 |
|
| Science Forum Index » Logic Forum » Counterintuitions and the well-ordering theorem... |
|
Page 8 of 18 Goto page Previous 1, 2, 3 ... 7, 8, 9 ... 16, 17, 18 Next |
|
| Author |
Message |
| Bill Taylor... |
 Posted: Thu Oct 29, 2009 5:42 pm |
|
|
|
Guest
|
Aatu Koskensilta <aatu.koskensi... at (no spam) uta.fi> wrote:
[quote]Why shouldn't we spend extraordinary amounts of energy on things we
don't take too seriously? Surely it's a dismal, truly teeth-gnashing
inducing idea, a revolting notion fit only to be rejected in a violent
explosion of metaphorical vomit, that we only spend considerable length
of time on stuff that we take very very seriously.
[/quote]
I agree totally with this!
As Bertie Russ said, "The time you enjoy wasting is not wasted time!"
[quote]I also find that I actually learn something of these electronic exchanges,
coming out of them illuminated with new wisdom ... ... not
necessarily anything of mathematical nature, but something of how
people, myself included, react to this or that mode of argumentation,
this or that way of putting this or that, this or that line of thought,
this or that level of formality, gathering in the process valuable
information about whether this or that way of putting an idea is
generally intelligible, gaining for myself many a curious factoids about
this or that English idiom and its use and abuse, what have you. Or this
or that. Bits and pieces, follies and human insight. That's what I reap
from these virtual encounters.
[/quote]
Exactly so! Very well put! Agree 100%. A keeper.
[quote]It is also my hope that, in spite of my sometimes needlessly aggressive
debating style, and peculiar and failed attempts at humour, those with
whom I battle wits leave these Usenetical battlefields slightly
improved, with a perspective on life just a whit expanded, ...
from what it was before.
[/quote]
I'm sure we all earnestly hope that.
[quote]It is customary in many newsgroups to include in otherwise off-topic
posts a nugget of topicality. Here goes:
[/quote]
And here's mine: In the absence of AC, CH can be seen to be
both true and false, depending on how it is worded:
1) There is a subset of R, of cardinality strictly between N and R.
2) There is a function on subsets of R whose values are bijections
between the argument and one of N_k, N, or R.
AC |- 1 == 2
But, definitionially speaking, 1 and 2 are both clearly false.
And they might both be false in models of ZF.
-- Troublesome Taylor |
|
|
| Back to top |
|
|
|
| Bill Taylor... |
| Posted: Thu Oct 29, 2009 5:49 pm |
|
|
|
Guest
|
Aatu Koskensilta <aatu.koskensi... at (no spam) uta.fi> wrote:
[quote]Rupert <rupertmccal... at (no spam) yahoo.com> writes:
When I first read about AD I thought it was an interesting hypothesis,
but I never had any feeling that it was intuitively plausible.
[/quote]
Different strokes, different folks. On my first hearing of it, it
struck
me as *obviously* true, and a game-theoretic necessity.
[quote]On the other hand as soon as I encountered AC
I was completely convinced that it was true.
[/quote]
As was I. Like believing in God. It was only much, much later that
I realized how I'd been cleverly hoodwinked!
[quote]Our agreement on these matters is most touching. Let's hug!
[/quote]
Please, there are children on this newsgroup!!
[quote]I soon learned determinacy was intended to apply to sets that behave,
sets that are in some intelligible manner built of basic, familiar and
cozy base by means of (transfinitely iterated) operations that make
(comfortable, familiar, cozy, mathematical) sense.
[/quote]
That must have been very cheering for you, like finding that heaven
was only intended for right-thinking folks, and not very bad boys!
To me, though, the nature of the sets it applies to is wholly
irrelevant.
The Lord welcomes even the most grievous sinners!
-- Welcoming William |
|
|
| Back to top |
|
|
|
| Bill Taylor... |
| Posted: Thu Oct 29, 2009 5:52 pm |
|
|
|
Guest
|
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
[quote]It's a bit like with the Jordan curve theorem: it's nice that we can
prove it, but had our definitions been such that it had come out as
false, we would only have concluded that our definitions needed
revision, not that the Jordan curve theorem is false.
