Another AC anomaly?
Herman Jurjus
I'm not sure yet what to conclude from it; that AC is horribly wrong, or
that WM is horribly right, or something else altogether.
In short the story goes like this:
A game is played, in which infinitely many coins are tossed, and there's
one player, who makes infinitely many guesses. Both are done over a
finite period of time. The tosses and guesses are not made faster and
faster, however, but slower and slower: at t = 1/n. There's no 'first'
move.
Claim:
There exists a strategy with which you're certain to guess all entries
correctly except for at most finitely many mistakes. Not 'certain' as in
'probability is 100%', but absolutely certain.
Reasoning:
On 2^w, consider the equivalence relation that makes x equivalent to y
when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a
set S that contains precisely one element from every equivalence class.
Strategy: at every move, you already know the results of the previous
tosses, which is an infinite tail of some sequence in 2^w. Now take the
unique element from S associated to that tail, take the n'th element of
that sequence from S, and deliver that as your move.
After some thinking, you will see that with this strategy, you're indeed
certain to guess wrong at most finitely many times.
Thanks, AC! Another nice mess you've gotten us into.
--
Cheers,
Herman Jurjus
David C. Ullrich
wrote:
>Has anyone seen this before?
>
>http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-result-of-infinitely-many-coin-tosses/
>
>I'm not sure yet what to conclude from it; that AC is horribly wrong, or
>that WM is horribly right, or something else altogether.
>
>In short the story goes like this:
>
>A game is played, in which infinitely many coins are tossed, and there's
>one player, who makes infinitely many guesses. Both are done over a
>finite period of time. The tosses and guesses are not made faster and
>faster, however, but slower and slower: at t = 1/n. There's no 'first'
>move.
In case I'm not the only one who couldn't figure out exactly what's
going on from that paragraph, the description on the page seems
more clear:
"For each n > 0, at \frac{1}{n} hours past 12pm the following is going
to happen: aware of the time, you are going to guess either heads or
tails, and then I am going to flip a coin and show you the result so
you can see if you are right or wrong. This process may have to be
done at different speeds to fit it all in to the hour between 12pm and
1pm."
>Claim:
>There exists a strategy with which you're certain to guess all entries
>correctly except for at most finitely many mistakes. Not 'certain' as in
>'probability is 100%', but absolutely certain.
>
>Reasoning:
>On 2^w, consider the equivalence relation that makes x equivalent to y
>when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a
>set S that contains precisely one element from every equivalence class.
>Strategy: at every move, you already know the results of the previous
>tosses, which is an infinite tail of some sequence in 2^w. Now take the
>unique element from S associated to that tail, take the n'th element of
>that sequence from S, and deliver that as your move.
>After some thinking, you will see that with this strategy, you're indeed
>certain to guess wrong at most finitely many times.
>
>Thanks, AC! Another nice mess you've gotten us into.
I think the moral is not that AC leads to the weirdness but that
this is a highly weird situation to begin with. At the time when
we make any given guess we've already been told the result of
infinitely many coin tosses...
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.)
William Hughes
>
> I'm not sure yet what to conclude from it; that AC is horribly wrong, or
> that WM is horribly right, or something else altogether.
>
> In short the story goes like this:
>
> A game is played, in which infinitely many coins are tossed, and there's
> one player, who makes infinitely many guesses. Both are done over a
> finite period of time. The tosses and guesses are not made faster and
> faster, however, but slower and slower: at t = 1/n. There's no 'first'
> move.
>
> Claim:
> There exists a strategy with which you're certain to guess all entries
> correctly except for at most finitely many mistakes. Not 'certain' as in
> 'probability is 100%', but absolutely certain.
>
> Reasoning:
> On 2^w, consider the equivalence relation that makes x equivalent to y
> when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a
> set S that contains precisely one element from every equivalence class.
> Strategy: at every move, you already know the results of the previous
> tosses, which is an infinite tail of some sequence in 2^w. Now take the
> unique element from S associated to that tail, take the n'th element of
> that sequence from S, and deliver that as your move.
> After some thinking, you will see that with this strategy, you're indeed
> certain to guess wrong at most finitely many times.
>
> Thanks, AC! Another nice mess you've gotten us into.
>
Note you can avoid AC, by defining
a_n(m)= s(m) m>n, a_n(k) heads otherwise.
The strategy is now to choose heads everytime.
Note that this does not work.
The problem is that although you use a_n
to choose your guess for n, you do not use
a_n for any guess for m>n. The fact that
such a guess would be correct is of no use.
If you could start the game with only a finite
number of wrong guesses you are done. However, the
game has no first move, so how do you start?
- William Hughes
Herman Jurjus
Yup - it's a game without a first move. So 'weird' is an accurate
qualification. Yet, it's not particularly difficult to give a
mathematical description of the situation, reason about it, and convince
ourselves that, mathematically, there's no problem with it. Ideally, it
should be much harder to make mathematical sense of a game like this.
That's why I included "WM is horribly right" as a possible moral.
Apparently there's something wrong with backward supertasks (and not
with ordinary, 'forward' supertasks). But why should that be?
--
Cheers,
Herman Jurjus
Herman Jurjus
What's s(m) ?
As a matter of fact, what's a_n(m) ?
The game involves tosses and guesses, i.e. two sequences with one index:
toss(n), guess(n).
--
Cheers,
Herman Jurjus
Rotwang
> David C. Ullrich wrote:
> > On Mon, 23 Nov 2009 12:22:17 +0100, Herman Jurjus <hjm...@hetnet.nl>
> > wrote:
>
>
> > I think the moral is not that AC leads to the weirdness but that
> > this is a highly weird situation to begin with. At the time when
> > we make any given guess we've already been told the result of
> > infinitely many coin tosses...
>
> Yup - it's a game without a first move. So 'weird' is an accurate
> qualification. Yet, it's not particularly difficult to give a
> mathematical description of the situation, reason about it, and convince
> ourselves that, mathematically, there's no problem with it. Ideally, it
> should be much harder to make mathematical sense of a game like this.
> That's why I included "WM is horribly right" as a possible moral.
>
> Apparently there's something wrong with backward supertasks (and not
> with ordinary, 'forward' supertasks). But why should that be?
I'm not sure what qualifies as a "backward" or "forward" supertask,
but this paradox is very similar to the AC solution to the infinite
prisoners-and-hats puzzle, which avoids any weirdness involving non-
well-ordered games:
http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle#Countably_Infinite-Hat_Solution
Jesse F. Hughes
> Apparently there's something wrong with backward supertasks (and not
> with ordinary, 'forward' supertasks). But why should that be?
Well, I'm not at all sure that there's no problem with forward
supertasks. Surely, it is not difficult to come up with a
problematic case.
For instance, take our favorite example: at each time t - 1/n, place
balls 10(n-1) to 10n - 1 in a vase and then remove ball n. At the end
the vase is empty.
Now alter the situation slightly. At each step, again place 10 balls
into the vase and then remove one ball, but remove the ball
*randomly*. At the end, the vase may contain any number of balls.
This strikes me as suitably counterintuitive to say that the forward
supertask has something wrong with it. Or, perhaps, with my
intuitions.
--
"To solve this problem, we define a security flag, known as the 'evil'
bit, in the IPv4 [RFC791] header. Benign packets have this bit set to
0; those that are used for an attack will have the bit set to 1."
-- RFC 3514
Herman Jurjus
Thanks for the reference.
(One of the comments on the OP's blog was "the Unexpected Hanging on
steroids" <g>)
--
Cheers,
Herman Jurjus
Herman Jurjus
> Herman Jurjus <hjm...@hetnet.nl> writes:
>
>> Apparently there's something wrong with backward supertasks (and not
>> with ordinary, 'forward' supertasks). But why should that be?
>
> Well, I'm not at all sure that there's no problem with forward
> supertasks. Surely, it is not difficult to come up with a
> problematic case.
>
> For instance, take our favorite example: at each time t - 1/n, place
> balls 10(n-1) to 10n - 1 in a vase and then remove ball n. At the end
> the vase is empty.
>
> Now alter the situation slightly. At each step, again place 10 balls
> into the vase and then remove one ball, but remove the ball
> *randomly*. At the end, the vase may contain any number of balls.
> This strikes me as suitably counterintuitive to say that the forward
> supertask has something wrong with it. Or, perhaps, with my
> intuitions.
More conclusive (at least for me): you switch a light bulb on and off;
after infinitely many steps, is the light on or off? (Or: you put one
and the same ball in the vase, out of the vase, in the vase, out of the
vase, etc. What's in the vase/where's the ball after infinitely many steps?)
If you can't trust supertask-reasoning in this case, why should you
trust it in other, seemingly less problematic cases?
For the sake of clarity: I'm not endorsing or advocating the
backward-supertask paradox as extremely important. Just wanted to share
it here, because I had never heard of it, and thought perhaps others
also hadn't.
--
Cheers,
Herman Jurjus
William Hughes
s is the sequence of tosses, so s(n) = toss(n)
As above, a_n(m) is equal to the toss if m>n, otherwise
is a head. Note that a_n is equal to s except perhaps on
a finite set.
For any n, a_n(n) is heads, so guess(n)=a_n(n) is heads.
I don't think the fact that time is reversed has any
real bearing on this. If the tosses are independent, perfect
knowledge of the future is of no help for the present.
From the quoted page
Now, since the representative sequence and the actual
sequence of heads and tails that occurs are in the same
equivalence class, they must only differ at finitely many
places.
Correct. Note this is true of sequence a_n.
So, if you guessed according to the representative
sequence, you have only made finitely many mistakes.
Correct. However, you did not guess acconding to the
representative sequence. You only used a_n for guess(n).
For other guesses you used other sequences.
- William Hughes
Jesse F. Hughes
> More conclusive (at least for me): you switch a light bulb on and
> off; after infinitely many steps, is the light on or off?
Yes, of course, that's much simpler and clearer.
--
Jesse F. Hughes
"You know, I -- when I speak, like right now, for example -- I'm
speaking to the American people, of course, and I want them to know
that I know how tough it is[...]" -- Reassuring words from Pres. Bush
Herman Jurjus
That would mean you always guess heads.
But if you always guess heads, and the tosses are random, you will make
infinitely many mistakes with probability 1.
