Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 Aatu Koskensilta Posts: 2,639 Registered: 6/28/08
Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted: Dec 3, 2012 11:33 AM

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.

--
Aatu Koskensilta (aatu.koskensilta@uta.fi)

"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

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