Yes, I agree completely in this case. I feel that if it is false,
then it means we have defined "continuous curve" incorrectly.
[/quote]
We all seeem to agree here.
[quote]As a matter of fact, I think it is provably *false* for some natural
ways of formulating continuity.
[/quote]
Whoa! That IS interesting! Would you do us all a kindness, Daryl,
and teach us about some of these natural alternatives.
Perspiring minds want to know!
-- Breathless Bill |
|
|
| Back to top |
|
|
|
| Bill Taylor... |
| Posted: Thu Oct 29, 2009 6:18 pm |
|
|
|
Guest
|
OK, these are my last 2 responses for today, bringing me up to date
for posts in this thread that appeared on or before yesterday.
Apologies if I'm repeating stuff, or if I'm making points that
have already been answered.
Aatu Koskensilta <aatu.koskensi... at (no spam) uta.fi> wrote:
[quote]But: either player 1 has a winning strategy or he hasn't. [etc]
I think it's a good idea to explicitly introduce quantifiers,
[/quote]
I agree! Wholeheartedly. It was the consideration of quantifiers
that convinced me, almost immediately, that AD just had to be true.
Of course, I like to consider sequential games *sequentially*,
and that means being able to extend quantifiers to omega depth,
as Daryl has already noted. Thanks Daryl (!) - you saved me
a lot of preliminary work. So here goes:-
Either
Ea Ab Ec Ad Ee Af.... <a, b, c, d, e, f ... > in WIN-1 (*)
is true, or it is false. This is the LEM that was adduced earlier.
If * is true, it says precisely that player-1 has a winning
strategy, now given sequentially (as seems most natural),
rather than simultaneously (as required by FOL-ZF)).
If it is false, then it is true that
~ Ea Ab Ec Ad Ee Af.... <a, b, c, d, e, f ... > in WIN-1;
and this is DeMorgan transformable
(quite legitimately by initial segements!) into...
Aa Eb Ac Ed Ae Ef.... <a, b, c, d, e, f ... > in WIN-1^C;
....which is precisely the statement that P-2 has a winning strategy.
Q.E.D.
.........
Now OC, we do not (yet) have a formalism in which we can treat
with such infinitely long statements, but attempts have been made
from time to time with varying success and domains of application.
I would claim, that these statements are of such AMAZINGLY simple
and "uniform" types, that they would be part of any reasonable
successful formalism, and that thus AD would be provable
in that context.
This may well be a pipe dream, as Aatu suggests, but it is
a clear-cut intuition in favour of AD, it uses quantifiers clearly
and concisely, (I have removed simplifying subscripts in favour
of a,b,c,d... for typographic clarity), and is far more believable
than considerations based on infinite sets of ridiculously
complex and inexpressible types.
-- Battling Bill. |
|
|
| Back to top |
|
|
|
| Bill Taylor... |
| Posted: Thu Oct 29, 2009 6:47 pm |
|
|
|
Guest
|
Aatu Koskensilta <aatu.koskensi... at (no spam) uta.fi> wrote:
[quote]This is a nice example of divergent intuitions!
[/quote]
It is clearly so.
[quote]My intuitions tell me nothing whatever about the existence of
winning strategies for chess or checkers.
[/quote]
You have been saying so all along. Perhaps game players have
better-informed intuitions. (Perhaps not.)
The rest of this article is irrelevantto the thread,
but may be interesting as a sidelight.
- - - - - -
[quote]I have played chess about five
times in my life, the plays consisting of my moving the pieces
essentially at random, the opponent declaring at some arbitrary,
or so it seemed to me, point that I'd lost.
[/quote]
Don't you just HATE opponents like that!? It's even worse at Go,
where the learner is expected to believe all his "dead" stones
can just be removed at the end, without any scoring penalty!
The proper thing is for the better player to ask, "You agree you're
hopelessly lost now?", and if there is any demurral or hesitation,
just play on till checkmate is actually thrashed out on the board.
