsistent, but not (2) arithmetic inconsistent
and analysis consistent. That strikes as a difference in
kind. Sure, there are other systems S which can substitute
for arithmetic here, or analysis here, and you can create an
ordering with these S. But there's still a difference in
kind between arithmetic and analysis. Maybe this difference
in kind isn't so important, and maybe it's not particularly
relevant--that's another story and another argument.
Basically, I was just trying to find a common point.
To be honest, I don't think we're so far apart on this
section, other than the semantic difference as to what
counts as "in kind", which doesn't strike me as so crucial,
and the logical necessity part, and that's in my other post.
Mike Oliver
Mike McCarty wrote:
>
> In article <3741D8C5...@math.ucla.edu>,
> Mike Oliver <oli...@math.ucla.edu> wrote:
> )That wasn't the structure of my argument. Still, name one such
> )counterexample. It seems to me that if you know that A is logically
> )necessary, then that in itself falsifies ~A, thereby showing directly
> )that ~A is in fact falsifiable.
>
> It is not difficult to show that if in a formal system A follows from
> the axioms, and that ~A follows from the axioms, then all propositions P
> follow from the axioms.
>
> So I believe you are assuming consistency when you make your statement.
I don't really follow. I'm assuming consistency of what, exactly?
The logical axioms? I suppose if the logical axioms are inconsistent,
then everything is logically necessary; but in that case the notion
of logical necessity is trivial, so it hardly seems a point against
my argument that you can't distinguish between PA and ZF on the basis
of logical necessity.
But in another post you seemed to indicate that you *agreed* with that
point of mine. So just what *are* you saying?
Bennett Standeven
On Mon, 17 May 1999, Mike Oliver wrote:
> In article <7hndc7$9mo$1...@wanadoo.fr>,
> "Andrew Boucher" <Helene....@wanadoo.fr> wrote:
> > I wasn't making claims of certainty, because I don't
> > find this (psychological) notion very useful.
>
> "Certainty" maybe wasn't the exact appropriate word. You were
> responding to what I thought was an excellent post by Mike
> McCarty wherein he showed how the generalization to Lebesgue
> integration--not needed to integrate actual functions that arise
> in practice, but needed to simplify the overall structure of
> the theory--was analogous to the use of arbitrarily large integers,
> which are *also* not needed to describe the observable world but
> which simplify the overall theory.
>
> You claimed that there was a qualitative difference here,
> and referred to "little itty-bitty trustworthy N". That was
> what I was calling a "claim of certainty". Maybe that wasn't
> the right word; my claim is that the two situations are in
> fact qualitatively very similar.
>
> > Secondly, I provided what I thought was a good--albeit a
> > short--proof.
> > [...]
> > Where do *you* think the argument falls down. Call me lazy,
> > but I don't want to do all the work here.
>
> All right, bear with me a little; there's nothing in your post
> to which my argument will be *directly* responsive but hopefully
> it'll show where I'm coming from (I quoted the above for some
> sort of context).
>
> First, for the duration of this post, let's interpret "possible"
> *not* to mean "true in some possible world," but rather "when
> assumed, does not lead in any *known* way to anything that directly
> contradicts observation of the physical world, nor make it impossible
> to reason usefully about that world".
>
> With this meaning of "possible" I agree that it's possible the universe
> is infinite, but I also think it's possible that it's *impossible*
> that the universe is infinite. Suspend disbelief for a moment
> and consider my example.
>
> Are you familiar with the theory Q, called Robinson arithmetic? It's
> a very weak theory indeed. I don't have the precise axioms at my
> fingertips, but I think that you take the first four Peano
> axioms, and then you violently restrict induction, so that not
> only does the predicate have to be Delta_0 (i.e. have only bounded
> quantifiers), but you're only allowed to conclude that this predicate
> holds *up*to*a*bound*. I.e. if P is Delta_0 with one free variable
> and you can prove P(0) and (all n<N)(P(n) --> P(n+1)), then
> you're allowed to conclude (all n<N)(P(n)). (Corrections welcomed
> if I've messed this up.)
>
> Now consider some simple definition of some outrageously large
> number, say Graham's number or Ackermann(100,100). I think
> it is a reasonable bet that Q does not prove the existence of
> either of these numbers via a proof that's short enough actually
> to be written down.
>
> Now suppose that someone were to find a proof of 0=1 from
> the theory "Q + Graham's number exists", but that the proof
> did not go through (at least via a proof short enough to be
> written down) to conclude Q |= 0=1.
>
> Would not the most reasonable conclusion be that Graham's number
> is simply too large to exist; that it is *logically*impossible*
> for there to be numbers that large?
>
Actually, there is a good reason to think that numbers which are too large
to be written down can still exist: I would estimate the number of atoms
in the Sun at around 10^53 or so. Assuming we write 30 marks a second
(using a machine, obviously), it would then take about 3*10^51 seconds, or
roughly 10^42 years (give or take an order of magnitude) to write down
that many marks. Clearly, the machine could not possibly survive long
enough to do this, even if we increased its speed by several orders of
magnitude. Thus, we still couldn't write down a proof that it exists. But
arithmetic still works on it, doesn't it?
>
> And yet, life would go on, wouldn't it? We'd still be able to
> use our old familiar numbers to count the number of panini rustici
> we need at the supermarket without worrying that there'll be
> a different number of them when we get home. The mathematics
> that designs our airplane wings would still work--not surprising,
> since the wing and the airflow can be modeled using only integers
> below a much smaller bound. And so on.
