Re: infinity

Randy Poe said:
>
> snapdragon31 wrote:
> > Dave Seaman wrote:
> > > On 7 Aug 2005 19:49:43 -0700, snapdragon31 wrote:
> > > You still have not answered my question, and you still have not provided
> > > a supporting argument of your own. I rest my case.
> > >
> >
> > Please don't do the dirty trick again. Your question is to ask for the
> > number of a ball remaining in the vase.
>
> It's not a dirty trick. The point is that every ball has a
> number n in 1, 2, ..., and the n-th ball is removed on the n-th step
> no matter what n you pick.
>
> That is the reasoning that says every ball is removed,
> therefore no balls remain at noon. The point of asking
> for which ball remains isn't to get you to name a
> specific ball, but to explain why there can be any
> ball which is not covered by "the n-th ball is removed
> on the n-th step".
The n+1th through the 10n+9th are present in the vase when the nth ball is
removed.
>
> > My answer is that I cannot
> > name the number of that ball.
>
> But what argument is there that such a ball exists?
The n+1th through the 10n+9th are present in the vase when the nth ball is
removed.
>
> > >From my last reply:
> > "Not be able to name the number of a ball remains in the vase does not
> > mean that the vase is empty. You can keep all the balls and only take
> > away whatever ball I name at the very last step.
>
> I don't understand.
The n+1th through the 10n+9th are present in the vase when the nth ball is
removed.
>
> > It does not prove
> > anything, right? I can prove that the number of balls grow
> > indefinitely, but you cannot prove that the vase is empty."
>
> Let n be the number of any ball put in the vase. Then
> n is not in the vase, because n was removed at step n.
The n+1th through the 10n+9th are present in the vase when the nth ball is
removed.
>
> That is a proof that the vase is empty.
Except that it's not.
>
> > > Yes, I am telling you that the vase becomes empty at noon, because B(0) =
> > > sum_{n=1^oo} B_n(0) = 0. Do the math.
> >
> > Please define what your symbols represent before I can do the math.
> > My guess would be:
> > B(0) - Number of balls in the vase at noon.
>
> No need to guess.
>
> Going back through the thread, I find this:
>
> >> If you had read my explanation elsewhere in the thread,
> >> you would have seen that I defined the function B(t) =
> >> sum_n B_n(t), where B_n is the characteristic function
> >> associated with ball n. Since B_n(0) = 0 for each n,
> >> it follows that B(0) = 0+0+0+... = 0.
>
> and this, also written by Dave Seaman:
>
> >> For each ball n, there is a time a_n when the ball is
> >> added to the vase, and a time b_n when the ball is removed,
> >> never to be returned. The exact values of a_n and b_n do
> >> not matter, but for the original version of the problem
> >> we may assume that a_n <= b_n < 0 for each n, and that the
> >> a_n form a nondecreasing sequence that approaches 0 (where
> >> t=0 represents noon).
> >>
> >> For each n there is a characteristic function B_n: R -> {0,1}
> >> such that B_n(t) is 1 if ball n is in the vase at time t, and
> >> 0 otherwise. In fact, we have
> >> Case I. Ball n is added at time n and removed at time b_n.
> >> -----------------------------------------------------------
> >> B_n(t) = 1, if a_n <= t < b_n,
> >> = 0, otherwise.
> >>
> >> We now define B(t) = sum_{n_1^oo} B_n(t). Thus B(t) expresses
> >> the number of balls in the vase at time t, and it is perfectly
> >> well defined for all t.
>
> > sum_(n=1^oo) - Your operation A, balls added to the vase before noon
> > B_n(0) - Your operation B, balls removed from the vase before noon
> >
> > If my guess is right,
>
> Your unnecessary "guess" is wrong.
Actually, it is almost anyone's guess what a given Cantorian is trying to say
at any given time.
>
> B_n(t) is an indicator of the presence of ball n in the
> vase at time t. B_n(0) is 1 if ball n is in the vase at
> time 0 (noon). But B_n(t) can only be 1 (ball n is
> in the vase at time t) if a_n <= t, i.e. the ball was
> put in before t, and b_n > t, i.e. the ball was removed
> after t.
>
> Since every ball is removed before noon at some time
> b_n < 0, no ball is present at noon.
The n+1th through the 10n+9th are present in the vase when the nth ball is
removed.
>
> > I gave a similar answer to someone. The detail is as follow
> > Let S1 be the set of balls added to the vase
> > S2 be the set of balls removed from the vase
> >
> > S1 = {1, 2, 3, ... , OO}
> > S2 = {1, 2, 3, ... , OO}
> >
> > Balls remains in the vase would be
> > S1 - S2 = OO - OO the result is undefined and is not necessary equal to
> > 0.
>
> That's because of the vagueness of the argument, not the
> vagueness of the answer.
Really? The how come you can't seem to agree on the answer?
>
> > The result can be anything. If you want the result to be 0 then you
> > need some kind of proof other than just saying it is.
>
> The above constitutes a proof: for any ball n, we can
> identify a time before noon at which it is removed.
Directly before which ten balls have been added.
>
> Which ball is not covered by that argument?
N+1 through 10n+9.
>
> - Randy
>
>

--
Smiles,

Tony
.

Relevant Pages

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