It may be that the better player was just trying to save you from
this apparent humiliation, but in the face of even the slightest
demurral, it must be done. Get your nose rubbed in it!
It might even make you learn quicker!
[quote]I'm also utterly tone deaf, and
have no aptitude for crossword puzzles; and mentally adding two
two-digit numbers takes about half a minute for me.
[/quote]
Yikes! These are VERY damaging admissions, IMHO.
[quote]I'm very proud of and pleased with all these shortcomings,
[/quote]
YIKES AGAIN!
You transform them from an *admission* into a CLAIM!
This reminds me of the perfect squelch, to those cocktail-party
matrons who sweetly declare with faux-superior innocence,
"Oh I was always TERRIBLE at math, at school!"
....as if it were a prideful thing. The proper reply is...
"Oh don't worry, I never learned to read!"
Still, Aatu, at least it seems you can read; unless you're
getting a reader-writer to do all the heavy lifting for you,
(like those pampered, allegedly dyslexic students we see
more and more of these days, who must think they're going
to have reader-writers throughout their working life!)
- - - - -
For my part:- I'm quite good at chess, even better at Go,
and many chess variants and a great many other board games;
have excellent ear for tunes and songs, recall a great many,
can play a few (very slightly), and sing most; am excellent
at crossword puzzles, especially in tandem with my wife;
also sudoku; am great at cards and Scrabble; and can do
mental arithmetic up to factorizing three-digit numbers
very quickly indeed, (due to long practice). I have also
a large, self-instructed familiarity with most forms of art
and philosophy, to the point where I can easily hold my own
with professionals in these topics, even though I regard
the topics as mostly rubbish.
There! How am I doing in the claims stakes?
Oh - I also have a very long-lasting marriage to a devoted wife,
with two very happy and societally effective children....
-- Well-rounded William. |
|
|
| Back to top |
|
|
|
| Rupert... |
| Posted: Thu Oct 29, 2009 6:54 pm |
|
|
|
Guest
|
On Oct 30, 12:54 pm, Bill Taylor <w.tay... at (no spam) math.canterbury.ac.nz>
wrote:
[quote]stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
I don't understand why you think that "all games are determined" is
intuitively true.
You gave the answer yourself, in terms of infinite-depth quantifiers.
It's not *obvious* that chess or checkers has a
winning strategy; it's *provable*.
It IS. (That is, a winning strategy or a drawing strategy for both.)
This just plain OBVIOUS to any game player.
It was obvious to me even before I started high school.
To prove it, you have to use the
fact that they are finite-length games
Yes, the proof requires finitude, but the intuition does NOT.
Again, consider your own i-d quantifiers!
What reason is there for believing that the principle
applies to games for which is not provable?
I think you are asking - whose intuitions about infinite-depth
games is more reliable. Is it the man in the street,
or the hardened game-player, or the hardened math-logician?
Obviously mileage will vary!
-- Board-gaming Bill
[/quote]
Sorry, I know you've addressed this elsewhere in the thread, but I
wonder if we could just go through this. I don't believe that AD is
true. I have some experience playing games, but maybe not as much as
you.
Do you want to have a go at convincing me that it's true? |
|
|
| Back to top |
|
|
|
| Bill Taylor... |
| Posted: Thu Oct 29, 2009 7:20 pm |
|
|
|
Guest
|
Bill Taylor <w.tay... at (no spam) math.canterbury.ac.nz> wrote:
[quote]And here's mine: In the absence of AC, CH can be seen to be
both true and false, depending on how it is worded:
1) There is a subset of R, of cardinality strictly between N and R.
2) There is a function on subsets of R whose values are bijections
between the argument and one of N_k, N, or R.
AC |- 1 == 2
But, definitionially speaking, 1 and 2 are both clearly false.
And they might both be false in models of ZF.
[/quote]
Oh what a giant cock-up!!!! Apologies to anyone misled!
Try these:
AC |- 1 == ~2
Models of ZF might satisfy both ~1 and ~2 ;
and both 1 and 2 are seen to be both false, definitionally speaking.