>
If it happened, I know I would worry about the argument above, and likely
so would others. Thus, sciences that make use of such large numbers would
be in trouble, even if most people wouldn't.
Mike Oliver
> Actually, there is a good reason to think that numbers which are too large
> to be written down can still exist:
I'm certainly not claiming that they *don't* exist. That would be silly.
I'm only questioning the notion that we can have some sort of a priori
knowledge of their logically necessary existence that is different in kind
from the justifications for analysis and set theory.
(Also, the examples I was talking about are quite a bit larger than merely
"too large to be written down" if by that you mean in decimal notation.)
james d. hunter
>
> On Mon, 17 May 1999, Mike Oliver wrote:
>
> > "Andrew Boucher" <Helene....@wanadoo.fr> wrote:
> > > I wasn't making claims of certainty, because I don't
> > > find this (psychological) notion very useful.
> >
> > responding to what I thought was an excellent post by Mike
> > McCarty wherein he showed how the generalization to Lebesgue
> > integration--not needed to integrate actual functions that arise
> > in practice, but needed to simplify the overall structure of
> > the theory--was analogous to the use of arbitrarily large integers,
> > which are *also* not needed to describe the observable world but
> > which simplify the overall theory.
> >
> > You claimed that there was a qualitative difference here,
> > and referred to "little itty-bitty trustworthy N". That was
> > what I was calling a "claim of certainty". Maybe that wasn't
> > the right word; my claim is that the two situations are in
> > fact qualitatively very similar.
> >
> > > short--proof.
> > > Where do *you* think the argument falls down. Call me lazy,
> > > but I don't want to do all the work here.
> >
> in the Sun at around 10^53 or so. Assuming we write 30 marks a second
> (using a machine, obviously), it would then take about 3*10^51 seconds, or
> roughly 10^42 years (give or take an order of magnitude) to write down
> that many marks. Clearly, the machine could not possibly survive long
> enough to do this, even if we increased its speed by several orders of
> magnitude. Thus, we still couldn't write down a proof that it exists. But
> arithmetic still works on it, doesn't it?
>
> >
> > And yet, life would go on, wouldn't it? We'd still be able to
> > use our old familiar numbers to count the number of panini rustici
> > we need at the supermarket without worrying that there'll be
> > a different number of them when we get home. The mathematics
> > that designs our airplane wings would still work--not surprising,
> > since the wing and the airflow can be modeled using only integers
> > below a much smaller bound. And so on.
No, airplanes would drop out of the sky like dead pigeons.
Integers don't model wings and airflow very well.
Mike Oliver
Does no one know how to edit? Bennett quoted basically my entire post
to address one small point, then you quoted basically Bennett's entire post
to respond, not to him, but to me, and again for a single small point.
My use of the notion of "modelling by integers" obviously elided quite
a bit of discussion, which I took for granted that the reader would be
able to fill in. Suffice it to say that any modelling you can do on
a (physical, commercially available) computer is in fact modelling by
integers, and in fact with a rather small upper bound to the integers
(say, the largest integer you can represent in 10^50 bytes; that should
cover me against a certain amount of future technical progress).
Now it's true that the integers are used *conceptually* to represent reals,
which are an immense conceptual convenience in modelling physical
phenomena. Moreover, at some point "convenience" is a bit of
a euphemism--you could say that if you want to go to the Moon
that it's "convenient" to have a rocket, but in principle you could
just jump.
But none of this changes the fact that when you model an airplane wing,
the answers you get are determined by the arithmetical relationships
integers below some bound, and that this bound is way way WAY below
Graham's number, and that if Q+"Graham's number exists" turned out
to be inconsistent, I would still consider the design of airplanes
to be roughly as trustworthy as I had before the shocking discovery.
David Petry
Mike Oliver wrote in message <3741D8C5...@math.ucla.edu>...
>David Petry wrote:
>> Mike Oliver wrote in message <374101A5...@math.ucla.edu>...
>> > Against someone arguing "A is logically necessary" I was
>> >arguing " ~A is possible ", but all I need is " ~A is not impossible ".
>> >Then if ~A is not falsifiable, you apparently agree that ~A is not
>> >impossible, which is all that is needed to refute "A is necessary".
>> You certainly can't conclude from "~A is not falsifiable" that
>> "A is logically necessary" is false. In fact, it's easy to provide
>> lots of counterexamples.
>That wasn't the structure of my argument.
It looks to me like the structure of your argument given in the above
paragraph.
> Still, name one such counterexample.
Let's look at an example from physics. Consider the statement:
Somewhere in the universe, there exists a particle with a
charge of pi times the charge of the electron.
That proposition is not falsifiable. That is, the proposition does
not make any prediction that could be proven wrong by
experiment.
The negation of that proposition is a theorem of standard
particle physics (i.e. it's logically necessary).
So if A is the negation of the proposition, then A is logically
necessary and ~A is not falsifiable.
>It seems to me that if you know that A is logically
>necessary, then that in itself falsifies ~A, thereby showing directly
>that ~A is in fact falsifiable.
If A is logically necessary, then that in itself falsifies the statement
"Our theory together with the statement ~A is logically consistent".
Mike Oliver
David Petry wrote:
>
> Mike Oliver wrote in message <3741D8C5...@math.ucla.edu>...
>> Still, name one such counterexample.
