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Sep 23, 2010 at 13:51 comment added Joel David Hamkins It was my pleasure! I think this precise topic is a good forum for practice with transfinite recursion and induction. It is also nice that the construction involves both $\aleph_1$ and $2^{\aleph_0}$, the first as the length of the iteration and the second as the size of the $\sigma$-algebra, and so one also learns their different natures.
Sep 23, 2010 at 13:43 comment added Joel David Hamkins No, I mean $\kappa^\omega$, or equivalently $\kappa^{\aleph_0}$. (The notational differences arise from viewing things as ordinals versus cardinals.) In your case, you have $\kappa=2^{\aleph_0}$, and so $\kappa^\omega=(2^{\aleph_0})^{\aleph_0}$, which is the same as $2^{\aleph_0}$. Although it is true that GCH simplifies cardinal arithmetic, and it is useful to consider a complex calculation first under the case of GCH as a rough estimation, it is the usual practice not to assume GCH in an argument without saying so.
Sep 23, 2010 at 13:39 vote accept zzzhhh
Sep 23, 2010 at 13:38 comment added zzzhhh Do you mean \kappa^{\omega_1} by \kappa^\omega? I studied set theory using Karel Hrbacek and Thomas Jech's "Introduction to Set Theory". In page 165 of this book, the author says "the Generalized Continuum Hypothesis greatly simplifies the cardinal exponentiation" so I alway take GCH for granted :-). The key is Lemma 3.6 and Theorem 3.8 in P166. Now I get it. I'm very sorry to forget to use these basic facts and 辜负了 this good book. Thanks you very much, Joel, and other people who read and reply, thank you all!
Sep 23, 2010 at 13:05 history edited Joel David Hamkins CC BY-SA 2.5
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Sep 23, 2010 at 12:56 comment added Joel David Hamkins I used the set-theorist notation $\omega_1$ for the first uncountable ordinal, which your text denotes $\Omega$.
Sep 23, 2010 at 12:54 history answered Joel David Hamkins CC BY-SA 2.5