Let's hope I've said what I mean this time....
b |
|
|
| Back to top |
|
|
|
| Daryl McCullough... |
| Posted: Thu Oct 29, 2009 8:30 pm |
|
|
|
Guest
|
Bill Taylor says...
[quote]
choice seems to strike many as evident, both in the sense people
explicitly state so
Might i humbly suggest that that is explainable by the fact that many
people are confronted with choice in the earliest stages of their
studies, and not with AD?
Another contributing factor being that almost all students are
introduced to it, motivated to it, by thoughts of making an infinite
sequence of choices. This convinces them of the common sense
of Countable Choice, which is a far less questionable assumption
than Wholesale Choice. Then the latter is slipped in more or
less surreptitiously alongside the former.
[/quote]
I don't think psychological or familiarity explains why people adopt
choice. People assume the axiom of choice because reasoning without
is an incredible pain. For example, without choice, we can't reason
about cardinality based on bigger/smaller. There are many different
notions of cardinality that are *inequivalent*: X < Y can be defined
by the existence of an injection from X to Y, or the existence of a
surjection from Y to X, and the two definitions are inequivalent.
You can't prove things for all elements of a set by well-ordering them
and using (ordinal) induction. Various definitions of "finite" become
inequivalent (Dedekind finite versus Tarski finite). We can't reason
from "A_i is a family of nonempty sets" to "the cross-product
A_0 x A_1 x A_2 ... is non-empty.
Reasoning without choice is a pain. Things become much complicated,
and seemingly not in an interesting way.
--
Daryl McCullough
Ithaca, NY |
|
|
| Back to top |
|
|
|
| Daryl McCullough... |
| Posted: Thu Oct 29, 2009 8:41 pm |
|
|
|
Guest
|
Bill Taylor says...
[quote]
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
I don't understand why you think that "all games are determined" is
intuitively true.
You gave the answer yourself, in terms of infinite-depth quantifiers.
[/quote]
But I really don't have an intuition about what infinitely many
alternations of quantifiers *mean*, unless I can do the Skolem
trick of replacing them by a single quantifier over strategies.
That trick is only valid if we assume choice.
[quote]It's not *obvious* that chess or checkers has a
winning strategy; it's *provable*.
It IS. (That is, a winning strategy or a drawing strategy for both.)
This just plain OBVIOUS to any game player.
[/quote]
As I said, I can see that it is obvious from the point
of view of "backing up" from a winning position, but
that only makes sense for finite games.
[quote]It was obvious to me even before I started high school.
[/quote]
I'm sure you never played infinite games before high school.
[quote]To prove it, you have to use the
fact that they are finite-length games
Yes, the proof requires finitude, but the intuition does NOT.
[/quote]
I don't agree. Once you introduce infinite games, the intuition
disappears completely for me. I know that some things that are
true for finite objects are not true for infinite objects, and
I have absolutely no reason to think it's not the case for
determinacy.
[quote]Again, consider your own i-d quantifiers!
[/quote]
That's a *notation*. Nothing follows from a notation. 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". It's not expressive enough
to make that distinction. But that doesn't mean anything, other
than we chose a notation that is only appropriate for determined
games.
--
Daryl McCullough
Ithaca, NY |
|
|
| Back to top |
|
|
|
| Daryl McCullough... |
| Posted: Fri Oct 30, 2009 6:29 am |
|
|
|
Guest
|
Bill Taylor says...
[quote]
stevendaryl3... at (no spam) yahoo.com (Daryl McCullough) wrote:
As a matter of fact, I think it is provably *false* for some natural
ways of formulating continuity.
Whoa! That IS interesting! Would you do us all a kindness, Daryl,
and teach us about some of these natural alternatives.
[/quote]
Thinking about it, I think the ambiguity is with how
"connected" is defined.
Okay, so here's a claim that I think is equivalent to the
Jordan curve theorem, or at least it's similarly intuitively
true: Consider the square with corners at (x,y)= (-1,-1),
(-1,1), (1,1) and (1,-1). Let C1 and C2 be two connected
curves such that C1 connects points (-1,1) and (1,-1), and C2
connects points (-1,-1) and (1,1). Then C1 and C2 must intersect.
