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Topic: Given a set , is there a disjoint set with an arbitrary cardinality?
Replies: 28   Last Post: Dec 4, 2012 5:50 PM

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 jaakov Posts: 11 Registered: 12/3/12
Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted: Dec 3, 2012 11:41 AM

On 03.12.2012 17:33, Aatu Koskensilta wrote:
> jaakov<removeit_jaakov@deleteit_ro.ru> writes:
>

>> Thank you. To finish this proof, we have to prove that your class, let
>> us call it Z, contains an initial segment of cardinality k. However,
>> informally speaking, X may hypothetically contain too many ordinals,
>> still being a set, such that too few ordinals remain in Z. Could you
>> please expand on why Z contains an initial segment of cardinality k?

>
> Assuming choice, the cardinality k of X is an (initial) ordinal, so
> take the set X' = {alpha | alpha< k} and let Y be the set {x + lambda |
> x in X'}, where lambda is an ordinal> any ordinal in X.
>

1. k is not related to the cardinality of X.

2. Your lambda need not exist. To show that lambda exists, one has to
show that a set may not contain arbitrarily large ordinals. Is it a
known fact or simply false?

Jaakov.

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Date Subject Author
12/3/12 jaakov
12/3/12 forbisgaryg@gmail.com
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Carsten Schultz
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Carsten Schultz
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 Butch Malahide
12/3/12 jaakov
12/3/12 Butch Malahide
12/4/12 jaakov
12/4/12 forbisgaryg@gmail.com
12/4/12 William Elliot
12/4/12 forbisgaryg@gmail.com
12/4/12 William Elliot
12/4/12 William Elliot
12/4/12 jaakov
12/4/12 William Elliot
12/4/12 jaakov
12/4/12 Shmuel (Seymour J.) Metz
12/4/12 Spammer
12/4/12 jaakov