
Re: Matheology § 300
Posted:
Jul 8, 2013 4:59 AM


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 subexpression 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 > wellfounded 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 > >