> I don't think the fact that time is reversed has any
> real bearing on this. If the tosses are independent, perfect
> knowledge of the future is of no help for the present.
>
> From the quoted page
>
> Now, since the representative sequence and the actual
> sequence of heads and tails that occurs are in the same
> equivalence class, they must only differ at finitely many
> places.
>
> Correct. Note this is true of sequence a_n.
Huh?
> So, if you guessed according to the representative
> sequence, you have only made finitely many mistakes.
>
> Correct. However, you did not guess acconding to the
> representative sequence. You only used a_n for guess(n).
> For other guesses you used other sequences.
What other guesses?
At t=1/n, you guess 'heads'/'tails' once, after which one toss is done.
And obviously the game lets you make each guess before the toss is done,
otherwise it would be rather easy to make no mistakes at all.
>
>
> - William Hughes
--
Cheers,
Herman Jurjus
William Hughes
Correct.
>
> > I don't think the fact that time is reversed has any
> > real bearing on this. If the tosses are independent, perfect
> > knowledge of the future is of no help for the present.
>
> > From the quoted page
>
> > Now, since the representative sequence and the actual
> > sequence of heads and tails that occurs are in the same
> > equivalence class, they must only differ at finitely many
> > places.
>
> > Correct. Note this is true of sequence a_n.
>
> Huh?
a_n differs from the sequence of tosses at finitely many places.
>
> > So, if you guessed according to the representative
> > sequence, you have only made finitely many mistakes.
>
> > Correct. However, you did not guess acconding to the
> > representative sequence. You only used a_n for guess(n).
> > For other guesses you used other sequences.
>
> What other guesses?
> At t=1/n, you guess 'heads'/'tails' once, after which one toss is done.
Take two different times, o and p. The guess for
time o is based on sequence a_o. The guess for time p
is based on sequence a_p [I think you may be confusing
the sequence a_n, with the nth element of sequence a_n,
a_n(n)]. For each guess you use a different sequence.
- William Hughes
Mike Terry
news:0210e1fa-23fa-4af9...@m33g2000vbi.googlegroups.com...
Yes, but there is not one sequence a_n, but rather one for every n...
In what follows, I'm using:
g := our guess sequence
r := the "representative" sequence.
Perhaps you're wanting me to have r_n instead of r? But then we would just
have
r_n = r_m for all n,m, (since they differ only in finitely many
places)
So there is just one r sequence... (I think you've missed this point?)
>
> So, if you guessed according to the representative
> sequence, you have only made finitely many mistakes.
>
> Correct.
Good... you're agreeing that we're only going to guess wrong a finite number
of times... (given the unachievable physical nature of the experiment in the
first place!)
> However, you did not guess acconding to the
> representative sequence.
Yes we did! That's our strategy: guess according to the representative
sequence. Note that there is only *one* representative sequence r that we
are using for ALL our guesses. (I.e. our g(n) = r(n) for all n.)
> You only used a_n for guess(n).
> For other guesses you used other sequences.
Well, that's why your approach doesn't work, while the one suggested in the
article does! (given the unachievable physical nature of the experiment in
the first place etc.)
In the article, r(n) is just one sequence, and our g(n) = r(n) for all n.
So g(n) = s(n) for all but finitely many n (by definition of r).
(So you can't claim to be eliminating the use of AC, since your approach
doesn't work.)
Regards,
Mike.
master1729
> Has anyone seen this before?
>
> http://possiblyphilosophy.wordpress.com/2008/09/22/gue
> ssing-the-result-of-infinitely-many-coin-tosses/
>
> I'm not sure yet what to conclude from it; that AC is
> horribly wrong, or
> that WM is horribly right, or something else
> altogether.
well AC is horribly wrong.
but i doubt that WM is horribly right.
what is WM suppose to be right about ?
typical AC.
>
> --
> Cheers,
> Herman Jurjus
regards
tommy1729
master1729
thompsons lamp.
i assumed people on sci.math knew about the term.
apparantly i was to optimistic again.
basicly thompsons lamp can be well expressed by ordinary number theory :
lim n -> oo n mod 2 = ???
there is no convergeance , so the answer is not 0 or 1 but ' div ' and in terms of my 3-valued logic :
'0' truth value.
btw all these things have been said a billion times on sci.math , both my own comments and analogues of thompsons lamp.
so i wonder why they are posted again , but i will look at the good side ; this gives me an opportunity to give a good reply and a good reminder of my ideas.
(and btw , yes an AC anomaly)
regards
tommy1729
Butch Malahide
>
> I'm not sure yet what to conclude from it; that AC is horribly wrong, or
> that WM is horribly right, or something else altogether.
>
> In short the story goes like this:
>
> A game is played, in which infinitely many coins are tossed, and there's
> one player, who makes infinitely many guesses. Both are done over a
> finite period of time. The tosses and guesses are not made faster and
> faster, however, but slower and slower: at t = 1/n. There's no 'first'
> move.
>
> Claim:
> There exists a strategy with which you're certain to guess all entries
> correctly except for at most finitely many mistakes. Not 'certain' as in
> 'probability is 100%', but absolutely certain.
>
> Reasoning:
> On 2^w, consider the equivalence relation that makes x equivalent to y
> when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a
> set S that contains precisely one element from every equivalence class.
> Strategy: at every move, you already know the results of the previous
> tosses, which is an infinite tail of some sequence in 2^w. Now take the
> unique element from S associated to that tail, take the n'th element of
> that sequence from S, and deliver that as your move.
> After some thinking, you will see that with this strategy, you're indeed
> certain to guess wrong at most finitely many times.
>
> Thanks, AC! Another nice mess you've gotten us into.
In plain mathematical terms: for any set X, there is a function f:X^N-
>X such that, for each sequence (x_1, x_2, ...} in X^N, the equality
x_n = f(x_{n+1}, x_{n+2}, ...) holds for all but finitely many n.
This is old hat: see Problem 5348, American Mathematical Monthly,
volume 72 (1965), p. 1136. (This has been discussed on sci.math in the
past.)
A
Am I misunderstanding this? Look at the following paragraph:
"The stategy you should adopt runs as follows. At 1/n hrs past 12 you
should be able to work out exactly which equivalence class the
completed sequence of heads and tails that will eventually unfold is
in. You have been told the result of all the previous tosses, and you
know there are only finitely many tosses left to go, so you know the
eventual completed sequence can only differ from what you know about
it at finitely many places. Given you know which equivalence class
you’re in, you just guess as if the representative of that equivalence
class was correct about the current guess. So at 1/n hrs past 12 you
just look at how the representative sequence says the coin will land
and guess accordingly."
Apparently we do not choose a strategy before time begins to pass, but
rather, we have only ever chosen a strategy for playing the game once
some nonzero amount of time (1/n hours past 12) has passed. But after
any nonzero amount of time has passed, we have made infinitely many
coin tosses already, and there are only finitely many remaining coin
tosses; so one of two things is true: either
A) we have already guessed infinitely many coin tosses incorrectly,
and hence there is no winning strategy for us, or
B) we have already guessed only finitely many coin tosses incorrectly,
in which case any strategy we choose will be a winning strategy, since
even if we guess all our remaining tosses wrong, we will still have
only guessed wrong a finite number of times.
In other words, the author of the original article on Wordpress does
not seem to be telling us how to choose a winning strategy from the
beginning of this game, but rather, how to choose a winning strategy
once some finite amount of time has already passed; and in that case,
either there exists no winning strategy, or any strategy we choose is
a winning strategy.
Perhaps I am misunderstanding something here, but this doesn't seem
like very novel stuff, and it doesn't seem to have anything to do with
the axiom of choice.
Butch Malahide
>
> In plain mathematical terms: for any set X, there is a function f:X^N->X such that, for each sequence (x_1, x_2, ...} in X^N, the equality
>
> x_n = f(x_{n+1}, x_{n+2}, ...) holds for all but finitely many n.
>
> This is old hat: see Problem 5348, American Mathematical Monthly,
> volume 72 (1965), p. 1136. (This has been discussed on sci.math in the
> past.)
Alan D. Taylor and C. Hardin have some recent papers on this subject:
http://www.math.union.edu/people/faculty/publications/taylora.html
A peculiar connection between the axiom of choice and predicting the
future (with C. Hardin), American Mathematical Monthly 115 (2008),
91-96.
An introduction to infinite hat problems (with C. Hardin),
Mathematical Intelligencer 30 (2008), 20-25.
Limit-like predictability for discontinuous functions (with C.
Hardin), Proceedings of the American Mathematical Society 137 (2009),
3123-3128.
Herman Jurjus
> On Nov 23, 5:22 am, Herman Jurjus <hjm...@hetnet.nl> wrote:
>> Has anyone seen this before?
>>
>> http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-resul...
>>
>
> In plain mathematical terms: for any set X, there is a function f:X^N-
>> X such that, for each sequence (x_1, x_2, ...} in X^N, the equality
> x_n = f(x_{n+1}, x_{n+2}, ...) holds for all but finitely many n.
>
> This is old hat: see Problem 5348, American Mathematical Monthly,
> volume 72 (1965), p. 1136. (This has been discussed on sci.math in the
> past.)
Good to know that!
When was that, approximately (I mean the sci.math discussion)?
And under what name was/is it known?
--
Cheers,
Herman Jurjus
Herman Jurjus
> On Nov 23, 6:22 am, Herman Jurjus <hjm...@hetnet.nl> wrote:
>> Has anyone seen this before?
>>
>> http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-resul...
>>
>> I'm not sure yet what to conclude from it; that AC is horribly wrong, or
>> that WM is horribly right, or something else altogether.
> Am I misunderstanding this? Look at the following paragraph:
>
> "The stategy you should adopt runs as follows. At 1/n hrs past 12 you
> should be able to work out exactly which equivalence class the
> completed sequence of heads and tails that will eventually unfold is
> in. You have been told the result of all the previous tosses, and you
> know there are only finitely many tosses left to go, so you know the
> eventual completed sequence can only differ from what you know about
> it at finitely many places. Given you know which equivalence class
> you�re in, you just guess as if the representative of that equivalence
> class was correct about the current guess. So at 1/n hrs past 12 you
> just look at how the representative sequence says the coin will land
> and guess accordingly."
