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| Butch Malahide... |
 Posted: Sun Nov 01, 2009 10:01 pm |
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On Nov 2, 12:29 am, Bill Taylor <w.tay... at (no spam) math.canterbury.ac.nz>
wrote:
[quote]stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
The reason I have little sympathy for this point of view is
that I don't see that traditional mathematics ever makes the
assumption that everything has an explicit name or description.
OC it doesn't. It never needed to. Before Cantor everything DID.
[/quote]
What objects of Euclid's geometry have explicit names or descriptions? |
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| Daryl McCullough... |
| Posted: Mon Nov 02, 2009 6:42 am |
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Bill Taylor says...
[quote]
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
The reason I have little sympathy for this point of view is
that I don't see that traditional mathematics ever makes the
assumption that everything has an explicit name or description.
OC it doesn't. It never needed to. Before Cantor everything DID.
[/quote]
That isn't true. Real numbers did not have names before Cantor.
Points and lines of Euclidean geometry didn't have names.
[quote]It's an amazing accomplishment of ZF and the concept of the
cumulative hierarchy that we can come so close to an explicit
taxonomy of all the objects that will ever come into play in
any mathematical argument, but I don't see that the existence
of such a standard model has any relevance to what mathematicians
*do* with sets.
I don't follow this sentence at all, sorry.
[/quote]
I'm saying that mathematicians do *not* need to assume that
there is a way to construct, or define, every mathematical
object. It is enough to be able to characterize them through
axioms. To do topology, you start off with "There is a set
with such and such properties", and you work out the consequences.
You don't need to ask: "What is the name of that set?" "What
are the names of its elements?"
[quote]Instead, they start with certain basic sets
that are *not* defined,
In pure set theory they all are. They are the empty set.
[/quote]
That's what I just said--in set theory, they *do* assume that
all sets can be built up from the empty set (but not through
applying definitions, though), but most of the *use* of
sets does not rely on their being built up this way.
[quote]To me, the astounding claim about set theory is that, whatever
your basic objects are --- paths of particles, wavefunctions,
That is all applied mat, applied set theory.
We are (I thought) speaking of pure set theory, as applied to
(at most) other pure math objects like N & R.
[/quote]
Yes. Pure mathematics should (in my opinion) be a *superset* of
applied mathematics. Whatever our foundation of mathematics should
be, it should at least include the ability to model the sorts of
problems that show up in physics and applied math. In applied math,
there is no assumption that every object is definable. There is
no reason to.
[quote]{ x | phi(x) }
should be taken as a model or paradigm for what a set actually IS.
I sort of agree with this,
Well! There's something.
but I don't feel that it is
appropriate to assume that the only possible conditions
phi are ones that are definable using pure set theory.
Set theory as applied to physical objects is completely trivial.
[/quote]
No, it's not.
[quote]If we are talking about sets of naturals, then "the set
of phone numbers of mathematicians" is a perfectly good
No, it's perfectly bad!
In a sense, the insistence that all mathematical objects
have an explicit description is more radical than constructivism.
It's NOT an "insistence", as you call it, but merely an observation
of standard (pre-Cantor) math.
[/quote]
It was *not* at all true, pre-Cantor. Where are you getting that
from? Before Cantor, there was no asumption that all Euclidean
points and lines had names. There was no assumption that every
real number had a finite description. The *possibility* of giving
every mathematical object a unique, finite description until after
set theory was being developed.
[quote]And I feel that set theory should
be conducted along the lines of those traditions.
[/quote]
There was no such tradition. I think you have things almost completely
backwards. Before attempts to formalize set theory, there was no attempt
to say that every mathematical object can be be given a unique description.
At least not since the discovery of transcendental reals.
--
Daryl McCullough
Ithaca, NY |
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| Daryl McCullough... |
| Posted: Mon Nov 02, 2009 6:46 am |
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Bill Taylor says...
[quote]
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
AC produces sets, by fiat of existence, ...
... - all these things simply don't exist,
in that no explicit set can be named, with those properties.
There is something a little strange about this philosophical
position: If you assume that all sets have an explicit
"membership criterion" Phi(x), then choice is *automatically*
true. You can well-order sets by their defining formulas.