>
> Let's look at an example from physics. Consider the statement:
>
> Somewhere in the universe, there exists a particle with a
> charge of pi times the charge of the electron.
>
> The negation of that proposition is a theorem of standard
> particle physics (i.e. it's logically necessary).
You think *quantum*mechanics* is logically necessary?
Boy, that sure is a joke on all those experimental folks running the
particle accelerators. They could have gotten all their results
from logic alone....
Torkel Franzen
> If you're talking about the Russell paradox, I believe that's a
> historical misconception. Russell's paradox was a problem for Russell
> and his Logicist buddies; it was never a problem AFAIK for Zermelo, who
> was starting from quite different intuitive motivations (the iterative
> hierarchy).
This isn't historically accurate. Zermelo doesn't suggest anything
like the iterative hierarchy in his original presentation - that came
twenty years later.
Mike Oliver
OK. I'm a bit surprised but I'll take your word for it. Are you
saying Zermelo *was* motivated by the Russell paradox? I thought
I had seen that idea debunked, but I'm a little vague on the details.
Russell Easterly
Andrew Boucher <Helene....@wanadoo.fr> wrote in message news:7hsmqj$683$1...@wanadoo.fr...
To repeat myself: arithmetic is
logically necessary, analysis and Cantorian set theory are
not.
Torkel Franzen
> OK. I'm a bit surprised but I'll take your word for it. Are you
> saying Zermelo *was* motivated by the Russell paradox? I thought
> I had seen that idea debunked, but I'm a little vague on the details.
Zermelo was motivated by the need to put set theory on a firm
axiomatic footing for mathematical purposes. In the preamble to his
1908 paper (see _From Frege to Godel_) he does mention Russell's
paradox, and he says (translated by Stefan Bauer-Mengelberg):
Now in the present paper I intend to show how the entire theory
created by Cantor and Dedekind can be reduced to a few
definitions and seven principles, or axioms, which appear to be
mutually independent. The further, more philosophical, question
about the origin of these principles and the extent to which they
are valid will not be discussed here.
Torkel Franzen
A PS about the translated passage from Zermelo. I think
Bauer-Mengelberg misunderstood the text at one point, so that "the
extent to which they are valid" should be rather "their range of
validity".
Andrew Boucher
----------
In article <7hqs4t$fji$1...@nntp3.atl.mindspring.net>, "David
Petry" <david...@mindspring.com> wrote:
>
>Mike Oliver wrote in message
<37409C6F...@math.ucla.edu>...
>>
>>
>>David Petry wrote:
>
>>> It boggles my mind that you see no need to incorporate
the concept
>>> of falsifiability into your definition of "possible".
That is, you allow
>>> that something is "possible" even if it has no
observable implications
>>> for the physical world.
>
>>But never ever *ever* have I heard anyone say that if a
proposition
>>is unfalsifiable then it is for that reason *impossible*.
>
>My claim is that it is neither meaningful to assert that an
unfalsifiable
>proposition is possible, nor that it is impossible.
>
>
>
>
I agree with you on this one, Mike. Apparently, we must
already know that a sentence is verifiable in order for it
to have a meaning. This is strange, because usually we must
know its meaning, in order to know what we have to verify.
Mike McCarty
)David Petry wrote:
)>
)> Mike Oliver wrote in message <374101A5...@math.ucla.edu>...
)>
)> > Against someone arguing "A is logically necessary" I was
)> >arguing " ~A is possible ", but all I need is " ~A is not impossible ".
)> >Then if ~A is not falsifiable, you apparently agree that ~A is not
)> >impossible, which is all that is needed to refute "A is necessary".
)>
)> You certainly can't conclude from "~A is not falsifiable" that
)> "A is logically necessary" is false. In fact, it's easy to provide
)> lots of counterexamples.
)
)That wasn't the structure of my argument. Still, name one such
)counterexample. It seems to me that if you know that A is logically
)necessary, then that in itself falsifies ~A, thereby showing directly
)that ~A is in fact falsifiable.
It is not difficult to show that if in a formal system A follows from
the axioms, and that ~A follows from the axioms, then all propositions P
follow from the axioms.
So I believe you are assuming consistency when you make your statement.
--
Mike McCarty
Andrew Boucher <Helene....@wanadoo.fr> wrote:
[snip]
)His was a good post. Still, don't touch N!
)
)I'll respond to specific things below, but again just to
)summarize my position in a nutshell: The situation is not
)qualitatively the same. The status of arithmetic is
)different than the status of set theory or analysis. The
)axioms of N (e.g. the Peano axioms) are logically necessary,
Stop right there. Please justify the claim that the Peano
axioms are logically necessary.
*I* certainly do not find them to be so. Certainly not any
more so than the the others you object to.
Mike
David Petry
Mike Oliver wrote in message <37423057...@math.ucla.edu>...
>You think *quantum*mechanics* is logically necessary?
Of course not. But given an axiomatic foundation for quantum
mechanics, there are logically necessary implications.
Nathan the Great
Richard Carr wrote:
> If one qualified
> subscribers to the theory as Cantorians also this still doesn't give any
> weight to Nathan who appears to think that all Cantorians are evil, if
> only because they can follow simple arguments that he can't.
He who thinks everything easy will end by finding everything difficult.
Therefore, the Sage, who regards everything as difficult, meets with no
difficulties in the end.
- TAO TEH CHING
Nathan the Great
Age 11
Mike Oliver
Fine, but that's not what's meant by a logically necessary proposition.
