Re: infinity


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".

> My answer is that I cannot
> name the number of that ball.

But what argument is there that such a ball exists?

> >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.

> 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.

That is a proof that the vase is empty.

> > 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.

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.

> 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.

> 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.

Which ball is not covered by that argument?

- Randy

.

Relevant Pages

  • Re: An uncountable countable set
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  • Re: infinity
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  • Re: An uncountable countable set
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  • Re: An uncountable countable set
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  • Re: infinity
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