(assuming that they never leave the square).
Now, the counterexample is this:
Let C1 be constructed as follows: C1 = the line segment
from (-1,1) to (0,1/2) together with the set of points
(x,y) such that 1 >= x > 0 and y = 1/2 + 1/2 sin(pi/(2x)).
Let C2 be constructed as follows: C2 = the line segment
from (-1,-1) to (0,0) together with the set of points (x,y)
such that 1 >= x > 0 and y = -1/2 + 1/2 sin(pi/(2x)),
together with the line segment from (1,0) to (1,-1).
C1 and C2 are connected curves in the sense that they
are one-dimensional, connected subsets of the plane,
and they don't intersect.
--
Daryl McCullough
Ithaca, NY |
|
|
| Back to top |
|
|
|
| Daryl McCullough... |
| Posted: Fri Oct 30, 2009 6:58 am |
|
|
|
Guest
|
Bill Taylor says...
[quote]And I meant to add - this was what was intended to happen with
the Banach-Tarski decomposition, or so I have read.
Allegedly, one of them hoped it would be so ludicrous that
mathies as a whole would thus drop AC, but that didn't happen.
[/quote]
Actually, the hard work of the Banach-Tarski doesn't involve
choice at all. If we parameterize the points on the sphere
by longitude and lattitude, consider the set of points such
that the longitude and lattitude are rational multiples of
pi radians. This object, a "holy" sphere, can be decomposed
into a finite number of pieces that can be reassembled by
translations and rotations to form two identical copies of
the original holy sphere.
That by itself is pretty amazing to me. And it doesn't
involve the axiom of choice (basically because choice
already holds for a countable set). Choice just allows
basically the same proof to go through for all points
on the sphere, not just a countable dense subset.
--
Daryl McCullough
Ithaca, NY |
|
|
| Back to top |
|
|
|
| Daryl McCullough... |
| Posted: Fri Oct 30, 2009 7:21 am |
|
|
|
Guest
|
Bill Taylor says...
[quote]
OHerman Jurjus <hjm... at (no spam) hetnet.nl> wrote:
But: either player 1 has a winning strategy or he hasn't.
Now what does it mean for player 1 to not have a winning strategy?
I'd say that amounts to 'player 2 has some way to prevent player 1 from
winning'.
That sound fairly unimpeachable.
[/quote]
It certainly doesn't to me. To say that player 1 has no winning
strategy means that, no matter how brilliant player 1 is, there is
a player 2 who can beat him. That's just different from saying
that there is a player 2 who can beat any player 1. The fact
that Bob as first player can always beat Carol, and Ted as
first player can always beat Alice, does *not* imply that
Bob can always beat Alice.
So the equivalence "Player 1 has no winning strategy"
does not *mean* the same thing as "Player 2 has a winning
strategy". It just doesn't. Now, they might be equivalent
for some games, but they don't *mean* the same thing, which
is what Herman was implying.
--
Daryl McCullough
Ithaca, NY |
|
|
| Back to top |
|
|
|
| James Burns... |
| Posted: Fri Oct 30, 2009 8:04 am |
|
|
|
Guest
|
Daryl McCullough wrote:
[quote]Bill Taylor says...
And I meant to add - this was what was intended to happen with
the Banach-Tarski decomposition, or so I have read.
Allegedly, one of them hoped it would be so ludicrous that
mathies as a whole would thus drop AC, but that didn't happen.
Actually, the hard work of the Banach-Tarski doesn't involve
choice at all. If we parameterize the points on the sphere
by longitude and lattitude, consider the set of points such
that the longitude and lattitude are rational multiples of
pi radians. This object, a "holy" sphere, can be decomposed
into a finite number of pieces that can be reassembled by
translations and rotations to form two identical copies of
the original holy sphere.
That by itself is pretty amazing to me. And it doesn't
involve the axiom of choice (basically because choice
already holds for a countable set). Choice just allows
basically the same proof to go through for all points
on the sphere, not just a countable dense subset.
[/quote]
Cool!