AHA! A very astute observation. This is the kind of thing
that had me bothered for a long time. And it is all bound up
with the UNDEFINABILITY of "DEFINABILITY",
as I have noted before.
All sets will be definable, but this notion cannot be delimited
in advance. So e.g. once you have a bounded set of reals,
it must have a least upper bound, BUT this lub will not be
definable at the same level - if the reals in the set are
definable at level "a" and below, then the lub will be most
likely be at level a+1. There is no end to the levels, they
can be notated exactly as any initial segment of the recursive
ordinals. But not of the whole set itself - that would be
declaring existences BY FIAT again. omega_1^CK itself is
purely a creature of ZF, with its all-encompassing power set
operation, (without which it cannot be defined.)
[/quote]
I think it's still true that the axiom of choice would hold
for such definable sets. If a set of nonempty sets is definable
at some level alpha, then there will be a definable choice
function at that level (or a few levels higher, maybe). The
only way to *falsify* the axiom of choice is to have a set
whose elements are *not* definable (at any level). At least,
that's the way it seems to me.
--
Daryl McCullough
Ithaca, NY |
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| Herman Jurjus... |
| Posted: Mon Nov 02, 2009 6:55 am |
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Bill Taylor wrote:
[quote]Herman Jurjus <hjm... at (no spam) hetnet.nl> wrote:
What are your /reasons/ for rejecting AC,
Well! That would be a whole thread in itself.
And I don't wish to bore the old hands YET AGAIN.
So I'll try to be brief.
AC produces sets, by fiat of existence, which simply DON'T EXIST!
Non-measurable sets in R,free ultrafilters on N, partitions of R^3
into non-parallel lines - all these things simply don't exist,
in that no explicit set can be named, with those properties.
[/quote]
And AD doesn't produce sets (strategies, certain large cardinals) that
also "simply don't exist" in this same sense?
--
Cheers,
Herman Jurjus |
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| Bill Taylor... |
| Posted: Mon Nov 02, 2009 5:59 pm |
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utch Malahide <fred.gal... at (no spam) gmail.com> wrote:
[quote]What objects of Euclid's geometry have explicit names or descriptions?
[/quote]
All of them. |
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| Bill Taylor... |
| Posted: Mon Nov 02, 2009 6:01 pm |
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On Nov 3, 12:46 am, stevendaryl3... at (no spam) yahoo.com (Daryl McCullough)
wrote:
[quote]Bill Taylor says...
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
AC produces sets, by fiat of existence, ...
... - all these things simply don't exist,
in that no explicit set can be named, with those properties.
There is something a little strange about this philosophical
position: If you assume that all sets have an explicit
"membership criterion" Phi(x), then choice is *automatically*
true. You can well-order sets by their defining formulas.
AHA! A very astute observation. This is the kind of thing
that had me bothered for a long time. And it is all bound up
with the UNDEFINABILITY of "DEFINABILITY",
as I have noted before.
All sets will be definable, but this notion cannot be delimited
in advance. So e.g. once you have a bounded set of reals,
it must have a least upper bound, BUT this lub will not be
definable at the same level - if the reals in the set are
definable at level "a" and below, then the lub will be most
likely be at level a+1. There is no end to the levels, they
can be notated exactly as any initial segment of the recursive
ordinals. But not of the whole set itself - that would be
declaring existences BY FIAT again. omega_1^CK itself is
purely a creature of ZF, with its all-encompassing power set
operation, (without which it cannot be defined.)
I think it's still true that the axiom of choice would hold
for such definable sets. If a set of nonempty sets is definable
at some level alpha, then there will be a definable choice
function at that level (or a few levels higher, maybe). The
only way to *falsify* the axiom of choice is to have a set
whose elements are *not* definable (at any level). At least,
that's the way it seems to me.
--
Daryl McCullough
Ithaca, NY[/quote] |
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| Bill Taylor... |
| Posted: Mon Nov 02, 2009 6:07 pm |
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O stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
[quote]I think it's still true that the axiom of choice would hold
for such definable sets.
[/quote]
It would, if you could gather enough together to talk about
them all simultaneously, but you can't.
[quote]The only way to *falsify* the axiom of choice is to have a set
whose elements are *not* definable (at any level). At least,
that's the way it seems to me.