>
> Apparently we do not choose a strategy before time begins to pass, but
> rather, we have only ever chosen a strategy for playing the game once
> some nonzero amount of time (1/n hours past 12) has passed. But after
> any nonzero amount of time has passed, we have made infinitely many
> coin tosses already, and there are only finitely many remaining coin
> tosses; so one of two things is true: either
>
> A) we have already guessed infinitely many coin tosses incorrectly,
> and hence there is no winning strategy for us, or
> B) we have already guessed only finitely many coin tosses incorrectly,
> in which case any strategy we choose will be a winning strategy, since
> even if we guess all our remaining tosses wrong, we will still have
> only guessed wrong a finite number of times.
>
> In other words, the author of the original article on Wordpress does
> not seem to be telling us how to choose a winning strategy from the
> beginning of this game, but rather, how to choose a winning strategy
> once some finite amount of time has already passed; and in that case,
> either there exists no winning strategy, or any strategy we choose is
> a winning strategy.
>
> Perhaps I am misunderstanding something here, but this doesn't seem
> like very novel stuff, and it doesn't seem to have anything to do with
> the axiom of choice.
That's also what I though, at first.
But the strategy description as given leads to a real strategy only
because/when you use AC as indicated. In general, not every
next-move-prescription leads to a full strategy.
Consider, for example, the next prescription:
if all your guesses so far have been 'heads', then guess 'tails' next,
in all other cases, guess 'heads' next.
Although this determines a 'next move' in every situation, there's no
overall strategy that fits this prescription, as is easy to see.
(This particular example resembles the 'Yablo paradox', btw.)
But by using AC in the way indicated, you do get a full strategy; that
is: a function that assigns to every sequence of tosses one unique
sequences of guesses, in such a way that every guess/move depends only
on the previous events (either previous tosses or previous guesses).
But as Butch Malahide has written, apparently this is already old hat in
sci.math.
--
Cheers,
Herman Jurjus
Herman Jurjus
> Herman Jurjus wrote :
>
>> Has anyone seen this before?
>>
>> http://possiblyphilosophy.wordpress.com/2008/09/22/gue
>> ssing-the-result-of-infinitely-many-coin-tosses/
>>
>> I'm not sure yet what to conclude from it; that AC is
>> horribly wrong, or
>> that WM is horribly right, or something else
>> altogether.
>
> well AC is horribly wrong.
>
> but i doubt that WM is horribly right.
>
> what is WM suppose to be right about ?
That current mathematics too easily allows weird things involving
infinity. (In this case the whole idea of a 'backwards infinite game'.)
--
Cheers,
Herman Jurjus
Butch Malahide
>
> But the strategy description as given leads to a real strategy only
> because/when you use AC as indicated. In general, not every
> next-move-prescription leads to a full strategy.
> Consider, for example, the next prescription:
> if all your guesses so far have been 'heads', then guess 'tails' next,
> in all other cases, guess 'heads' next.
> Although this determines a 'next move' in every situation, there's no
> overall strategy that fits this prescription, as is easy to see.
> (This particular example resembles the 'Yablo paradox', btw.)
>
> But by using AC in the way indicated, you do get a full strategy; that
> is: a function that assigns to every sequence of tosses one unique
> sequences of guesses, in such a way that every guess/move depends only
> on the previous events (either previous tosses or previous guesses).
In the present context, I think it is best to define a "strategy" as a
function which depends only on the *opponent's* previous choices,
i.e., on the previous tosses but *not* on the previous guesses.
master1729
WM is correct about that.
Regards
tommy1729
Butch Malahide
It has probably come up more than once. I found some stuff by googling
on "problem 5348".
Butch Malahide
OK, I find two sci.math threads: "Ill-defined puzzle" and "An
astonishing application of the AC". The first one has no mathematical
content. Problem 5348 is also mentioned (with an application) in the
sci.math.research thread "Trick property".
William Hughes
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "William Hughes" <wpihug...@hotmail.com> wrote in message
My notation is confusing as often "sequence a_n" means
sequence a with nth term a_n. However by sequence a_n
I mean a sequence (nth term a_n(n)). Sequences a_n and a_m
may not be the same for n =/= m. I.e. there is a different
sequence for every n.
<snip>
>
> In the article, r(n) is just one sequence, and our g(n) = r(n) for all n.
Yes and no. It does in fact follow that there is only one
sequence r(n) and this is the key to the "paradox".
However, my reading of the article was that it implied
it was enough
to take at each step, a sequence which differed from
the sequence of tosses at only a finite number of points.
As my example shows, this can be done without using
AC and is not sufficient.
However, to get a single sequence, r, we need to associate
a representative element to each equivalence class.
For this we need AC. However, I do not see any problem here
and though the result of being able to choose
(nonconstructively) a representative sequence may
be seen as counterintuitive it is certainly no
more so than the fact that a infinite set
can be put in one to one correspondence with a subset.
I am not sure to what extent the "reverse supertask"
is of interest. Since it is impossible to start
a game with no first move, talk of strategy is
not very interesting.
- William Hughes
Tim Little
> http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-result-of-infinitely-many-coin-tosses/
Yes; with weird premises of things like infinite supertasks and
decision-making with uncountably infinite amounts of information, AC
(among other things) implies some odd results.
> I'm not sure yet what to conclude from it; that AC is horribly
> wrong, or that WM is horribly right, or something else altogether.
I'd conclude that weird premises result in weird conclusions.
- Tim
Tim Little
> If you could start the game with only a finite number of wrong
> guesses you are done. However, the game has no first move, so how
> do you start?
Indeed. That's one point where the premise is screwy to begin with.
The same paradox is much better expressed in the "prisoners and hats"
version. Although it has the same uncountably infinite information
load, at least it has a beginning.
- Tim
David C. Ullrich
wrote:
>David C. Ullrich wrote:
>> On Mon, 23 Nov 2009 12:22:17 +0100, Herman Jurjus <hjm...@hetnet.nl>
>> wrote:
>>
>>> Has anyone seen this before?
>>>
>>>
>>> I'm not sure yet what to conclude from it; that AC is horribly wrong, or
>>> that WM is horribly right, or something else altogether.
>>>
>>> In short the story goes like this:
>>>
>>> A game is played, in which infinitely many coins are tossed, and there's
>>> one player, who makes infinitely many guesses. Both are done over a
>>> finite period of time. The tosses and guesses are not made faster and
>>> faster, however, but slower and slower: at t = 1/n. There's no 'first'
>>> move.
>>
>> In case I'm not the only one who couldn't figure out exactly what's
>> going on from that paragraph, the description on the page seems
>> more clear:
>>
>> "For each n > 0, at \frac{1}{n} hours past 12pm the following is going
>> to happen: aware of the time, you are going to guess either heads or
>> tails, and then I am going to flip a coin and show you the result so
>> you can see if you are right or wrong. This process may have to be
>> done at different speeds to fit it all in to the hour between 12pm and
>> 1pm."
>>
>>> Claim:
>>> There exists a strategy with which you're certain to guess all entries
>>> correctly except for at most finitely many mistakes. Not 'certain' as in
>>> 'probability is 100%', but absolutely certain.
>>>
>>> Reasoning:
>>> On 2^w, consider the equivalence relation that makes x equivalent to y
>>> when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a
>>> set S that contains precisely one element from every equivalence class.
>>> Strategy: at every move, you already know the results of the previous
>>> tosses, which is an infinite tail of some sequence in 2^w. Now take the
>>> unique element from S associated to that tail, take the n'th element of
>>> that sequence from S, and deliver that as your move.
>>> After some thinking, you will see that with this strategy, you're indeed
>>> certain to guess wrong at most finitely many times.
>>>
>>> Thanks, AC! Another nice mess you've gotten us into.
>>
>> I think the moral is not that AC leads to the weirdness but that
>> this is a highly weird situation to begin with. At the time when
>> we make any given guess we've already been told the result of
>> infinitely many coin tosses...
>
>Yup - it's a game without a first move. So 'weird' is an accurate
>qualification. Yet, it's not particularly difficult to give a
>mathematical description of the situation, reason about it, and convince
>ourselves that, mathematically, there's no problem with it. Ideally, it
>should be much harder to make mathematical sense of a game like this.
Why should this be hard? You wouldn't expect that being given
infinitely much information would be very powerful?
>That's why I included "WM is horribly right" as a possible moral.
>
>Apparently there's something wrong with backward supertasks (and not
>with ordinary, 'forward' supertasks). But why should that be?
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.)
George Greene
> Yup - it's a game without a first move. So 'weird' is an accurate
> qualification. Yet, it's not particularly difficult to give a
> mathematical description of the situation, reason about it, and convince
> ourselves that, mathematically, there's no problem with it.
The whole inference paradigm for standard classical FOL
assumes that all rules of inference have FINITE number of
premises. Even ONE attempt to make something depend
on an infinite amount of prior information is a violation of the
whole paradigm. This applies just as strongly to computing
terms from arguments as it does to deriving conclusions from
premises.
George Greene
> More conclusive (at least for me): you switch a light bulb on and off;
> after infinitely many steps, is the light on or off?
Well, there is more than one infinity.
Ordinally, there are even successor ordinals after
the first infinity. The reversed order (with no first element) can
also have a last one. Basically, any infinity of steps that has
a last element will have an answer to this question.
Any infinite sequence that does not have a last element
needs to get its "answer" from some NON-standard convention.
George Greene
> wrote:
> >Yup - it's a game without a first move. So 'weird' is an accurate
> >qualification. Yet, it's not particularly difficult to give a
> >mathematical description of the situation, reason about it, and convince
> >ourselves that, mathematically, there's no problem with it. Ideally, it
> >should be much harder to make mathematical sense of a game like this.
On Nov 24, 6:39 am, David C. Ullrich <dullr...@sprynet.com> wrote:
> Why should this be hard? You wouldn't expect that being given
> infinitely much information would be very powerful?
Or even inadmissibly powerful; I mean,
the definition of both a first-order language and
the standard classical first-order inference paradigm
PROHIBIT BOTH 1) functions from taking infinitely many arguments,
AND 2) inference rules from taking infinitely many premises.