A proposition is logically necessary if it follows from logic alone,
with no assumptions at all.
james d. hunter
>
> Mike Oliver wrote in message <37423057...@math.ucla.edu>...
>
> >You think *quantum*mechanics* is logically necessary?
>
> Of course not. But given an axiomatic foundation for quantum
> mechanics, there are logically necessary implications.
That's true, but the implications from QM aren't always
necessarily the "best" implication. QM and Relativity
are mutually inconsistent, so there is some "pre-physics"
logic concerning if the model you're using is appropriate.
james d. hunter
>
> "james d. hunter" wrote:
> >
> > Bennett Standeven wrote:
> > > On Mon, 17 May 1999, Mike Oliver wrote:
> >>> And yet, life would go on, wouldn't it? We'd still be able to
> >>> use our old familiar numbers to count the number of panini rustici
> >>> we need at the supermarket without worrying that there'll be
> >>> a different number of them when we get home. The mathematics
> >>> that designs our airplane wings would still work--not surprising,
> >>> since the wing and the airflow can be modeled using only integers
> >>> below a much smaller bound. And so on.
>
> > No, airplanes would drop out of the sky like dead pigeons.
> > Integers don't model wings and airflow very well.
>
> Does no one know how to edit? Bennett quoted basically my entire post
> to address one small point, then you quoted basically Bennett's entire post
> to respond, not to him, but to me, and again for a single small point.
Yes, everyone knows how to edit. On Usenet though, if you edit
a post you stand a 50-50 chance of hearing somebody
whine about their copyrighted words being trashed, so it's
better just to let it go.
> My use of the notion of "modelling by integers" obviously elided quite
> a bit of discussion, which I took for granted that the reader would be
> able to fill in. Suffice it to say that any modelling you can do on
> a (physical, commercially available) computer is in fact modelling by
> integers, and in fact with a rather small upper bound to the integers
> (say, the largest integer you can represent in 10^50 bytes; that should
> cover me against a certain amount of future technical progress).
Are you sure you know how computers work? You don't think
that just because ultra-finitists think that computers are
discrete finite objects that they actually are, do you?
Integers don't model computers very well either.
There's a big difference between integers and sampled continuous
objects.
> Now it's true that the integers are used *conceptually* to represent
reals,
> which are an immense conceptual convenience in modelling physical
> phenomena. Moreover, at some point "convenience" is a bit of
> a euphemism--you could say that if you want to go to the Moon
> that it's "convenient" to have a rocket, but in principle you could
> just jump.
>
> But none of this changes the fact that when you model an airplane
wing,
> the answers you get are determined by the arithmetical relationships
> integers below some bound, and that this bound is way way WAY below
> Graham's number, and that if Q+"Graham's number exists" turned out
> to be inconsistent, I would still consider the design of airplanes
> to be roughly as trustworthy as I had before the shocking discovery.
Airplane design has almost -nothing- do with integers.
What you're obviously talking about is the number of particles
in the universe. Whatever that number is, if it even exists,
doesn't affect the 'infiniteness' of the universe.
Mike McCarty
)>
)> In article <3741D8C5...@math.ucla.edu>,
)> Mike Oliver <oli...@math.ucla.edu> wrote:
)> )That wasn't the structure of my argument. Still, name one such
)> )counterexample. It seems to me that if you know that A is logically
)> )necessary, then that in itself falsifies ~A, thereby showing directly
)> )that ~A is in fact falsifiable.
)>
)> It is not difficult to show that if in a formal system A follows from
)> the axioms, and that ~A follows from the axioms, then all propositions P
)> follow from the axioms.
)>
)> So I believe you are assuming consistency when you make your statement.
)
)I don't really follow. I'm assuming consistency of what, exactly?
)The logical axioms? I suppose if the logical axioms are inconsistent,
)then everything is logically necessary; but in that case the notion
)of logical necessity is trivial, so it hardly seems a point against
)my argument that you can't distinguish between PA and ZF on the basis
)of logical necessity.
)
)But in another post you seemed to indicate that you *agreed* with that
)point of mine. So just what *are* you saying?
Did you read my post? I'm confused by your question. You state
If A is logically necessary, then that falsifies ~A.
I know of no such requirement on formal systems, so I asked whether you
were assuming consistency.
Is that hard to understand?
I'm not being sarcastic, or anything like that.
Mike McCarty
james d. hunter <jim.h...@spam.free.jhuapl.edu.> wrote:
)Bennett Standeven wrote:
[snip]
)> > a different number of them when we get home. The mathematics
)> > that designs our airplane wings would still work--not surprising,
)> > since the wing and the airflow can be modeled using only integers
)> > below a much smaller bound. And so on.
)
) No, airplanes would drop out of the sky like dead pigeons.
) Integers don't model wings and airflow very well.
I don't folllow this at all. Computers are only capable of dealing with
integers. I speak as one who has written floating point packages, BTW.
In particular, I have implemented the math library for the C language.
Computers store integers with a maximum size, not even all integers. All
computer models eventually devolve into operations on integers.
So what are you trying to say?
Mike McCarty
)>
)> Mike Oliver wrote in message <37423057...@math.ucla.edu>...
)>
)> >You think *quantum*mechanics* is logically necessary?
)>
)> Of course not. But given an axiomatic foundation for quantum
)> mechanics, there are logically necessary implications.
)
)Fine, but that's not what's meant by a logically necessary proposition.
)A proposition is logically necessary if it follows from logic alone,
)with no assumptions at all.
