Re: infinity

Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:

> Jesse F. Hughes said:
>> Half the time, you claim that the vase can't be empty because we never
>> remove a "last ball"[1]. So, I fix it so that we remove a last ball.
>> How can you complain?
>>
>> Is it that we didn't remove a second-to-last? And third-to-last? No
>> sweat. I'll fix it some more.
>>
>> 12:00 - 2^(-n) min. Put balls 10*n - 9, 10 * 9 - 8, ..., 10 * n
>> in. Take out ball 2n.
>> 12:00 + 2^(-n) min Take out ball 2n - 1.
>>
>> How about that? Now we remove a first, second, third ball...
>> and at noon we start removing the ... fourth from last, third from
>> last, second from last and last ball. Now surely the vase is empty.
>>
>> Right, Tony?
>>
>> Footnotes:
>> [1] Of course, you say about another thought experiment that the vase
>> *is* empty even though there is no last ball removed.
>>
>>
> You can't demonstrate that the balls we started with originally are all gone
> when you remove this "last" ball, so it's really a kludge, and an irrelevant
> waste of time. Nice try.

On the contrary, it is utterly trivial to demonstrate that every ball
put into the vase has been taken out in this and every other case.
Indeed, it's what we have been demonstrating over and over from the
beginning. Here: I'll do it now.

Every ball put in is labeled by a (positive) natural number and thus
can be written 2k or 2k-1 for some (positive) natural k . Ball 2k is
removed at 12:00 - 2^(-k) and ball 2k-1 is removed at 12:00 + 2^(-k),
so every ball put in is removed prior to 12:01.

Up until now, your response has been that if we don't remove a last
ball it can't be empty. This made little sense to me, but since I'm
such an accommodating guy, I revised the thought experiment to remove
a last ball, a next-to-last ball, a next-to-next-to-last ball and so
on.

Of course, now you claim (here and in another post) that the last ball
isn't relevant. You claim this because (1) I can tell you which ball
is last in this example and (2) in another example, you say the vase
is empty but can't identify a last ball removed.

Really, your response is pathetically unprincipled.

--
Jesse F. Hughes

"I want to really eat myself, so then I'll be a coalgebra."
-- Quincy P. Hughes, Age 3 1/2
.

Relevant Pages

  • Re: An uncountable countable set
    ... The only critical time dependency is that each ball to be inserted shall ... the vase is empty at noon of anything of any balls ... An affirmative answer confirms that the vase is empty at noon. ... given the times of insertions and removals. ...
    (sci.math)
  • Re: An uncountable countable set
    ... the vase, is consistent with the fact that no balls are removed at noon? ... The only relevant question is "According to the rules set up in the ... is each ball inserted before noon also removed before noon?" ... An affirmative answer confirms that the vase is empty at noon. ...
    (sci.math)
  • Re: An uncountable countable set
    ... Does Han claim that there is any ball put in that is not taken out? ... the vase empties". ... But in order for the vase to transition from not-empty ... If the vase ever became empty, ...
    (sci.math)
  • Re: infinity
    ... Number of balls in the vase at noon is f= OO. ... Unfortunately, if infinity gets involved, this statement alone is not sufficient to claim the vase is empty before noon. ... ball) is in the vase. ... You also keep ignoring my question. ...
    (sci.math)
  • Re: infinity
    ... "the vase is empty at noon". ... the vase at noon, let Rbe true if ball n was removed before noon. ... You got a concrete proof. ...
    (sci.math)