You can legitimately work partially around this by using
infinite sets as arguments IF they are finitarily definable.
But when, as with AC, they are providing an infinite amount
of information, then, yes, precisely as DCU says, you would
EXPECT "problematic" situations to arise from the combination
of "use of an infinite amount of information" in a context where
that is AGAINST the "usual" policies.
Jesse F. Hughes
No, that's not enough.
Suppose that the light begins "on", at each step, I toggle the state.
It seems that you agree that after omega-many steps, we do not know
whether the light is on or off. But if we do not know at omega
whether the light is on or off, then surely we do not know whether it
is on or off at omega + 1.
Right?
To put it differently, you claim "any infinity of steps that has
a last element will have an answer to this question." w + 1 is an
"infinity of steps" with a last element, but if we have an answer at
w + 1, then we also have an answer at w.
--
Jesse F. Hughes
"Yes, I'm one of those arrogant people who tries to be quotable.
There is actually at least one person who quotes me often."
-- James Harris
Virgil
"Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> George Greene <gre...@email.unc.edu> writes:
>
> > On Nov 23, 9:48�am, Herman Jurjus <hjm...@hetnet.nl> wrote:
> >> More conclusive (at least for me): you switch a light bulb on and off;
> >> after infinitely many steps, is the light on or off?
> >
> > Well, there is more than one infinity.
> > Ordinally, there are even successor ordinals after
> > the first infinity. The reversed order (with no first element) can
> > also have a last one. Basically, any infinity of steps that has
> > a last element will have an answer to this question.
> > Any infinite sequence that does not have a last element
> > needs to get its "answer" from some NON-standard convention.
>
> No, that's not enough.
>
> Suppose that the light begins "on", at each step, I toggle the state.
> It seems that you agree that after omega-many steps, we do not know
> whether the light is on or off. But if we do not know at omega
> whether the light is on or off, then surely we do not know whether it
> is on or off at omega + 1.
>
> Right?
I'm not sure that going from omega to omega + 1 is a "step" in the same
sense as going from, say, 5 to 6 is a step.
Jesse F. Hughes
> In article <87pr77z...@phiwumbda.org>,
> "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> George Greene <gre...@email.unc.edu> writes:
>>
>> > On Nov 23, 9:48 am, Herman Jurjus <hjm...@hetnet.nl> wrote:
>> >> More conclusive (at least for me): you switch a light bulb on and off;
>> >> after infinitely many steps, is the light on or off?
>> >
>> > Well, there is more than one infinity.
>> > Ordinally, there are even successor ordinals after
>> > the first infinity. The reversed order (with no first element) can
>> > also have a last one. Basically, any infinity of steps that has
>> > a last element will have an answer to this question.
>> > Any infinite sequence that does not have a last element
>> > needs to get its "answer" from some NON-standard convention.
>>
>> No, that's not enough.
>>
>> Suppose that the light begins "on", at each step, I toggle the state.
>> It seems that you agree that after omega-many steps, we do not know
>> whether the light is on or off. But if we do not know at omega
>> whether the light is on or off, then surely we do not know whether it
>> is on or off at omega + 1.
>>
>> Right?
>
> I'm not sure that going from omega to omega + 1 is a "step" in the same
> sense as going from, say, 5 to 6 is a step.
Well, that seems like a step to me, whereas I'm not happy referring to
the omega'th step. But let's see what we're talking about.
Let's use the usual trick: For n in w, step n occurs at t - 1/n.
Let's also let step w + n occur at t + 1 - 1/n (for n > 0), so step
w + 1 occurs at time t. Thus, we have actions occurring at
t - 1, t - 1/2, t - 1/3, ... t, t + 1/2, t + 2/3, ....
Seems to me if we're willing to grant the original supertask, then
there's nothing less plausible about the new supertask.
At t - 1, the bulb is on. Immediately after t - 1 (and up to and
including t - 1/2), the bulb is off. Immediately after t - 1/2, it's
on again, and so on.
Now, according to George, we should have no trouble figuring out what
happens immediately after time t, if I read him right. But in the
original task (when nothing was done *at* t), we didn't know the state
of the bulb at t. I don't see how we know the state of the bulb
immediately after t either.
--
Jesse F. Hughes
"Marriage.. ..is the union of two persons of different sex for
life-long reciprocal possession of their sexual faculties."
-- Immanuel Kant, who died an unmarried virgin
master1729
> -- James
> -- James Harris
indeed.
and that one person who quotes JSH is JFH.
tommy1729
Jesse F. Hughes
>> "Yes, I'm one of those arrogant people who tries to
>> be quotable.
>> There is actually at least one person who quotes me
>> often."
>> --
>> --
>> -- James
>> -- James Harris
>
> indeed.
>
> and that one person who quotes JSH is JFH.
You know what makes a good joke a great joke?
When someone takes the time to explain it.
--
Jesse F. Hughes
"We will run this with the same kind of openness that we've run
Windows." Steve Ballmer, speaking about MS's new ".Net" project.
Daryl McCullough
>Let's use the usual trick: For n in w, step n occurs at t - 1/n.
>Let's also let step w + n occur at t + 1 - 1/n (for n > 0), so step=20
>w + 1 occurs at time t. Thus, we have actions occurring at
>
>t - 1, t - 1/2, t - 1/3, ... t, t + 1/2, t + 2/3, ....
>
>Seems to me if we're willing to grant the original supertask, then
>there's nothing less plausible about the new supertask.
>
>At t - 1, the bulb is on. Immediately after t - 1 (and up to and
>including t - 1/2), the bulb is off. Immediately after t - 1/2, it's
>on again, and so on.
>
>Now, according to George, we should have no trouble figuring out what
>happens immediately after time t, if I read him right. But in the
>original task (when nothing was done *at* t), we didn't know the state
>of the bulb at t. I don't see how we know the state of the bulb
>immediately after t either.
It's not specified by the "theory" of light switches. What
we are assuming about light switches is that if it is switched
on at time t_1, and it is not switched off between t_1 and t_2, then
it is still on at time t_2. If it is switched off at time t_1,
and is not switched on between t_1 and t_2, then it is still
off at time t_2.
Now, we add to this the assumption that the light switch is
turned on at times t - 1, t - 1/3, t - 1/5, ..., and that
is switched off at times t - 1/2, t - 1/4, ...
So we have a bunch of statements about what the state of
the light is at various times, and those statements don't
determine what the state of the light is at time t.
--
Daryl McCullough
Ithaca, NY
T.H. Ray
> Jesse F. Hughes wrote:
> > Herman Jurjus <hjm...@hetnet.nl> writes:
> >
> >> Apparently there's something wrong with backward
> supertasks (and not
> >> with ordinary, 'forward' supertasks). But why
> should that be?
> >
> > Well, I'm not at all sure that there's no problem
> with forward
> > supertasks. Surely, it is not difficult to come up
> with a
> > problematic case.
> >
> > For instance, take our favorite example: at each
> time t - 1/n, place
> > balls 10(n-1) to 10n - 1 in a vase and then remove
> ball n. At the end
> > the vase is empty.
> >
> > Now alter the situation slightly. At each step,
> again place 10 balls
> > into the vase and then remove one ball, but remove
> the ball
> > *randomly*. At the end, the vase may contain any
> number of balls.
> > This strikes me as suitably counterintuitive to say
> that the forward
> > supertask has something wrong with it. Or,
> perhaps, with my
> > intuitions.
>
> More conclusive (at least for me): you switch a light
> bulb on and off;
> after infinitely many steps, is the light on or off?
> (Or: you put one
> and the same ball in the vase, out of the vase, in
> the vase, out of the
> vase, etc. What's in the vase/where's the ball after
> infinitely many steps?)
> If you can't trust supertask-reasoning in this case,
> why should you
> trust it in other, seemingly less problematic cases?
>
Infinity isn't a number. Infinitely many steps have no
context in this problem. The problem description fixes
the domain at countably infinite; however, even that term
has no meaning in the absence of a higher cardinality.
One has to mean, in this context, an arbitrarily large
number of steps. In order to answer the question, though,
one requires infinite information. That is, in an
arbitrarily long string of 0s and 1s representing the
state of the lamp at a discrete moment, one must be able
to identify a segment of a certain magnitude and inspect
the endpoint for the state at that singular moment. A
little reflection informs us that information is not
discrete--even though moments of time are countable; i.e.,
however short or long the segment, the initial condition
(off or on) determines the final state (odd or even
number). If there is no final state, there is no certain
answer.
The deep implications of discretely countable moments
containing infinite information have led some researchers
(Lev Goldfarb for one, I for another--though by very
different routes) to the published conclusion that time
and information are identical. A geometric, or pre-
geometric approach, as she would say, leads Fotini
Markopoulou to a theory that _only_ time, and not space,
exists.
In any case, though, the axiom of choice has no relevance
to a question in which time plays the central role,
because even though one may choose arbitrary initial
conditions, such a choice compels no information that
lies outside that model and which imposes a non-arbitrary
moment on both our physical and mathematical experience,
no matter whether we deem such a moment arbitrarily short
or arbitrarily long.
Tom
LauLuna
>
> I'm not sure yet what to conclude from it; that AC is horribly wrong, or
> that WM is horribly right, or something else altogether.
>
> In short the story goes like this:
>
> A game is played, in which infinitely many coins are tossed, and there's
> one player, who makes infinitely many guesses. Both are done over a
> finite period of time. The tosses and guesses are not made faster and
> faster, however, but slower and slower: at t = 1/n. There's no 'first'
> move.
>
> There exists a strategy with which you're certain to guess all entries
> correctly except for at most finitely many mistakes. Not 'certain' as in
> 'probability is 100%', but absolutely certain.
>
> Reasoning:
> On 2^w, consider the equivalence relation that makes x equivalent to y
> when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a
> set S that contains precisely one element from every equivalence class.
> Strategy: at every move, you already know the results of the previous
> tosses, which is an infinite tail of some sequence in 2^w. Now take the
> unique element from S associated to that tail, take the n'th element of
> that sequence from S, and deliver that as your move.
> After some thinking, you will see that with this strategy, you're indeed
> certain to guess wrong at most finitely many times.
>
> Thanks, AC! Another nice mess you've gotten us into.