But, nothing follows from logic alone, with no assumptions at all.
Have you studied First Order Formal Logic?
james d. hunter
>
> In article <37420FBD...@jhuapl.edu>,
> james d. hunter <jim.h...@spam.free.jhuapl.edu.> wrote:
> )Bennett Standeven wrote:
>
> [snip]
>
> )> > a different number of them when we get home. The mathematics
> )> > that designs our airplane wings would still work--not surprising,
> )> > since the wing and the airflow can be modeled using only integers
> )> > below a much smaller bound. And so on.
> )
> ) No, airplanes would drop out of the sky like dead pigeons.
> ) Integers don't model wings and airflow very well.
>
> I don't folllow this at all. Computers are only capable of dealing with
> integers. I speak as one who has written floating point packages, BTW.
> In particular, I have implemented the math library for the C language.
That's not really true. Computers deal with integrals, derivatives,
trig, and and even evil Cantorian math just find. I think I have
read sci.math more than once and have seen mathematicians
referring to Mathmatica, Maple, etc.
> Computers store integers with a maximum size, not even all integers.
All
> computer models eventually devolve into operations on integers.
>
> So what are you trying to say?
I'm saying that if you're doing number crunching that's true.
Also I'm saying that Usenet posters rarely use their computers
for number crunching.
---
Jim
Mike McCarty
)>
)> Mike Oliver wrote in message <37423057...@math.ucla.edu>...
)>
)> >You think *quantum*mechanics* is logically necessary?
)>
)> Of course not. But given an axiomatic foundation for quantum
)> mechanics, there are logically necessary implications.
)
) necessarily the "best" implication. QM and Relativity
) are mutually inconsistent, so there is some "pre-physics"
) logic concerning if the model you're using is appropriate.
Don't say that out loud around here. I made that statment a year or so
ago, and got lambasted fourteen ways from Sunday, by "would be"
physicists. For some reason, the recent graduates in physics have been
told "QM is a relativistic theory", and translated that into their
minds into "QM is completely compatible with relativity", when Mr. Bell
showed that the two theories are *incompatible*.
So, let's not bring up that subject again, or you'll find yourself
engaged in a war with people saying things like
Well, I just got my *degree* in physics, and I *know* they are
compatible. What are *your* qualifications?
Mike
ruper...@my-dejanews.com
Mike Oliver <oli...@math.ucla.edu> wrote:
>
>
> Richard Carr wrote:
>
> > On Sat, 8 May 1999, Mike Oliver wrote:
>
> > :Well, superficially at least there's some tiny question about what
> > :it means to be an inaccessible cardinal in the absence of Choice.
> > :That is, an inaccessible cardinal is one that's regular and
strongly
> > :limit, where kappa is strongly limit iff for every lambda < kappa,
> > :2^lambda < kappa.
> > :
> > :If we don't have Choice, what does this "<" mean? If A<B means
there's
> > :an injection from A to B and none from B to A (the usual meaning
without
> > :choice), then if kappa is inaccessible then Choice must hold a
fortiori
> > :up to rank kappa. In that case L(V_kappa) is an inner model of
> > :ZFC+"there exists an inaccessible", so we would have Con(ZF+IC) --
> Con(ZFC+IC).
>
> > OK.
Here is a definition of inaccessible cardinal which can be used in the
absence of AC: kappa is inaccessible if whenever x is in V_kappa, and y
is a subset of V_kappa, and there's a mapping from x onto y, y is also
in V_kappa. This definition is equivalent to the usual one in the
presence of AC. This definition comes from The Mathematical Society of
Japan's EDM 2. (EDM = Encyclopaedic Dictionary of Mathematics).
If kappa is inaccessible in this sense, it is weakly inaccessible, so
it is strongly inaccessible in L. So Con(ZF+IC)=>Con(ZFL+IC)
--== Sent via Deja.com http://www.deja.com/ ==--
---Share what you know. Learn what you don't.---
Mike Oliver
Mike McCarty wrote:
> Did you read my post? I'm confused by your question. You state
>
> If A is logically necessary, then that falsifies ~A.
>
> I know of no such requirement on formal systems, so I asked whether you
> were assuming consistency.
Where do you see a formal system? I didn't mention any formal system
at all. That was why I couldn't figure out what I was supposed to
be assuming the consistency *of*.
Mike Oliver
Mike McCarty wrote:
>
> In article <3743122C...@math.ucla.edu>,
> Mike Oliver <oli...@math.ucla.edu> wrote:
> )Fine, but that's not what's meant by a logically necessary proposition.
> )A proposition is logically necessary if it follows from logic alone,
> )with no assumptions at all.
>
> But, nothing follows from logic alone, with no assumptions at all.
Well, if that's true, then nothing is logically necessary. That's
OK by me; I'm generally quite skeptical of claims of necessity and
prefer to view dam' near everything as contingent.
However, traditionally, the so-called "logical axioms" are not considered
to be "assumptions" at all, but rather to be part of logic itself. So
for example "if p then p" is considered to be a logical truth which
does not need to be assumed, and is therefore logically necessary.
There is no well-defined and agreed-upon demarcation between what constitutes
pure logic and what requires making assumptions. But I think hardly
anyone would put quantum mechanics into pure logic (apparently not Petry
either; he just either didn't know or didn't use the standard meaning
for "logically necessary").
Mike Oliver
"james d. hunter" wrote:
>
> Mike Oliver wrote:
> >
> You don't think that just because ultra-finitists think that
> computers are discrete finite objects that they actually are, do you?
