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Topic: § 516 On Ducks and Bathtubs
Replies: 1   Last Post: Jun 11, 2014 8:07 AM

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 karl Posts: 397 Registered: 8/11/06
Re: § 516 On Ducks and Bathtubs
Posted: Jun 11, 2014 8:07 AM
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Am 11.06.2014 12:40, schrieb WM:
> On Wednesday, 11 June 2014 11:42:28 UTC+2, karl wrote:
>> 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]
>>
> Have you read and understood this?
>>
>
>> If this enumeration does not contain every positive rational, give an example of such a number.

No response, you can't.
>
>

>>
> s a
>>
>> k_0 such that for all k>k_0
>>
>> always t \in s_k.
>>
>> This set is empty. If you disagree, please name an element of this set.

No response, you can't.
>
>
> I can see only examples of numbers that leave the set s_n as enumerated rational q_n

>whereas infinitely many not enumerated rationals are added at the same time.

>If you claim that all can be enumerated,

>then you must find at least one, that leaves whereas none is added. Can you understand that?

No. you claim that the set lim s_n is not empty. The only thing you have to show is an element which is in it.
)
Your demand to find a set s_n - n finite!- with the properties above has nothing to do which the cardinality of the
limit set. *)

I repeat for all the intellectually challenged here:

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.

It is up to you to show that your claim is true !!!!!!!
So, no contradiction at all.

*) I repeat again :
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!

>

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