User Bjørn Kjos-Hanssen

1,672 Actions

answered

Aug 18, 2010 at 6:29

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accepted

Mar 15, 2010 at 21:10

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Nice to see you guys, and thanks for the good answers.

Mar 15, 2010 at 20:48

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Okay, so the proof uses the class $\omega$ but is noncommittal as to whether it is a set, that makes sense.

Mar 15, 2010 at 20:40

awarded

Teacher

Mar 15, 2010 at 9:50

answered

Mar 15, 2010 at 8:17

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awarded

Scholar

Mar 15, 2010 at 1:05

awarded

Supporter

Mar 15, 2010 at 1:05

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And the orbit exists because it is the intersection of all subsets of $A$ that are closed under $f$... nice.

Mar 15, 2010 at 0:56

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Francois points to maybe the key thing I was missing - how obvious is it that if there is a Dedekind infinite set then $\omega$ exists?

Mar 14, 2010 at 19:06

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Student

Mar 13, 2010 at 8:05

asked

Mar 13, 2010 at 7:35

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