> Integers don't model computers very well either. There's a big
> difference between integers and sampled continuous objects.
The only sense that I can make out of this is that there are actual
messy physical properties going on inside a computer, so that even
when the abstract description of the computer says that a certain
bit ought to be a 1, in actual practice it might be a 0.
This, of course, is undeniable. The situation even has a
name. It's called a "hardware error".
Mike Oliver
I, Mike Oliver wrote:
> messy physical properties going on inside a computer, so that even
^^^^^^^^^^
For "properties" read "processes".
Mike McCarty
)>
)> If A is logically necessary, then that falsifies ~A.
)>
)> I know of no such requirement on formal systems, so I asked whether you
)> were assuming consistency.
)
)Where do you see a formal system? I didn't mention any formal system
)at all. That was why I couldn't figure out what I was supposed to
)be assuming the consistency *of*.
Ok, please define the meaning of
A is logically necessary.
I have encountered the term "logically necessary" only in the context of
formal logic. I suppose you must be using it in some other context or
way that I haven't encountered.
Mike McCarty
)> Mike Oliver <oli...@math.ucla.edu> wrote:
)> )Fine, but that's not what's meant by a logically necessary proposition.
)> )A proposition is logically necessary if it follows from logic alone,
)> )with no assumptions at all.
)>
)> But, nothing follows from logic alone, with no assumptions at all.
)
)Well, if that's true, then nothing is logically necessary. That's
)OK by me; I'm generally quite skeptical of claims of necessity and
)prefer to view dam' near everything as contingent.
Logic is a collection of rules governing deduction. It tells one how to
reason from a collection of premises to conclusions. It does not tell
you what the premises are. It is also the study of logical systems, and
how to build consistent sets of rules.
)However, traditionally, the so-called "logical axioms" are not considered
)to be "assumptions" at all, but rather to be part of logic itself. So
If you mean "logical axioms" as opposed to the "non-logical axioms" of a
first order formal system, then yes, they are usually *categorized*
differently.
)for example "if p then p" is considered to be a logical truth which
)does not need to be assumed, and is therefore logically necessary.
In this sense, we take Logic to mean "that which must be true in all
possible (one may substitute humanly conceivable) worlds". IOW, the
deductions which one makes are (assuming consistency) all tautologies.
)There is no well-defined and agreed-upon demarcation between what constitutes
)pure logic and what requires making assumptions. But I think hardly
I believe that there is.
)anyone would put quantum mechanics into pure logic (apparently not Petry
)either; he just either didn't know or didn't use the standard meaning
)for "logically necessary").
Who says that physics must be logical? Who says the physical world must
cater to the needs of human mental processes?
Mike McCarty
In article <3743438A...@math.ucla.edu>,
Mike Oliver <oli...@math.ucla.edu> wrote:
)
)
)"james d. hunter" wrote:
)>
)> Mike Oliver wrote:
)> >
)> You don't think that just because ultra-finitists think that
)> computers are discrete finite objects that they actually are, do you?
No, computer scientist *define* computers to be finite objects, hence
they are. If you mean something other than discrete finite objects, then
fine, just don't call them computers.
)> Integers don't model computers very well either. There's a big
)> difference between integers and sampled continuous objects.
)
)The only sense that I can make out of this is that there are actual
)messy physical properties going on inside a computer, so that even
)when the abstract description of the computer says that a certain
)bit ought to be a 1, in actual practice it might be a 0.
)
)This, of course, is undeniable. The situation even has a
)name. It's called a "hardware error".
)
)--
)Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love")
)
)Finger for PGP public key, or visit http://www.math.ucla.edu/~oliver.
)1500 bits, fingerprint AE AE 4F F8 EA EA A6 FB E9 36 5F 9E EA D0 F8 B9
james d. hunter
>
> I would have responded in a different way.
>
> In article <3743438A...@math.ucla.edu>,
> Mike Oliver <oli...@math.ucla.edu> wrote:
> )
> )
> )"james d. hunter" wrote:
> )>
> )> Mike Oliver wrote:
> )> >
> )> You don't think that just because ultra-finitists think that
> )> computers are discrete finite objects that they actually are, do you?
>
> No, computer scientist *define* computers to be finite objects, hence
> they are. If you mean something other than discrete finite objects, then
> fine, just don't call them computers.
Well, that definition is OK with me. Now we just have to
find a computer scientist that owns one of them. Just
like we have to find an economist who knows what
money, energy, or entropy is.
Mike McCarty
)> I don't folllow this at all. Computers are only capable of dealing with
)> integers. I speak as one who has written floating point packages, BTW.
)> In particular, I have implemented the math library for the C language.
)
) That's not really true. Computers deal with integrals, derivatives,
) trig, and and even evil Cantorian math just find. I think I have
) read sci.math more than once and have seen mathematicians
) referring to Mathmatica, Maple, etc.
No, they do not. They deal with arrays of integers used to represent
strings of information, and perform operations on the elements of these
arrays to derive new arrays. These new arrays when displayed, are
interpretable as being "symbolic math". I have written programs to do
exactly what you describe, so I speak from knowledge.
It is becoming rapidly apparent that you do not know what a computer is,
nor what it does, nor how it does it.
) > Computers store integers with a maximum size, not even all integers.
)All
) > computer models eventually devolve into operations on integers.
) >
) > So what are you trying to say?