>
> Cheers,
> Herman Jurjus
In fact, you don't need the axiom of choice.
At any 1/n hour past 12pm it is already determinate what equivalence
class the eventual sequence is in. Since you know what the previous
results are and there are only finitely many outstanding results, you
can complete the sequence at random and take it as your
representative. Which means that from any 1/n hr past 12pm on you can
guess at random.
That is, you can make all your guesses at random and still be certain
to guess wrong only finitely many times.
I'd say the paradox arises from the assumption that an infinite
sequence of past events is terminated. So it seems akin to some
versions of Benardete's and Yablo's paradoxes.
The following is a version of Benardete's. Consider an infinite past
with a gong peal occurring at each day and a hearer being deafened by
it iff no previous gong peal has deafened it. The hearer must be deaf
from eternity, which, paradoxically, implies that no gong peal deafens
it.
The problem is that at any day it is already determinate whether the
hearer is deaf or not and in such a way that no event at no day can
contribute to the fact.
Similarly, what the equivalence class of the eventual sequence is, is
determinate at any 1/n, no matter how big n is; hence no toss at no 1/
n contributes to the fact. Of course, this is paradoxical.
Regards.
Mike Terry
news:36cb4819-04fd-4dbe...@p8g2000yqb.googlegroups.com...
..but then how would you prove that there have been only a finite number of
incorrect guesses? If the guesses always follow the representative sequence
(whose existence follows from AC) it's this that allows us to deduce at the
end that we've only made finitely many mistakes.
>
> That is, you can make all your guesses at random and still be certain
> to guess wrong only finitely many times.
>
Your proof for this doesn't work...
Mike.
Tim Little
> At any 1/n hour past 12pm it is already determinate what equivalence
> class the eventual sequence is in. Since you know what the previous
> results are and there are only finitely many outstanding results,
> you can complete the sequence at random and take it as your
> representative. Which means that from any 1/n hr past 12pm on you
> can guess at random.
That strategy fails for the same reason as the previous non-choice
strategy. The choice sequence works only because you provably used
the same sequence in the *past*, and hence at the time of your
decision had already made only finitely many errors.
> That is, you can make all your guesses at random and still be
> certain to guess wrong only finitely many times.
No, in your case you can't prove anything at all about how many of
your past guesses were correct.
> The following is a version of Benardete's. Consider an infinite past
> with a gong peal occurring at each day and a hearer being deafened by
> it iff no previous gong peal has deafened it. The hearer must be deaf
> from eternity, which, paradoxically, implies that no gong peal deafens
> it.
Yes, mixing in other types of paradox is one reason why I prefer other
formulations of this kind of AC problem.
- Tim
Bill Taylor
> Well, I'm not at all sure that there's no problem with forward
> supertasks. Surely, it is not difficult to come up with a
> problematic case.
Yes, the worth of supertasks as indicators of philosophical
concerns is very much up in the air. Some seem relevant, others
just stupid. Perhaps (temporally) well-ordered supertasks are
more sensible than most. But I doubt that's all there is to it.
One of my favourites is this; for naturals n:- compare...
a) At each time 1 - 1/n, add balls numbered
2^(n-1) to 2^n - 1 to the pot, and remove ball number n.
b) At each time 1 - 1/n, add 2^(n-1) - 1 balls to the pot, and
replace the numbering stickers in agreement with case (a).
After time 1:
the final situation in case (a) is that the pot is empty.
in case (b), the pot has infinitely many balls with no stickers!
And yet at any intermediate time the two cases are indistinguishable!
This sort of example shows that even omega-supertasks
can be remarkably silly!
> Now alter the situation slightly. At each step, again place 10 balls
> into the vase and then remove one ball, but remove the ball
> *randomly*. At the end, the vase may contain any number of balls
Actually NOT. The pot will be empty(!) [with probability 1]
For any ball, the probability of it being removed is like a harmonic
series, which sums to oo, which by Borel-Cantelli means it will
happen for sure. [meaning probability one, as always here]
HOWEVER - if you add (say) 1, 4, 9, 16... balls per turn,
and again remove one at random, each turn, then (Borel-Cantelli)
each ball has a positive probability of being left behind.
It is an interesting problem in probability generating functions
to work out the individual proabilities for each ball,
and the expected number, left in the pot at the end!
-- Borellic Bill
Bill Taylor
> with ordinary, 'forward' supertasks). But why should that be?
So not only can Achilles not catch up with the tortoise,
but he can't even get off the starting line!! :)
-- Bionic Bill
Daryl McCullough
>One of my favourites is this; for naturals n:- compare...
>
>a) At each time 1 - 1/n, add balls numbered
> 2^(n-1) to 2^n - 1 to the pot, and remove ball number n.
>
>b) At each time 1 - 1/n, add 2^(n-1) - 1 balls to the pot, and
> replace the numbering stickers in agreement with case (a).
>
>After time 1:
> the final situation in case (a) is that the pot is empty.
> in case (b), the pot has infinitely many balls with no stickers!
>
>And yet at any intermediate time the two cases are indistinguishable!
>
>This sort of example shows that even omega-supertasks
>can be remarkably silly!
Actually, I think that they are fun to think about.
What this example (and similar ones) show is that
for tasks involving a transfinite number of steps,
you have to be more precise about what you are doing.
In order for mathematics to tell us what the result
is of performing some supertask, you have to describe
the supertask using standard mathematical objects
(typically sets). There are modeling choices that
have to be made, and the answer can depend on the
modeling choice. If it does depend on the modeling
choice, that means that the original supertask was
insufficiently specified.
The big ambiguity is how to compute limit states.
Associated with each ordinal alpha, there is a
corresponding state of the system, S_alpha. The
statement of the supertask explains how to go
from S_alpha to S_{alpha+1}. But that tells us
nothing about the state S_alpha when alpha is
a limit ordinal.
There are certain assumptions about the limit
states that are so "obvious" that they seem
to go without saying. For instance: "If a
ball is added at stage alpha_1, and is never
removed at any stage beta such that
alpha_1 < beta < alpha_2, and alpha_2 is a
limit ordinal, then the ball is present at
stage alpha_2. But that's an assumption about
the limit state. To really reason about supertasks,
you have to state all the assumptions about
limit states.
>
>> Now alter the situation slightly. At each step, again place 10 balls
>> into the vase and then remove one ball, but remove the ball
>> *randomly*. At the end, the vase may contain any number of balls
>
>Actually NOT. The pot will be empty(!) [with probability 1]
>
>For any ball, the probability of it being removed is like a harmonic
>series, which sums to oo, which by Borel-Cantelli means it will
>happen for sure. [meaning probability one, as always here]
>
>HOWEVER - if you add (say) 1, 4, 9, 16... balls per turn,
>and again remove one at random, each turn, then (Borel-Cantelli)
>each ball has a positive probability of being left behind.
Very interesting.
WM
> It is an interesting problem in probability generating functions
> to work out the individual proabilities for each ball,
> and the expected number, left in the pot at the end!
Here is another interesting task: Use balls representing the positive
rationals. The first time fill in one ball. Then fill in always 100
balls and remove 100 balls, leaving inside the ball representing the
smallest of the 101 rationals. If you get practical experience, you
will accomplish every Centuria in half time. So after a short while
you will have found the smallest positive rational.
Instead of 100 balls you can use only 80 balls as well. The Romans
also used to have 80 men in a Centuria- --- That pointed already to
the final period of the empire.
Of course the game is only possible, if there is an actual infinity
waiting to be finished. --- And that points to the final period of
mathematics.
Regards, WM
William Hughes
> Here is another interesting task: Use balls representing the positive
> rationals. The first time fill in one ball. Then fill in always 100
> balls and remove 100 balls, leaving inside the ball representing the
> smallest of the 101 rationals.
[at random with any measure that gives a positive probability
to each rational]
> If you get practical experience, you
> will accomplish every Centuria in half time. So after a short while
> you will have found the smallest positive rational.
Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim
you will have an empty set.
- William Hughes
WM
> On Nov 26, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Here is another interesting task: Use balls representing the positive
> > rationals. The first time fill in one ball. Then fill in always 100
> > balls and remove 100 balls, leaving inside the ball representing the
> > smallest of the 101 rationals.
>
> [at random with any measure that gives a positive probability
> to each rational]
Simply take the first, seconde, third ... Centuria according to
Cantor's well-ordering of the positive rationals. Then there is no
need for considering any probabilities.
>
> > If you get practical experience, you
> > will accomplish every Centuria in half time. So after a short while
> > you will have found the smallest positive rational.
>
> Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim
> you will have an empty set.
Besides your assertion, you have arguments too, don't you?
In particular you can explain, how the empty set will emerge while
throughout the whole time the minimum contents of the vase is 1 ball?
Regards, WM
LauLuna
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "LauLuna" <laureanol...@yahoo.es> wrote in message
>
> - Show quoted text -
At any 1/n point you are guessing according a representative of the
equivalence class of the eventual sequence, if you just complete the
sequence of the previous results with random guesses for the finite
number of the future ones.
LauLuna
> On 2009-11-25, LauLuna <laureanol...@yahoo.es> wrote:
>
> > At any 1/n hour past 12pm it is already determinate what equivalence
> > class the eventual sequence is in. Since you know what the previous
> > results are and there are only finitely many outstanding results,
> > you can complete the sequence at random and take it as your
> > representative. Which means that from any 1/n hr past 12pm on you
> > can guess at random.
>
> That strategy fails for the same reason as the previous non-choice
> strategy. The choice sequence works only because you provably used
> the same sequence in the *past*, and hence at the time of your
> decision had already made only finitely many errors.
>
> > That is, you can make all your guesses at random and still be
> > certain to guess wrong only finitely many times.
>
> No, in your case you can't prove anything at all about how many of
> your past guesses were correct.
In order to reveal the similarity between this and Benardete's
paradox, put it this way:
First note that what the relevant equivalence class is, is already
determinate at all 1/n points.
Then assume that at any 1/n point I construct it -by guessing at
random the future results- iff I haven't constructed it at any 1/m
with m>n. Certainly, I possess a representative at all 1/n points and
certainly I construct or choose it at no 1/n point: so I inherit it,
just as Benardete's hearer inherits his deafened ears.