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.
Jim Burns |
|
|
| Back to top |
|
|
|
| Herman Jurjus... |
| Posted: Fri Oct 30, 2009 10:25 am |
|
|
|
Guest
|
Daryl McCullough wrote:
[quote]Bill Taylor says...
OHerman Jurjus <hjm... at (no spam) hetnet.nl> wrote:
But: either player 1 has a winning strategy or he hasn't.
Now what does it mean for player 1 to not have a winning strategy?
I'd say that amounts to 'player 2 has some way to prevent player 1 from
winning'.
That sound fairly unimpeachable.
It certainly doesn't to me. To say that player 1 has no winning
strategy means that, no matter how brilliant player 1 is, there is
a player 2 who can beat him. That's just different from saying
that there is a player 2 who can beat any player 1. The fact
that Bob as first player can always beat Carol, and Ted as
first player can always beat Alice, does *not* imply that
Bob can always beat Alice.
So the equivalence "Player 1 has no winning strategy"
does not *mean* the same thing as "Player 2 has a winning
strategy". It just doesn't. Now, they might be equivalent
for some games, but they don't *mean* the same thing, which
is what Herman was implying.
[/quote]
I didn't (intend to) imply that they conceptually /mean/ the same thing,
but only that one 'should' imply the other, in a sense similar to the
sense in which the Jordan curve theorem 'should' be true.
(Something like "no doubt you can make a system in which the two are not
equivalent, but then that system does not represent the 'real' universe
of sets, the 'real' notion of sequence of numbers, etc.")
As regards AD+AC being 'true' simultaneously:
When pushed to the extreme, i can't take on a 'realist' point of view
regarding sets, without either ignoring parts of my intuitions (and
making an arbitrary choice about which parts), or accepting both AD and
AC (which is not really an option, of course, unless we get extreme and
drop FOL or so - but this would in itself constitute a rejection of
realism).
Hence, i can't be a 'realist' when it comes to 'the real universe of
sets' without being intellectually dishonest to myself to some degree. I
must take *some* part of it with a grain of salt.
--
Cheers,
Herman Jurjus |
|
|
| Back to top |
|
|
|
| Bill Taylor... |
| Posted: Fri Oct 30, 2009 7:06 pm |
|
|
|
Guest
|
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*,
[/quote]
Of course you do!
(Else I have a quite different understanding of the words you use.)
The mere fact that you *came up* with the idea, independently of me,
shows you have a fair idea of what it means. Naturally we don't
understand *every* nuance, implication, relation and what-not
about them, but that could be said of most math things.
[quote]unless I can do the Skolem
trick of replacing them by a single quantifier over strategies.
[/quote]
Sure you *can* do that if it helps. It's a parallel way of looking
at the same thing, but not identical. As you say, Skolem's
automatically raise questions of simultaneity and choice.
[quote]It's not *obvious* that chess or checkers has a
winning strategy;
It IS.(That is, a winning strategy or a drawing strategy for both)
It was obvious to me even before I started high school.
I'm sure you never played infinite games before high school.
[/quote]
I's talking about CHESS, you goose, when I mentioned high school! :)
[quote]I don't agree. Once you introduce infinite games, the intuition
disappears completely for me.
[/quote]
Well, I could say exactly the same thing about choice,
which you seem to find no problems with.
How are we to resolve this quagmire!?
[quote]I know that some things that are
true for finite objects are not true for infinite objects,
[/quote]
Apparently these do not include choice...
[quote]Again, consider your own i-d quantifiers!
That's a *notation*. Nothing follows from a notation.
[/quote]
OCN. It is the concept that the notations (attempt to) represent
that things follow from.
[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".
[/quote]
OCN! Because there isn't one, (or so I claim), for tie-free games.
[quote]It's not expressive enough to make that distinction.
[/quote]
Nothing is expressive enough to make distinctions that don't exist!
-- Wriggling William
** Set theory is a shotgun marriage -
between well-ordering and power-set.
** The two parties get along OK;
but they hardly seem made for each other. |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Tue Dec 15, 2009 4:08 am
|
|