)
) I'm saying that if you're doing number crunching that's true.
) Also I'm saying that Usenet posters rarely use their computers
) for number crunching.
No matter what you use a computer for, it boils down to operations on
integers, since that's all they can store and manipulate. Letters are
stored as small (usually 8 bit) integers, for example. Any operations on
strings are performed on arrays of these small integers.
Mike McCarty
Staff Engineer
Alcatel USA
james d. hunter
>
> "james d. hunter" wrote:
> >
> > Mike Oliver wrote:
> > >
>
That happens frequently, but it's almost always a software problem.
> This, of course, is undeniable. The situation even has a
> name. It's called a "hardware error".
I know what they're called. Usually the bigger problem
is deciding if a computer does or does not have a
hardware problem.
james d. hunter
>
> In article <3743293A...@jhuapl.edu>,
> james d. hunter <jim.h...@spam.free.jhuapl.edu.> wrote:
> )Mike McCarty wrote:
>
> )> I don't folllow this at all. Computers are only capable of dealing with
> )> integers. I speak as one who has written floating point packages, BTW.
> )> In particular, I have implemented the math library for the C language.
> )
> ) That's not really true. Computers deal with integrals, derivatives,
> ) trig, and and even evil Cantorian math just find. I think I have
> ) read sci.math more than once and have seen mathematicians
> ) referring to Mathmatica, Maple, etc.
>
> No, they do not. They deal with arrays of integers used to represent
> strings of information, and perform operations on the elements of these
> arrays to derive new arrays. These new arrays when displayed, are
> interpretable as being "symbolic math". I have written programs to do
> exactly what you describe, so I speak from knowledge.
Real computers do -nothing- with integers.
You're talking about a finite state machines, which are different
than computers.
Mike Oliver
> In this sense, we take Logic to mean "that which must be true in all
> possible (one may substitute humanly conceivable) worlds".
Well, almost. That's the usual definition of "necessary," not "logically
necessary". I think of "logically necessary" as being, at least potentially,
a stronger notion; it's imaginable that something could be true in all
possible worlds, but not for any reason that can be deduced by pure logic.
But I don't propose to give an example of something that's necessary but
not logically necessary -- I just note that it's not logically necessary
that everything that's necessary is logically necessary :-)
> IOW, the
> deductions which one makes are (assuming consistency) all tautologies.
Well, the word "tautology" is usually used for things that are *propositionally*
valid; i.e. you can see that they're always true merely by exhausting the
truth table. Not everything that's considered a logical truth is of this
form. For example, let P and Q be one-place predicates, and consider
the proposition:
[(all x)(Px --> Qx) & (exists x)( Px )] --> (exists x)( Qx )
This is a validity of the lower predicate calculus and is considered
by most to be logically necessary, but it's not a tautology in the usual
sense.
> Who says that physics must be logical? Who says the physical world must
> cater to the needs of human mental processes?
Not I.
David Petry
Mike Oliver wrote in message <3743122C...@math.ucla.edu>...
>David Petry wrote:
>> Mike Oliver wrote in message <37423057...@math.ucla.edu>...
>> >You think *quantum*mechanics* is logically necessary?
>> Of course not. But given an axiomatic foundation for quantum
>> mechanics, there are logically necessary implications.
>Fine, but that's not what's meant by a logically necessary proposition.
>A proposition is logically necessary if it follows from logic alone,
>with no assumptions at all.
I considered the possibility that that is what you meant, but I had
trouble believing that such a notion could be deemed relevant to a
discussion of falsifiability and possibility. So I guess I'm lost in
this sub-thread.
However, my original assertion still stands, as far as I can see.
It doesn't make sense to assert that something is "possible", if
that something doesn't have observable implications.
Bennett Standeven
On Tue, 18 May 1999, Mike Oliver wrote:
> Bennett Standeven wrote:
>
> > Actually, there is a good reason to think that numbers which are too large
> > to be written down can still exist:
>
> I'm certainly not claiming that they *don't* exist. That would be silly.
> I'm only questioning the notion that we can have some sort of a priori
> knowledge of their logically necessary existence that is different in kind
> from the justifications for analysis and set theory.
A priori? Then I guess an empirical argument wouldn't help. Do you
consider it to be logically necessary that, say, 8 exists?
>
> (Also, the examples I was talking about are quite a bit larger than merely
> "too large to be written down" if by that you mean in decimal notation.)
>
Actually, I meant tally notation (to represent the process of counting up
to the number). Of course, decimal notation allows numbers much larger
than any physical quantity to be represented (such as 10^1000, for
example).
Mike Oliver
David Petry wrote:
> However, my original assertion still stands, as far as I can see.
> It doesn't make sense to assert that something is "possible", if
> that something doesn't have observable implications.
Do you think it can make sense at least to assert "A is not necessary"
even if ~A is not falsifiable?
If not, then you seem to be establishing a rule whereby if you can
establish that ~A is not falsifiable, then you automatically conclude
that A is necessary. I'd love to see how *that* plays out.
But if so, then by most understandings of modal logic you seem
to be allowing the conclusion "~A is possible" since ~A is possible
_if_and_only_if_ A is not necessary.
Jeremy Boden
<david...@mindspring.com> writes
>
>Mike Oliver wrote in message <3743122C...@math.ucla.edu>...
>>David Petry wrote:
>>> Mike Oliver wrote in message <37423057...@math.ucla.edu>...
>>> >You think *quantum*mechanics* is logically necessary?