So, having no need to construct or choose the representative at any 1/
n point, I don't need AC.
As I see it, the essential paradox here is that the relevant
equivalence class is already fixed at each 1/n point, so that no toss
contributes to its determination.
Regards.
LauLuna
>
> - Show quoted text -
This answer of mine is wrong. That does not ensure anything. My point
is made in my Nov 27 response to Tim Little's Nov 26 post. I reproduce
it for courtesy:
"In order to reveal the similarity between this and Benardete's
paradox, put it this way:
First note that what the relevant equivalence class is, is already
determinate at all 1/n points.
Then assume that at any 1/n point I construct it -by guessing at
random the future results- iff I haven't constructed it [in that same
way]* at any 1/m with m>n. Certainly, I possess a representative at
all 1/n points and certainly I construct or choose it at no 1/n point:
so I inherit it, just as Benardete's hearer inherits its deafened
ears.
So, having no need to construct or choose the representative at any 1/
n point I don't need AC.
As I see it, the essential paradox here is that the relevant
equivalence class is already fixed at each 1/n point, so that no toss
contributes to its determination."
*The [] phrase was added.
Best.
LauLuna
> Of course the game is only possible, if there is an actual infinity
> waiting to be finished. --- And that points to the final period of
> mathematics.
Yes, but that existence is not enough for the game to become possible.
Regards.
Tim Little
> First note that what the relevant equivalence class is, is already
> determinate at all 1/n points.
>
> Then assume that at any 1/n point I construct it -by guessing at
> random the future results- iff I haven't constructed it at any 1/m
> with m>n.
Hence you contradict your original strategy which specified that you
guess at random at every 1/n point. This new strategy is exactly the
AC strategy. It cannot be proven in ZF alone that there exists a set
of representatives covering all equivalence classes, so the strategy
can fail if AC is not assumed.
> As I see it, the essential paradox here is that the relevant
> equivalence class is already fixed at each 1/n point, so that no
> toss contributes to its determination.
Have you seen the "infinite line of prisoners with hats" version
illustrating the same use of AC? In this version, there is an queue
of prisoners, having a back of the queue but extendingly infinitely
forward. Each prisoner can see all the colours of hats worn by those
ahead.
Their task is to guess their own hat colour; if only finitely many
guess incorrectly, they all go free. They get to confer beforehand,
but cannot communicate with each other in any way after the hats are
placed on their heads, and they only get one guess.
In this formulation, there is no question of inheritance or tasks that
cannot be begun.
- Tim
William Hughes
Since outside of Wolkenmuekenheim there is no reason to
expect the number of balls to be continuous at infinity
outside of Wolkenmuekenheim there is no problem.
Inside of Wolkenmuekenheim we can use "logic" like
All FISONS have a fixed last element.
N does not have a fixed last element.
N is a FISON
and prove anything.
- William Hughes
A
Let S denote a set with exactly 101 elements. Let Q+ denote the
positive rational numbers. Let inj(S,Q+) denote the set of injective
functions from S to Q+. Let {x_n} denote a sequence of elements of inj
(S,Q+) with the following properties:
1. Let im x_n denote the image of x_n. Then the union of im x_n for
all n is all of Q+.
2. For any n, the intersection of im x_n with im x_(n+1) consists of
exactly one element, which is the minimal element (in the standard
ordering on Q+) in im x_n.
Let X denote the subset of Q+ defined as follows: a positive rational
number x is in X if and only if there exists some positive integer N
such that, for all M > N, x is in the image of x_M.
We are talking about X, right? This is the set of balls which I, and
others, claim is empty, and WM claims is non-empty? Or does WM dispute
one of the definitions above?
WM
But it is enough for the game to be discussed as I did. The result is
that the result is impossible (must be empty set but cannot be empty
set). And that is the same as in case of any bijection including an
infinite set like N or Q or both.
Regards, WM
WM
We are talking about a vase which is never emptied completely!
Hence it cannot be empty unless "infinity" is identical to "never".
But this describes potential infinity and excludes phantasies like
Cantor's finished diagonal number.
Regards, WM
Regards, WM
WM
> On Nov 26, 4:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 26 Nov., 19:22, William Hughes <wpihug...@hotmail.com> wrote:
> > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim
> > > you will have an empty set.
>
> > Besides your assertion, you have arguments too, don't you?
> > In particular you can explain, how the empty set will emerge while
> > throughout the whole time the minimum contents of the vase is 1 ball?
>
> Since outside of Wolkenmuekenheim there is no reason to
> expect the number of balls to be continuous at infinity
Why then do you expect the digits of Cantor's diagonal number to be
"continuous" at infinity (contrary to being *not* at infinity)?
Regards, WM
Virgil
<aa9e46c0-56da-4510...@b2g2000yqi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Why would anyone ever expect a numerical digit to be continuous?
All the ones I am aware of are members of a finite set of discrete
objects.
And why would you expect to find a digit of any sort "at infinity", when
there is no such a position as "at infinity".
Virgil
<9d0132ac-2c6b-447f...@s20g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Maybe in your muecked up worlds, but, fortunately, mathematics does not
occur in such worlds.
WM
> In article
> <aa9e46c0-56da-4510-8345-8dee84745...@b2g2000yqi.googlegroups.com>,
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Nov., 02:50, William Hughes <wpihug...@hotmail.com> wrote:
> > > On Nov 26, 4:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > On 26 Nov., 19:22, William Hughes <wpihug...@hotmail.com> wrote:
> > > > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim
> > > > > you will have an empty set.
>
> > > > Besides your assertion, you have arguments too, don't you?
> > > > In particular you can explain, how the empty set will emerge while
> > > > throughout the whole time the minimum contents of the vase is 1 ball?
>
> > > Since outside of Wolkenmuekenheim there is no reason to
> > > expect the number of balls to be continuous at infinity
>
> > Why then do you expect the digits of Cantor's diagonal number to be
> > "continuous" at infinity (contrary to being *not* at infinity)?
>
> Why would anyone ever expect a numerical digit to be continuous?
>
> All the ones I am aware of are members of a finite set of discrete
> objects.
And there is none that does not belong to a rational number.
>
> And why would you expect to find a digit of any sort "at infinity", when
> there is no such a position as "at infinity".
If there is no "at infinity", then there cannot be a "behind
infinity", so there is no omega and no omega + 1.
In fact you are right - as so often. There is no "at infinity". The
vase is never empty. There is no smallest positive rational, There are
not all rationals.
Regards, WM
Alan Smaill
> We are talking about a vase which is never emptied completely!
>
> Hence it cannot be empty unless "infinity" is identical to "never".
> But this describes potential infinity and excludes phantasies like
> Cantor's finished diagonal number.
But you lose control at infinity!
So your "hence" doesn't work.
I have that on good authority.
> Regards, WM
>
> Regards, WM
--
Alan Smaill
William Hughes
> On 27 Nov., 02:50, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Nov 26, 4:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 26 Nov., 19:22, William Hughes <wpihug...@hotmail.com> wrote:
> > > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim
> > > > you will have an empty set.
>
> > > Besides your assertion, you have arguments too, don't you?
> > > In particular you can explain, how the empty set will emerge while
> > > throughout the whole time the minimum contents of the vase is 1 ball?
>
> > Since outside of Wolkenmuekenheim there is no reason to
> > expect the number of balls to be continuous at infinity
>
<snip attempted change of subject>
The argument why the change is the number
of balls is not problem is simple. Outside Wolkenmuekenheim
There is a contradiction at the last step
There is no last step
There is no contradiction.
Of course inside Wolkenmuekenheim ANYTHING can change
(lists, sets, last elements, fixed last elements etc.)
There is a contradiction at the last step.
There is a last step (it is something
that changes)
There is a contradiction.
- William Hughes
WM
> WM <mueck...@rz.fh-augsburg.de> writes:
> > We are talking about a vase which is never emptied completely!
>
> > Hence it cannot be empty unless "infinity" is identical to "never".
> > But this describes potential infinity and excludes phantasies like
> > Cantor's finished diagonal number.
>
> But you lose control at infinity!
>
> So your "hence" doesn't work.
It works if there is anyone who does not lose control at infinity.
That's enough.
Regards, WM
WM
*********************
you will have an empty set.
********************
>
> > > > Besides your assertion, you have arguments too, don't you?
> > > > In particular you can explain, how the empty set will emerge while
> > > > throughout the whole time the minimum contents of the vase is 1 ball?
>
> > > Since outside of Wolkenmuekenheim there is no reason to
> > > expect the number of balls to be continuous
*****************
at infinity
*****************
>
> The argument why the change is the number
> of balls is not problem is simple. Outside Wolkenmuekenheim
>
> There is a contradiction at the last step
> There is no last step
> There is no contradiction.
Therefore the vase is never empty - and the only error is your
assertion it would be empty after the last step and there was a state
of the vase "at infinity".
Regards, WM
Alan Smaill
> On 27 Nov., 11:43, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > We are talking about a vase which is never emptied completely!
> >
> > > Hence it cannot be empty unless "infinity" is identical to "never".
> > > But this describes potential infinity and excludes phantasies like
> > > Cantor's finished diagonal number.
> >
> > But you lose control at infinity!
> So does Cantor.
Good to know you agree that *you* lose control.
Let's leave Cantor out of it.
> > So your "hence" doesn't work.
>
> It works if there is anyone who does not lose control at infinity.
> That's enough.
If you are not in control, it doesn't work for you;
you can't say anything about what does or doesn't happen at infinity,
except by pretending that it must be the same as the finite case.
That pretence is worthless when you admit you have lost control.
A
The set X described above is certainly the empty set, as one can
easily prove, despite im x_n being nonempty for each n. This seems to
be a rigorous statement of what you are describing by talking about
balls, vases, etc. I have never seen any mention of "potential
infinity" or "completed infinity" which is precise enough to even be
considered mathematics, and talking about these ideas generally leads
to nothing better than confusion--it might be better to speak
precisely about the limits or sets that you mean, in any given
situation, rather than worrying whether they represent "potential
infinity" or "completed infinity."
Dik T. Winter
...
> Therefore the vase is never empty -
Similar in the sequence 1/n the elements value is never zero.