[/quote]
Well, were not trying to falsify it formally, in any sense; that
would probably be impossible. We're just noticing that it's
not true, of the "sets" that people use it on. The real problem,
OC, is not AC itself, but the attempt to talk about
"all real numbers at once" - this is where ZF breaks down.
However, we are stuck with ZF for various reasons, and that
is OK; but if we have to swallow Powerset, we should not go
the whole hog, and throw good money after bad, and swallow
AC itself - it just ISN"T true, of sets of subsets of reals of
the type it is used for. Swallowing the C after the ZF will
just painfully stick in our craws.
-- Bronchial Bill |
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| Bill Taylor... |
| Posted: Mon Nov 02, 2009 6:13 pm |
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Herman Jurjus <hjm... at (no spam) hetnet.nl> wrote:
[quote]AC produces sets, by fiat of existence, which simply DON'T EXIST!
Non-measurable sets in R,free ultrafilters on N, partitions of R^3
into non-parallel lines - all these things simply don't exist,
in that no explicit set can be named, with those properties.
And AD doesn't produce sets (strategies, certain large cardinals) that
also "simply don't exist" in this same sense?
[/quote]
Well, it's odd, isn't it, but it doesn't seem to!?
I would be delighted to see a counterexample.
That is, can you produce a well-defined set, A, such that
there ISN'T, on the face of it, any clear winning strategy for game A?
As I say, I'd LOVE to see one. I haven't been able to find one.
-- Baffled Bill |
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| David C. Ullrich... |
| Posted: Tue Nov 03, 2009 6:44 am |
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On Mon, 2 Nov 2009 20:13:15 -0800 (PST), Bill Taylor
<w.taylor at (no spam) math.canterbury.ac.nz> wrote:
[quote]Herman Jurjus <hjm... at (no spam) hetnet.nl> wrote:
AC produces sets, by fiat of existence, which simply DON'T EXIST!
Non-measurable sets in R,free ultrafilters on N, partitions of R^3
into non-parallel lines - all these things simply don't exist,
in that no explicit set can be named, with those properties.
And AD doesn't produce sets (strategies, certain large cardinals) that
also "simply don't exist" in this same sense?
Well, it's odd, isn't it, but it doesn't seem to!?
I would be delighted to see a counterexample.
[/quote]
This is your lucky day. AD implies the existence of a model of ZF.
That model "simply doesn't exist", if we take that to mean
"no explicit set can be named which is a model of ZF" as above.
[quote]That is, can you produce a well-defined set, A, such that
there ISN'T, on the face of it, any clear winning strategy for game A?
As I say, I'd LOVE to see one. I haven't been able to find one.
-- Baffled Bill
[/quote]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| Daryl McCullough... |
| Posted: Tue Nov 03, 2009 6:49 am |
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Bill Taylor says...
[quote]
O stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
I think it's still true that the axiom of choice would hold
for such definable sets.
It would, if you could gather enough together to talk about
them all simultaneously, but you can't.
[/quote]
I'm not sure what that means. In the case of reals, presumably
if every real is definable at some level, then the collection
of all such levels has a supremum, alpha. In which case, there
would be a definable choice function on reals at level alpha+1.
The alternative, it seems to me, is to say that there is no *set*
of all reals (because every level produces new reals, and the
levels are never completed).
[quote]The only way to *falsify* the axiom of choice is to have a set
whose elements are *not* definable (at any level). At least,
that's the way it seems to me.
Well, were not trying to falsify it formally, in any sense; that
would probably be impossible. We're just noticing that it's
not true, of the "sets" that people use it on.
[/quote]
I'm noting just the opposite: *if* people only use definable
sets, *then* there is a definable well-ordering of those sets.
We can well-order the reals as follows:
Let level(r) = the first level at which r becomes definable.
Then we can say r1 < r2 if level(r1) < level(r2) or if
level(r1) = level(r2) and the formula defining r1 is
less complicated than the formula defining r2.
[quote]The real problem, OC, is not AC itself, but the attempt to talk about
"all real numbers at once" - this is where ZF breaks down.
[/quote]
Then you *are* saying that the reals don't form a set. Well,
then AC is definitely not the problem, separation is. Whenever
we talk about the set of all reals r such that Phi(r), that's
an illegitimate operation if the reals don't form a set. It
would be a proper class, perhaps.
