LudovicoVan
Posts:
2,971
From:
London
Registered:
2/8/08
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Re: Matheology § 062
Posted:
Jul 14, 2012 1:48 PM
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"LudovicoVan" <julio@diegidio.name> wrote in message
news:jts46m$fd5$1@speranza.aioe.org...
> "LudovicoVan" <julio@diegidio.name> wrote in message
> news:jto0e6$4ah$1@speranza.aioe.org...
>> "LudovicoVan" <julio@diegidio.name> wrote in message
>> news:jtftlb$ala$1@speranza.aioe.org...
>>> "LudovicoVan" <julio@diegidio.name> wrote in message
>>> news:jtftcf$a23$1@speranza.aioe.org...
>>>> "William Hughes" <wpihughes@gmail.com> wrote in message
>>>> news:abd1cfc8-e1de-4504-bd7f-bac8fa5efc7a@h9g2000yqi.googlegroups.com...
>>>> <snip>
>>>>
>>>>> Balls only enter the vase at a step
>>>>> from the statement of the problem
>>>>>
>>>>> The only steps are n in N
>>>>> from the statement of the problem
>>>>>
>>>>> At every step n in N, the only balls that
>>>>> enter the vase are labeled 1+10(n-1) to 10n
>>>>> from the statement of the problem
>>>>
>>>> The statement of the problem talks about an "infinite supply of balls",
>>>> which justifies a broader approach in terms of ordinals (and the
>>>> distinction balls vs. finite/non-finite labels). Anyway, I am not any
>>>> good at arithmetic with ordinals, so here it is over N, i.e. your
>>>> setting. Of course, the "structure" we get is not as reach as that
>>>> given by ordinals and we can only conclude that the limit set is
>>>> countably infinite. (My apologies for any trivial mistakes.)
>>>>
>>>> Let N be the set of balls (w.l.o.g. as each ball is uniquely labeled by
>>>> a natural number here).
>>>>
>>>> Let n, m in N.
>>>>
>>>> Let V(n) be the vase at step n, defined by the rules of the game as:
>>>>
>>>> V(0) := { }
>>>> V(n) := V(n-1) U { m | 10n-9 <= m <= 10n } \ { m | m = n }
>>>>
>>>> Let V(w) be the limit set (the vase at time 0), defined as: V(w) :=
>>>> U_{n<w}.
>>>
>>> Should read: V(w) := U_{n<w} V(n).
>>
>> Also, should be: the vase at time t->0.
>>
>>>> Using some arithmetic, this set can be expressed as:
>>>>
>>>> V(w) = N \ { m in N | (m-1) = 0 (mod 10) }
>>>>
>>>> We can now define a bijection between V(w) and N by the functions:
>>>>
>>>> f(m) := m - ceil(m/10)
>>>> F(n) := n + ceil(n/9)
>>>>
>>>> QED.
>
> For more clarity, denote the limit set as V(n->w) (as opposed to V(n=w),
> which would be the extended setting).
>
> The proof is incorrect, namely in the arithmetic. Per the usual
> formulation of the problem, where it is ball n that exits at step n,
> directly expressing V(n->w) (by transfinite recursion) becomes impossible
> as the sequence V is not well-founded (or, I cannot see it).
>
> To work around this problem, we can first map the sequence V to a sequence
> W such that:
>
> For all n, |W(n)| = |V(n)|,
> and W is a well-founded sequence of sets.
>
> We can then express W(n->w), say as:
>
> W(n->w) = N \ { m | m-1 = 0 (mod 10) }
>
> As seen already, we then have that W(n->w) is countable.
>
> Finally, we conclude that V(n->w) is countable, too. QED?
>
> The objection, as I now seem to get it, is that:
>
> "For all n, |W(n)| = |V(n)|" does not entail "|W(n->w)| = |V(n->w)|", i.e.
> it does not entail that the two limit sets have the same cardinality.
The missing ingredient (I suppose, I am not a pro):
"Exists a mapping such that, for all n: |W(n)| = |V(n)| by that mapping"
does entail
"|W(n->w)| = |V(n->w)|"
QED.
> I still need to think and digest this, but I can note again that such
> objection is not a proof that |V(n->w)| = 0, or that |V(n->w)|->0, or of
> any such absurdity.
-LV
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