> and the only error is your
> assertion it would be empty after the last step and there was a state
> of the vase "at infinity".
Never heard about limits? We may consider the following for limits of
sets (X_n etc are sets):
lim sup X_n:
x in lim sup X_n if and only if x in infinitely many X_n
lim inf X_n:
x in lim inf X_n if there is some n0 such that x in X_n for n > n0
lim X_n:
exists when lim sup X_n = lim inf X_n and in that case is equal to
them.
You may verify that in the case you proposed lim X_n does exist and is equal
to the empty set.
Note:
1 = lim |X_n| != |lim X_n| = 0
but that should not come as a surprise.
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
William Hughes
> On 27 Nov., 14:48, William Hughes <wpihug...@hotmail.com> wrote:
>
> *********************
> you will have an empty set.
> ********************
>
>
>
> > > > > Besides your assertion, you have arguments too, don't you?
> > > > > In particular you can explain, how the empty set will emerge while
> > > > > throughout the whole time the minimum contents of the vase is 1 ball?
>
> > > > Since outside of Wolkenmuekenheim there is no reason to
> > > > expect the number of balls to be continuous
>
> *****************
> at infinity
> *****************
>
>
>
> > The argument why the change is the number
> > of balls is not problem is simple. Outside Wolkenmuekenheim
>
> > There is a contradiction at the last step
> > There is no last step
> > There is no contradiction.
>
> Therefore the vase is never empty
Nope, only in Wolkenmuekenheim, where we cannot
have all steps without a last step, do we have
The vase can only be empty if there is
a last step.
Ouside of Wolkenmeukenheim we can have all steps
with no last step so
The vase is empty after all steps.
- William Hughes
WM
So all steps have been done whereas the last is pending.
And if we leave the vase with a ball inside and without touching it?
After all none-steps have been done: the vase will be empty.
Matheology at its best.
Regards, WM
WM
Completed infinity is the same as actual infinity. You need it to
obtain a diagonal number. Without that the number would never get
finished.
Potential infinity is the infinity of mathematics. It has always been
used until matheology started.
> and talking about these ideas generally leads
>> to nothing better than confusion--it might be better to speak
> precisely about the limits or sets that you mean, in any given
> situation, rather than worrying whether they represent "potential
> infinity" or "completed infinity
You can talk about limits but you cannot talk about Cantor's diagonal
without actual infinity. If doing so, you get numbers that cannot be
named. This is the best proof showing that actual infinity is not to
be considered mathematics.
Regards, WM
WM
> In article <ac583807-eb9d-4ae4-86f2-37f329692...@k17g2000yqh.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> ...
> > Therefore the vase is never empty -
>
> Similar in the sequence 1/n the elements value is never zero.
Correct. But in the sequence 1,1,1... the limit is 1 with no doubt.
> You may verify that in the case you proposed lim X_n does exist and is equal
> to the empty set.
>
> Note:
> 1 = lim |X_n| != |lim X_n| = 0
> but that should not come as a surprise.
Let one ball rest in the vase in eternity. The limit will be 1.
Consider the sequence 1, 101, 1, 101, 1, 101, ... There is no limit.
Nevertheless the sequence of minima
1,1,1,... has limit 1 as before.
Therefore the result of set theory shows that set theory is not
mathematics.
Regards, WM
William Hughes
> So all steps have been done whereas the last is pending.
Nope. Outside of Wolkenmeukenheim
we can do all steps without doing a last step So all steps
have been done, however there is no last step so the last
is not pending.
- William Hughes
A
I do not know what "matheology" is, nor what "completed infinity,"
"actual infinity," and "potential infinity" are. Do these things
actually have rigorous definitions?
>
> > and talking about these ideas generally leads
> >> to nothing better than confusion--it might be better to speak
> > precisely about the limits or sets that you mean, in any given
> > situation, rather than worrying whether they represent "potential
> > infinity" or "completed infinity
>
> You can talk about limits but you cannot talk about Cantor's diagonal
> without actual infinity. If doing so, you get numbers that cannot be
> named. This is the best proof showing that actual infinity is not to
> be considered mathematics.
>
> Regards, WM
I do not know what "Cantor's diagonal" you are speaking of, but the
method of proving that the real numbers are uncountable which is
usually called "Cantor's diagonal argument" does not ever mention any
"actual infinity."
Perhaps you have some constructivist objections to mathematics as it
is practiced using ZFC set theory and the nonconstructive proofs it
allows. If that's so, then fine; Cantor's diagonal argument works fine
in ZFC but a strict constructivist, who insists on a more restrictive
set theory or a more restrictive logic, can find reasons to object to
it.
Virgil
<ee22fd89-2e1f-464d...@j19g2000yqk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Nov., 08:41, Virgil <Vir...@home.esc> wrote:
> > In article
> > <aa9e46c0-56da-4510-8345-8dee84745...@b2g2000yqi.googlegroups.com>,
> >
> >
> >
> >
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 27 Nov., 02:50, William Hughes <wpihug...@hotmail.com> wrote:
> > > > On Nov 26, 4:51�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > On 26 Nov., 19:22, William Hughes <wpihug...@hotmail.com> wrote:
> > > > > > Only in Wolkenmuekenheim. �Outside of Wolkenmuekenheim
> > > > > > you will have an empty set.
> >
> > > > > Besides your assertion, you have arguments too, don't you?
> > > > > In particular you can explain, how the empty set will emerge while
> > > > > throughout the whole time the minimum contents of the vase is 1 ball?
> >
> > > > Since outside of Wolkenmuekenheim there is no reason to
> > > > expect the number of balls to be continuous at infinity
> >
> > > Why then do you expect the digits of Cantor's diagonal number to be
> > > "continuous" at infinity (contrary to being *not* at infinity)?
> >
> > Why would anyone ever expect a numerical digit to be continuous?
> >
> > All the ones I am aware of are members of a finite set of discrete
> > objects.
>
> And there is none that does not belong to a rational number.
There are several that do not belong to 1/3.
> >
> > And why would you expect to find a digit of any sort "at infinity", when
> > there is no such a position as "at infinity".
>
> If there is no "at infinity", then there cannot be a "behind
> infinity", so there is no omega and no omega + 1.
In the set of rationals, for example, there can be "before" and an
"after" without there being an "at", e.g. before and after sqrt(2). So
your claim requires proofs whch tou do not have.
>
> In fact you are right - as so often. There is no "at infinity". The
> vase is never empty.
The case starts empty, so WM is again trivially wrong.
> There is no smallest positive rational, There are
> not all rationals.
Which ones are not there?
Virgil
<d7d04c6c-84f7-4bc7...@c3g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Nov., 11:43, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > We are talking about a vase which is never emptied completely!
> >
> > > Hence it cannot be empty unless "infinity" is identical to "never".
> > > But this describes potential infinity and excludes phantasies like
> > > Cantor's finished diagonal number.
> >
> > But you lose control at infinity!
> >
> So does Cantor.
Cantor maintains a good deal more control with infiniteness that WM does.
>
> > So your "hence" doesn't work.
>
> It works if there is anyone who does not lose control at infinity.
Since you have just, in effect, claimed no one has, your "hence" fails
to work by your own argument.
> That's enough.
It is way too much of your nonsense.
>
> Regards, WM
Virgil
<ac583807-eb9d-4ae4...@k17g2000yqh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> Therefore the vase is never empty - and the only error is your
> assertion it would be empty after the last step and there was a state
> of the vase "at infinity".
While there is no 'last step' in this infinite sequence of steps, there
is an 'after all steps' because there is a time at which every step has
been completed.
So WM is WRONG
AGAIN
AS USUAL.
WM
Whatever you may think. If you wish to argue that the sequence of
balls has limt 0 then you are outside of mathematics. And there is no
need to further answer your "arguments".
Regards, WM
WM
modern logic and set theory, a mixture between mathematics and
theology. The latter is prevailing.
, nor what "completed infinity,"
> "actual infinity," and "potential infinity" are. Do these things
> actually have rigorous definitions?
>
That depends on your understanding of rigor.
>
> > > and talking about these ideas generally leads
> > >> to nothing better than confusion--it might be better to speak
> > > precisely about the limits or sets that you mean, in any given
> > > situation, rather than worrying whether they represent "potential
> > > infinity" or "completed infinity
>
> > You can talk about limits but you cannot talk about Cantor's diagonal
> > without actual infinity. If doing so, you get numbers that cannot be
> > named. This is the best proof showing that actual infinity is not to
> > be considered mathematics.
>
> > Regards, WM
>
> I do not know what "Cantor's diagonal" you are speaking of, but the
> method of proving that the real numbers are uncountable which is
> usually called "Cantor's diagonal argument" does not ever mention any
> "actual infinity."
Nevertheless it requires actual infinity.
>
> Perhaps you have some constructivist objections to mathematics as it
> is practiced using ZFC set theory and the nonconstructive proofs it
> allows. If that's so, then fine; Cantor's diagonal argument works fine
> in ZFC but a strict constructivist, who insists on a more restrictive
> set theory or a more restrictive logic, can find reasons to object to
> it.-
A strict constructivist denies actaul infinity and, hence, uncountable
sets.
With only potential, i.e., not finished infinity, i.e., reasonable
infinity, the diagonal number (exchanging 0 by 1) of the following
list can be found in the list as an entry:
0.0
0.1
0.11
0.111
...
Thus Cantor's method fails - here and everywhere.
Regards, WM
WM
> In article
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > Therefore the vase is never empty - and the only error is your
> > assertion it would be empty after the last step and there was a state
> > of the vase "at infinity".
>
> While there is no 'last step' in this infinite sequence of steps, there
> is an 'after all steps' because there is a time at which every step has
> been completed.
Whatever you may think. If you wish to argue that the sequence of
balls has limit 0 then you are outside of mathematics. And there is
Virgil
<a02569c0-3b73-4c60...@b15g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Since Mueckenheim so seldom manages to get himself inside of
mathematics, he is hardly an authority on what goes on there.
N mathematics, it is quite possible, even common, for an infinite
sequence to be completed and have a limit which is not a member of the
sequence itself.
And if the problem of the balls and urn can be stated unambiguously, and
it has been, then its analysis can be done inside of mathematics, even
though it does not appear to be possible in Wolkenmeukenheim.