[quote]However, we are stuck with ZF for various reasons, and that
is OK; but if we have to swallow Powerset, we should not go
the whole hog, and throw good money after bad, and swallow
AC itself - it just ISN"T true, of sets of subsets of reals of
the type it is used for. Swallowing the C after the ZF will
just painfully stick in our craws.
[/quote]
That doesn't make a bit of sense to me. You are saying that AC
s bad because it produces undefinable sets of reals. But it
doesn't *unless* there are already undefinable reals. AC would
not be the *source* of the undefinability. It's the innocent
bystander.
--
Daryl McCullough
Ithaca, NY |
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| Daryl McCullough... |
| Posted: Tue Nov 03, 2009 6:51 am |
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Bill Taylor says...
[quote]utch Malahide <fred.gal... at (no spam) gmail.com> wrote:
What objects of Euclid's geometry have explicit names or descriptions?
All of them.
[/quote]
No, they do not. Lines and points don't have names in Euclidean geometry.
--
Daryl McCullough
Ithaca, NY |
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| Jesse F. Hughes... |
| Posted: Tue Nov 03, 2009 8:04 am |
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James Burns <burns.87 at (no spam) osu.edu> writes:
[quote]It might be interesting to point this out to
the next person to prove that the reals are countable --
that /if/ the reals are countable, then Banach-Tarski
is unavoidable.
[/quote]
You really think that such persons are likely to believe that's a
consequence?
--
Jesse F. Hughes
"Well, I guess that's what a teacher from Oklahoma State University
considers proper as Ullrich has said it, and he is, in fact, a teacher
at Oklahoma State University." -- James S. Harris presents a syllogism |
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| Bill Taylor... |
| Posted: Tue Nov 03, 2009 6:28 pm |
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stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
[quote]The reason I have little sympathy for this point of view is
that I don't see that traditional mathematics ever makes the
assumption that everything has an explicit name or description.
OC it doesn't. It never needed to. Before Cantor everything DID.
That isn't true. Real numbers did not have names before Cantor.
[/quote]
Oh heavens, of course they did!
All the integers and ratiobnals had them, and also all the reals
we ever use in applications, like pi, e, sin^(-1)(2/7) and so on.
[quote]Points and lines of Euclidean geometry didn't have names.
[/quote]
Oh heavens, of course they did!
I'll ignore the trivial fact that all Euclid's diagrams had letters
Wed Nov 4 17:27:50 NZDT 2009
/users/math/wft13/News> m r
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
[quote]The reason I have little sympathy for this point of view is
that I don't see that traditional mathematics ever makes the
assumption that everything has an explicit name or description.
OC it doesn't. It never needed to. Before Cantor everything DID.
That isn't true. Real numbers did not have names before Cantor.
[/quote]
Oh heavens, of course they did!
All the integers and ratiobnals had them, and also all the reals
we ever use in applications, like pi, e, sin^(-1)(2/7) and so on.
[quote]Points and lines of Euclidean geometry didn't have names.
[/quote]
Oh heavens, of course they did!
I'll ignore the trivial fact that all Euclid's diagrams had letters
on the points and lines, and go straight to Descartes. No-one doubts
he was doing Euclidean geometry, and extending it, and all his
points were real number pairs, see just above.
[quote]To do topology, you start off with "There is a set
with such and such properties", and you work out the consequences.
You don't need to ask: "What is the name of that set?" "What
are the names of its elements?"
[/quote]
Oh yes, that's true. Abstract math like topology and group theory
aand so on, start off with no names; but as soon as you get to
particular examples, they do.
[quote]Yes. Pure mathematics should (in my opinion) be a *superset* of
applied mathematics. Whatever our foundation of mathematics should
be, it should at least include the ability to model the sorts of
problems that show up in physics and applied math.
[/quote]
OUCH!! Now I recall, that we have had differences of opinion
before, on this matter. In effect, it involves which takes
"supremacy", physics or math. You claim the former,
I claim the latter.
[quote]In applied math,
there is no assumption that every object is definable. There is
no reason to.
[/quote]
EXACTLY! They simply *are* all definable, so thus, there is no
reason to, no reason to make any particular assumption about it!
