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Topic: Matheology � 300
Replies: 27   Last Post: Jul 9, 2013 2:50 PM

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Posts: 3,792
From: London
Registered: 2/8/08
Re: Matheology § 300
Posted: Jul 7, 2013 2:04 PM
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"Michael Klemm" <m_f_klemm@t-online.de> wrote in message
> Julio Di Egidio wrote:

>>>>> lim_{n->oo} 1/Card(N\Fison(n)) =
> = 1/Card(N\(lim_{n->oo} Fison(n))) =
> = 1/Card(N\N)=
> = 1/Card({}) =
> = 1/0
> = oo

>>>>> What am I doing wrong?
>>>> The first lim_{n->oo} refers to a sequence of naturals
>> But that is not a sequence of naturals:

> The expression lim_{n->oo} 1/Card(N\Fison(n))
> means:
> Limes of the sequence
> (1/Card(N\Fison(1)), 1/Card(N\Fison(2)),1/Card(N\Fison(3)),....) =
> (0,0,0,....), the value n = oo excluded.
> This limes is 0.

Again, that for all n in N the value is 1/oo just does not entail that the
limit is 1/oo.

> The expression
> Card(N\(lim_{n->oo} Fison(n)))
> means:
> Limes of the sequence
> ({2,3,4,...},{3,4,5,...},{4,5,6,...},...),
> {oo} not contained in the sequence and
> oo not contained in any of its members.

That's not an argumen against the identity I am assuming. On the contrary,
you have not supported your:

lim_{n->oo} 1/Card(N\F(n)) = lim_{n->oo} 1/oo

I.e. your "pre-emptive" substitution of Card(N\F(n)) with Card(N) = oo seems
unwarranted. That equality is true for all n in N, but it is not true in
the limit, where n->oo, and F(n) = N.

Indeed, here is how I'd print the sequence and its limit:

1/Card({1, 2, 3, 4, 5, ...})
1/Card({2, 3, 4, 5, ...})
1/Card({3, 4, 5, ...})

> This limit is by definition the set of all
> naturals contained in infinitively many members
> of the sequence. Thus, the limit is the empty set {}.
> Hence one obtains
> 1/Card(N\lim_{n->oo} Fison(n)) = oo.
> Therefore the error in the calculation is the first equation lim_{n->oo}
> 1/Card(N\Fison(n)) =
> = 1/Card(N\(lim_{n->oo} Fison(n))).

I'd insist that it's rather your

lim_{n-->oo} 1/Card(|N\FIS(n)) = lim_{n-->oo} 1/oo

that is incorrect. In fact, where's 'n' gone?


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