William Hughes
> On 27 Nov., 18:50, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Nov 27, 1:32 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > So all steps have been done whereas the last is pending.
>
> > Nope. Outside of Wolkenmeukenheim
> > we can do all steps without doing a last step So all steps
> > have been done, however there is no last step so the last
> > is not pending.
>
> Whatever you may think
<snip unsupported assertion>
An unsupported assertion by WM
will work in Wolkenmuekenheim,
but is of no use outside Wolkenmeukenheim.
- William Hughes
William Hughes
> With only potential, i.e., not finished infinity, i.e., reasonable
> infinity, the diagonal number (exchanging 0 by 1) of the following
> list can be found in the list as an entry:
>
> 0.0
> 0.1
> 0.11
> 0.111
> ...
Only in Wolkenmuekenheim where the argument goes
Every entry in the list has a fixed last 1
The diagonal number does not have a fixed last 1
The diagonal number is an entry in the list
- William Hughes
WM
> On Nov 27, 3:33 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > With only potential, i.e., not finished infinity, i.e., reasonable
> > infinity, the diagonal number (exchanging 0 by 1) of the following
> > list can be found in the list as an entry:
>
> > 0.0
> > 0.1
> > 0.11
> > 0.111
> > ...
>
> Only in Wolkenmuekenheim where the argument goes
>
> Every entry in the list has a fixed last 1
> The diagonal number does not have a fixed last 1
> [...]
Every diagonal number is in the list.
Regards, WM
William Hughes
> On 27 Nov., 21:17, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Nov 27, 3:33 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > With only potential, i.e., not finished infinity, i.e., reasonable
> > > infinity, the diagonal number (exchanging 0 by 1) of the following
> > > list can be found in the list as an entry:
>
> > > 0.0
> > > 0.1
> > > 0.11
> > > 0.111
> > > ...
>
> > Only in Wolkenmuekenheim where the argument goes
>
> > Every entry in the list has a fixed last 1
> > The diagonal number does not have a fixed last 1
>
> There is not a fixed last entry
So, every entry in the list has a fixed last 1.
(We don't need a fixed last entry to say this)
We still have
Every entry in the list has a fixed last 1
The diagonal number does not have a fixed last 1
> Every diagonal number is in the list.
Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim
there is only one diagonal number and it is not
in the list.
- William Hughes
Virgil
<e998c756-bb05-4370...@a21g2000yqc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
It is a area which does not exist outside of Wolkenmeukenheim, and does
not include much of either logic or of set theory
>
> , nor what Wolkenmeukenheim are. Do these things
> > actually have rigorous definitions?
> >
> That depends on your understanding of rigor.
Those who understand and apply rigor have no need for any of
Muekenheim's perverted notions of "completed infinity," "actual
infinity," and "potential infinity"
> >
> > > > and talking about these ideas generally leads
> > > >> to nothing better than confusion--it might be better to speak
> > > > precisely about the limits or sets that you mean, in any given
> > > > situation, rather than worrying whether they represent "potential
> > > > infinity" or "completed infinity
> >
> > > You can talk about limits but you cannot talk about Cantor's diagonal
> > > without actual infinity. If doing so, you get numbers that cannot be
> > > named. This is the best proof showing that actual infinity is not to
> > > be considered mathematics.
> >
> > > Regards, WM
> >
> > I do not know what "Cantor's diagonal" you are speaking of, but the
> > method of proving that the real numbers are uncountable which is
> > usually called "Cantor's diagonal argument" does not ever mention any
> > "actual infinity."
>
> Nevertheless it requires actual infinity.
It only requires a very simple an understanding of "infinite" which is
still way beyond Muekenheim's capabilities.
A non-empty set is finite if for any total ordering of its elements
there is a last element, and is infinite otherwise, i.e., is totally
orderable ssso that it does NOT have a last element.
> >
> > Perhaps you have some constructivist objections to mathematics as it
> > is practiced using ZFC set theory and the nonconstructive proofs it
> > allows. If that's so, then fine; Cantor's diagonal argument works fine
> > in ZFC but a strict constructivist, who insists on a more restrictive
> > set theory or a more restrictive logic, can find reasons to object to
> > it.-
>
> A strict constructivist denies actaul infinity and, hence, uncountable
> sets.
But mathematics is not in thrall to strict constructionalists.
>
> Thus Cantor's method fails - here and everywhere.
Only within the highly constraining boundaries of Wolkenmuekenheim.
Those who are not so constrained, like Cantor, do not all fail.
Virgil
<ddb474bb-2113-4cff...@d21g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Given any endless list of endless sequences from a two element set, such
as Cantor's {m,w}, at which position in that list is the "anti-diagonal'
to that list, Muekenheim?
Unless Muekenheim has a specific and correct answer, he gets a failing
grade.
WM
How do you know, unless you have seen the last?
Regards, WM
William Hughes
> > there is only one diagonal number and it is not
> > in the list.
>
> How do you know, unless you have seen the last?
>
You use induction to show that every entry
in the list has a final 1. So you don't have
to see an entry to know that it has a final 1
(there is no last entry in any case).
There is a constructive proof that the
diagonal number does not have a final 1.
- William Hughes
Virgil
<f4e15df0-a3c0-48e4...@j4g2000yqe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Given a specific list of endless binary sequences, the so called Cantor
diagonal is the result of a specific and unambiguous algorithm applied
to that list, so it is, for any given list, unique, and not a member of
the list from which it is constructed.
Which WM would have known if he had any sense.
Ross A. Finlayson
> In article
Binary sequences aren't unique representations of real numbers.
(Binary and ternary (trinary) anti-diagonal cases require refinement.)
For example, the list contains .1 then all zeros, the anti-diagonal
is .0111... = .100..., anti-diagonal is on the list.
Ross F.
Virgil
<e70292e5-e110-48dc...@g1g2000pra.googlegroups.com>,
The original Cantor diagonal argument did not deal with real numbers
either, so what is your point?
> (Binary and ternary (trinary) anti-diagonal cases require refinement.)
But as neither I nor Cantor were not dealing with numbers in any base,
your objections are, as usual, irrelevant.
>
> For example, the list contains .1 then all zeros, the anti-diagonal
> is .0111... = .100..., anti-diagonal is on the list.
But, in the Cantor argument, the lists in question are not of functions
from N to range {0,1} but of functions from N to range {m,w} with no
assumption that such a function corresponds to any sort of number.
So Ross is, as usual, in over his head.
>
> Ross F.
WM
Use induction to show that the diagonal number cannot have more digits
than every entry of the list.
There is a constructive proof that the list does not have a final
entry.
Reagrds, WM
WM
> In article
They correspond to sequences of w's m's. And WM has shown, that every
initial sequence of the diagonal is in a Cantor's list.
Now, you can believe that Cantor's argument (the diagonal is not in
the list) is as powerful as my argument (every initial sequence of the
diagonal is in the list - and the diagonal has no more symbols than
every initial sequence). Then you are a matheologian.
Or you can even believe that Cantor's argument outperforms my argument
and that there are more real numbers than can be identified. Then you
are what I call a Fool Of Matheology.
Regards, WM
Jesse F. Hughes
> Use induction to show that the diagonal number cannot have more
> digits than every entry of the list.
Great! Er, but how does induction *do* that?
--
Jesse F. Hughes
"Usenet is demonstrably dangerous. It needs to be regulated."
--James S. Harris, voice of reason and moderation
William Hughes
WM has conceded that you can use induction
to show that every element of the list has
a final 1, and that there is a constructive
proof that the diagonal number does not have a
final 1.
WM has a new argument.
> Use induction to show that the diagonal number cannot have more digits
> than every entry of the list.
This cannot be done. All you do is show that
every one of an infinite number of different
numbers, none of which is the diagonal number,
cannot have more digits than every entry of the list.
[Outside of Wolkenmuekenheim where the
diagonal number does not change.
Inside of Wolkenmuekenheim the diagonal
number changes, and everything it changes to has
fewer digits than some entry in the list]
- William Hughes
William Hughes
> WM <mueck...@rz.fh-augsburg.de> writes:
> > Use induction to show that the diagonal number cannot have more
> > digits than every entry of the list.
>
> Great! Er, but how does induction *do* that?
It is simple, Just move to Wolkenmuekenheim
where the diagonal number changes. Use induction
to show that whatever it changes to has fewer
digits than some entry in the list.
- William Hughes
Virgil
<d2007b71-d653-48fd...@m26g2000yqb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Any argument by Cantor is at least as "powerful", and usually a good
deal more logical than any argument by muekenheim.
Then you are a matheologian.
If Muekenheim despises being a matheologian, it must be a good thing to
be.
> Or you can even believe that Cantor's argument outperforms my argument
Any of Cantor's arguments "outperform" all of Muekenheim's.
Cantor was a mathematician, but muekenheim is merely a putzer at math,
and probably with everything else.
> Then you are what I call
Since Muekenheim's judgement on matters mathematical is so inept and
self contradictory and incoherent, being in his bad graces is a
compliment.
Ross A. Finlayson
> In article
The point was that your hasty overgeneralization was false and that it
represents in your non-acknowledgment hypocritical criticism.
> > (Binary and ternary (trinary) anti-diagonal cases require refinement.)
>
> But as neither I nor Cantor were not dealing with numbers in any base,
> your objections are, as usual, irrelevant.
>
No, it was just noted a specific constructive counterexample to that
lists of (expansions representing) real numbers don't contain their
antidiagonals.
>
>
> > For example, the list contains .1 then all zeros, the anti-diagonal
> > is .0111... = .100..., anti-diagonal is on the list.
>
> But, in the Cantor argument, the lists in question are not of functions
> from N to range {0,1} but of functions from N to range {m,w} with no
> assumption that such a function corresponds to any sort of number.
>
It was simply an example that a list of real numbers contains its anti-
diagonal because of dual representation of some standard Eudoxus/
Dedekind/Cauchy fixed radix expansion expressions of a real number.
> So Ross is, as usual, in over his head.
>
>
No. Why are you trying to bait and switch instead of simply
acknowledging that particular lambast was mistaken?
EF is a CDF.
Ross F.