And I extend this view to set theory, sets ought to be definable,
so there is no need to, (it may well be counter-productive to)
assert this as a formal axiom. It just IS so. We only then need
to avoid implicitly (or explicitly, together with claims of reality)
other axioms that declare undefinable sets.
[quote]Set theory as applied to physical objects is completely trivial.
No, it's not.
[/quote]
Again, the primacy of math over physics. The set theory used
in science is utterly trivial compared to that developed in math.
[quote]It was *not* at all true, pre-Cantor. Where are you getting that
from? Before Cantor, there was no asumption that all Euclidean
points and lines had names. There was no assumption that every
real number had a finite description.
[/quote]
There was no overt assumption, because it was obviosul;y true.
See my remarks above about Z, Q, coordinate geometry etc).
It was only when peole started to come up withh undefinable
objects, (a la Cantor/Zermelo), that others began to notice
this profound departure.
[quote]At least not since the discovery of transcendental reals.
[/quote]
Transcendence has no influence - pi is still perfectly definable.
-- Battling Bill |
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| Bill Taylor... |
| Posted: Tue Nov 03, 2009 7:01 pm |
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stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
[quote]You gave the answer yourself,
in terms of infinte-depth quantifiers
But I really don't have an intuition about what infinitely many
alternations of quantifiers *mean*,
Of course you do!
No, I don't.
[/quote]
Well, as I say, we are clearly using the above words
with differing meanings.
[quote]So the whole concept seems insufficiently "tied down".
[/quote]
Indeed so. If it had been tied down, we would all be using them now.
But such formalization has never been completed. (AFAIK)
[quote]Might mean the infinite conjunction of the following statements:
....
Another meaning is given in terms of strategy functions.
....
Then in terms of quasistrategies,
[/quote]
Your example translations are excellent attempts, but I suspect
none of them will capture the precise meaning of what we
would want an infinite-depth quantifer to mean, alas.
These formalizations are not quite "on", as some previous
examples have shown.
[quote]My point is that you simultaneously developed 2 different intuitions
about chess (or whatever): (1) Eventually, the game always ends.
(2) There is a winning strategy for one player or the player.
So I don't see how chess gives you any intuitions about games
that *don't* end.
[/quote]
I might well charge you with the same observations about
choosing from infinite numbers of sets!
[quote]I am curious, though, about the *form* that your intuition
about the existence of a winning strategy took. You really
thought in terms of *functions* from board positions to
your next move?
[/quote]
Yes, absolutely.
(OC I didn't mentally use the word "function" at that stage of
my early high school years,but it would amount to that.)
[quote]That's very remarkable to me, if that's
the case. When I played chess, I would pick a certain number
of moves to "think ahead" (and my pitiful brain could never
think more than two or three). On the basis of looking ahead,
I could decide that certain moves are losers. If a win
was sufficiently close at hand, then I could decide that
certain moves were winners. But in the general case, the
best I could say was that certain moves were neither
obviously winners, nor obviously losers.
[/quote]
OC this is all very true. But it is observation about
the *practical* side of actually playing chess, rather than the
theoretical side of what types of positions & moves there might be.
[quote]I'm curious as to how the idea that there *must* be a winning
move occurred to you. Presumably, if the win was too many
moves in the future, then you didn't actually *find* this
winning move.
[/quote]
Indeed not! Not being a terabyte genius!
[quote]So how did you figure that it must exist?
[/quote]
Like I say, it just seemed obvious, from considerations of
what actually happens in a game; any game.
[quote]Well, I would *not* say that my intuition about choice
has anything to do with generalization from finite sets.
[/quote]
Well, you obviously believe that, so I doubt anything
I could say will budge you. But it is an admission,
or claim, which tells its own story.
[quote]Now, suppose I have a box that contains a bunch of other
nonempty boxes (it doesn't matter how many). I can imagine
doing the following: for each box, I assign a helper
(that might mean infinitely many helpers,
[/quote]
OK OK, I get the picture. It is a mental picture we all
had at one time. I daresay I might be able to come up with
a picture of "infinitely many game players" moving
"infinitely fast" and so on, but what's the point.
[quote]There is nothing special about finiteness of the sets
in this intuition.
