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Topic: § 516 On Ducks and Bathtubs
Replies: 1   Last Post: Jun 11, 2014 5:42 AM

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karl

Posts: 397
Registered: 8/11/06
Re: § 516 On Ducks and Bathtubs
Posted: Jun 11, 2014 5:42 AM
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Am 11.06.2014 11:12, schrieb WM:
> On Wednesday, 11 June 2014 10:55:54 UTC+2, karl wrote:
>
>

>>> Not my problem. It is set theory that claims that steps like omega + 1 exist and that all rationals can be enumerated.
>>
>
>> Please give a reference for this claim! This is only your imagination.
>
> [...] die ersten transfiniten Zahlen Cantors, die Zahlen der zweiten
> Zahlklasse, wie sie Cantor nennt. Zu ihnen gelangen wir also einfach
> durch ein Hinüberzahlen über das gewöhnliche abzählbare Unendlich, d.
> h. durch eine ganz naturgemäße und eindeutig bestimmte, konsequente
> Fortsetzung des gewöhnlichen Zählens im Endlichen. [D. Hilbert (1925),
> Quelle s. KB 090813]
>
>

>> Therefore all rational will appear in the list sometimes.
>
> The rationals are not "all" rationals. Cantor accomplishes only to enumerate the first few rationals


>as is proved by the fact that the not enumerated rationals cannot be enumerated

> in finite steps and cannot be enumerated thereafter.
>>


You are contradicting yourself. You said:

> s_1 = { q | 0 < q =< 1 } \ {q_1}

> s_2 = (s_1 U { q | 1 < q =< 2 }) \ {q_2}

> ...

> s_(n+1) = (s_n U { q | n < q =< n+1 }) \ {q_(n+1)}

> ... where the positive rationals q are enumerated a la Cantor.

If this enumeration does not contain every positive rational, give an example of such a number.


> the set of not enumerated rationals grows in every finite step and cannot shrink thereafter.

Who claims this?
You because you assume that somehow s_n is approaching lim s_n, which is not correct.
lim s_n is the set of all rational numbers which remain forever in the sets s_n, i.e. an element t for which there is a
k_0 such that for all k>k_0
always t \in s_k.
Tis set is empty. If you disagree, please name an element of this set.
So, no contradiction at all.

*) I repeat:
In a famous work of German romanticism (I leave it to you to find out which, since you like riddles) one episode is in
madhouse and it is written there:

Nro. 18 ist ein Rechenmeister, der die letzte Zahl finden will.

So you too are looking for the last number, all n's have gone, Hilberts hotel is full, no further guests can be
accommodated!

>
> Regards, WM
>





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