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

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Michael Klemm

Posts: 50
Registered: 11/13/12
Re: Matheology § 300
Posted: Jul 8, 2013 4:59 AM
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Julio Di Egidio wrote:

>> If you accept that 1/oo is a natural not
>> depending on the natural n, then the limit n->oo of
>> the constant sequence (1/oo, 1/oo, ...) is 1/oo.
>> The way how the number 1/oo is obtained
>> doesn't matter.

>
> Well, I do not see how that can be acceptable. The original expression
> has a sub-expression dependent on n under the limit, not just a constant
> value.


The original expression is lim_{n->oo} 1/Card(N\Fison(n)),
and n = oo ist not evaluated. Therefore all members of the considered
sequence
equal c = 1/oo = 0.

> Again, that the sequence is constant for all n in N does not entail that
> the limit for n->oo must be equal to that constant value.


Why not? You may state the lemma:
Let c be a real. Then the the sequence (c, c, c, ...) has the limit c,
in short lim_{n->oo} c = c.

In a second step you apply this lemma with c = 1/Card(N\Fison(n)), n < oo.


> Indeed, by your token:
>
> lim_{n->oo} Card(N\F(n)) = oo (1)


correct

> yet:
>
> lim_{n->oo} N\F(n) = {} (2)


correct

> or is that supposed to be N, too??

no

> At least, I suppose we'd agree that:
>
> lim_{n->oo} F(n) = N (3)


yes


> Of course, I am assuming that the limit of the cardinality of our
> well-founded sequence is equal to the cardinality of the limit of the
> sequence.


Why?

> I still do not see how we could define the limit otherwise (I may be
> missing formal details), but even less I can see (1) compatible with (2).


This is not only a question of defining the limit but also of the
cardinality.
The general set convergence is explained in
http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior
In the present case the outer limit equals the inner.

Concerning the cardinality one has:
Card(A) <= Card(B) iff there exists an injective function A -> B
Card(A) = Card(B) iff there exists a bijective function A -> B
and Card({}) = 0,
see
http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem

Regards
Michael
>
> Julio
>
>





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