[/quote]
You hope.
[quote]Using infinite quantifiers, we can't (as far as I know) make
the distinction between "Player 1 has no winning strategy" and
"Player 2 has a winning strategy".
OCN! Because there isn't one,(or so I claim), for tie-free games.
You don't see that what your saying is circular?
[/quote]
Yes, I see your point. But I'm not trying to formally prove
anything here, I'm just giving a description of my intuitions,
in more than one way. It is VERY like the situation whereby
AC was (said to be) justified by reference to "combinatorial"
ideas about sets; which (as I noted) turned out to be the same
thing said in a different way. Possibly sociologically/didactically
useful, but probably no logical help.
-- Wittering William |
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| Daryl McCullough... |
| Posted: Wed Nov 04, 2009 7:10 am |
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Bill Taylor says...
[quote]
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
You gave the answer yourself,
in terms of infinte-depth quantifiers
But I really don't have an intuition about what infinitely many
alternations of quantifiers *mean*,
Of course you do!
No, I don't.
Well, as I say, we are clearly using the above words
with differing meanings.
[/quote]
Which words? "infinitely many quantifiers"?
[quote]So the whole concept seems insufficiently "tied down".
Indeed so.
[/quote]
To say that it's not "tied down" to me is synonymous
with "we don't know (precisely) what it means".
[quote]Might mean the infinite conjunction of the following statements:
....
Another meaning is given in terms of strategy functions.
....
Then in terms of quasistrategies,
Your example translations are excellent attempts, but I suspect
none of them will capture the precise meaning of what we
would want an infinite-depth quantifer to mean, alas.
[/quote]
Then what *DOES* it mean? That's why I say I don't
know what infinitely many quantifiers mean.
[quote]My point is that you simultaneously developed 2 different intuitions
about chess (or whatever): (1) Eventually, the game always ends.
(2) There is a winning strategy for one player or the player.
So I don't see how chess gives you any intuitions about games
that *don't* end.
I might well charge you with the same observations about
choosing from infinite numbers of sets!
[/quote]
I gave you an answer. My intuition about choice has nothing
to do with the number of elements in the set.
[quote]Well, I would *not* say that my intuition about choice
has anything to do with generalization from finite sets.
Well, you obviously believe that, so I doubt anything
I could say will budge you.
[/quote]
I'm talking about *my* intuitions, here. Yes, I doubt
that you know more about them than I do.
What is special about finite sets is that for such sets
a choice function is always *definable*. But definability
as a criterion for set existence is *your* thing, not mine.
[quote]Now, suppose I have a box that contains a bunch of other
nonempty boxes (it doesn't matter how many). I can imagine
doing the following: for each box, I assign a helper
(that might mean infinitely many helpers,
OK OK, I get the picture. It is a mental picture we all
had at one time. I daresay I might be able to come up with
a picture of "infinitely many game players" moving
"infinitely fast" and so on, but what's the point.
[/quote]
I certainly am willing to consider such godlike players, and
I don't see how the claim that all games are determined follows
from it. If your conclusion follows from such considerations,
then I would consider that a plausibility argument for AD. As
it is, I haven't heard anything that comes close to explaining
why you have such an intuition.
[quote]There is nothing special about finiteness of the sets
in this intuition.
You hope.
[/quote]
I *know*.
[quote]Using infinite quantifiers, we can't (as far as I know) make
the distinction between "Player 1 has no winning strategy" and
"Player 2 has a winning strategy".
OCN! Because there isn't one,(or so I claim), for tie-free games.
You don't see that what your saying is circular?
Yes, I see your point. But I'm not trying to formally prove
anything here, I'm just giving a description of my intuitions,
[/quote]
But you *haven't* given a description of your intuitions.
You have simply *stated* that such and such follows from
my intuitions, but why it follows, you haven't given any
clue.
[quote]in more than one way. It is VERY like the situation whereby
AC was (said to be) justified by reference to "combinatorial"
ideas about sets;
[/quote]
There is certainly a coherent notion of sets for which AC is
true. Are they the real sets, or are they somehow a "nonstandard"
notion of sets? I don't know how to answer that. I'm not sure
whether the question has a definite meaning.
--
Daryl McCullough
Ithaca, NY |
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