>>> Of course not. But given an axiomatic foundation for quantum
>>> mechanics, there are logically necessary implications.
>>Fine, but that's not what's meant by a logically necessary proposition.
>>A proposition is logically necessary if it follows from logic alone,
>>with no assumptions at all.
>
>I considered the possibility that that is what you meant, but I had
>trouble believing that such a notion could be deemed relevant to a
>discussion of falsifiability and possibility. So I guess I'm lost in
>this sub-thread.
>
>It doesn't make sense to assert that something is "possible", if
>that something doesn't have observable implications.
>
statement?
Since you aren't able to carry out any "observations" on me - at least I
hope not! - then you should assert that I can't possibly exist.
Personally, I don't find this argument very convincing.
Isn't it a bit contradictory to talk about "logical necessity"?
If "logical necessity" is to be a feature of logic or metaphysics or
whatever then don't we need a definition of logical necessity - in terms
of logic?
We could then have fun having circular arguments about whether our
particular definition of logical necessity is *logically necessary*
--
Jeremy Boden mailto:jer...@jboden.demon.co.uk
Jeremy Boden
>"james d. hunter" wrote:
>> Mike Oliver wrote:
>> >
>> You don't think that just because ultra-finitists think that
>> computers are discrete finite objects that they actually are, do you?
>> Integers don't model computers very well either. There's a big
>> difference between integers and sampled continuous objects.
>
>The only sense that I can make out of this is that there are actual
>messy physical properties going on inside a computer, so that even
>when the abstract description of the computer says that a certain
>bit ought to be a 1, in actual practice it might be a 0.
>
>name. It's called a "hardware error".
>Love")
>
Assuming that the computer is a digital computer, observe that all
digitisations are essentially integers. A digital computer models all
continuous phenomena by collections of integers.
james d. hunter
The computer doesn't do the modelling, the modeller does.
The 'essential integers' of course aren't integers.
Mike McCarty
)> possible (one may substitute humanly conceivable) worlds".
)
)Well, almost. That's the usual definition of "necessary," not "logically
)necessary". I think of "logically necessary" as being, at least potentially,
)a stronger notion; it's imaginable that something could be true in all
)possible worlds, but not for any reason that can be deduced by pure logic.
)But I don't propose to give an example of something that's necessary but
)not logically necessary -- I just note that it's not logically necessary
)that everything that's necessary is logically necessary :-)
)
)> IOW, the
)> deductions which one makes are (assuming consistency) all tautologies.
)
)Well, the word "tautology" is usually used for things that are *propositionally*
)valid; i.e. you can see that they're always true merely by exhausting the
)truth table. Not everything that's considered a logical truth is of this
)form. For example, let P and Q be one-place predicates, and consider
)the proposition:
)
) [(all x)(Px --> Qx) & (exists x)( Px )] --> (exists x)( Qx )
)
)This is a validity of the lower predicate calculus and is considered
)by most to be logically necessary, but it's not a tautology in the usual
)sense.
I dunno what definition you use for tautology. But what you stated
above is a tautology by mine.
)> Who says that physics must be logical? Who says the physical world must
)> cater to the needs of human mental processes?
)
)Not I.
Ok. Then you shouldn't bring in physical concepts to your arguments.
Mike McCarty
)>
)> In article <3743438A...@math.ucla.edu>,)> )
)> )"james d. hunter" wrote:
)> )>
)> )> Mike Oliver wrote:
)> )> >
)> )> computers are discrete finite objects that they actually are, do you?
)>
)> No, computer scientist *define* computers to be finite objects, hence
)> they are. If you mean something other than discrete finite objects, then
)> fine, just don't call them computers.
)
) Well, that definition is OK with me. Now we just have to
) find a computer scientist that owns one of them. Just
) like we have to find an economist who knows what
) money, energy, or entropy is.
So far, in this thread, I haven't seen a need to find anyone who owns
anything. What I've seen is epistomology being argued. And for that, no
one needs any machines of any sort. But when the argument begins to
involve definitions of words which have commonly accepted definitions,
then it is important not to get mired in arguments where people are
using the same words, but not meaning the same thing.
Mike McCarty
)> james d. hunter <jim.h...@spam.free.jhuapl.edu.> wrote:
)> )Mike McCarty wrote:
)>
)> )> integers. I speak as one who has written floating point packages, BTW.
)> )> In particular, I have implemented the math library for the C language.
)> )
)> ) That's not really true. Computers deal with integrals, derivatives,
)> ) read sci.math more than once and have seen mathematicians
)> ) referring to Mathmatica, Maple, etc.
)>
)> No, they do not. They deal with arrays of integers used to represent
)> strings of information, and perform operations on the elements of these
)> arrays to derive new arrays. These new arrays when displayed, are
)> interpretable as being "symbolic math". I have written programs to do
)> exactly what you describe, so I speak from knowledge.
)
) Real computers do -nothing- with integers.
) You're talking about a finite state machines, which are different
) than computers.
While we are being ultra picky, the correct grammar is
different from
not
different than
Pray tell, how many computers have you designed? How many have you
constructed? How many have you specified?
In fact, please give a definition of a computer. (I mean a *real*
computer.
Please do not use the word "machine", unless you intend that your
computer satisfy the definition of a machine. The definition of
machine, as you may recall, is any combination of simple machines. The
simple machines are
lever
inclined plane
etc.
I forget the exact list, but I recall that there were 5 or 7 of them on
the list.