This question comes from notes to section 1.2 in page 4041 of Folland's "real analysis: modern techniques and their applications", 2nd edition. At the end of this note, the author asserts that card(M(E))=c, i.e., aleph_1, if E has cardinality aleph_0 or aleph_1, using Proposition 0.14. But in order for Proposition 0.14 to be applicable, each E_alpha in Proposition 1.23 must has cardinality less than or equal to aleph_1. I don't know how to obtain it from its construction. could you please help me with this problem? Thanks! This part of note is pasted below, together with Proposition 0.14. alt text http://i54.tinypic.com/2iiy2u.png
2 Answers
First, as Asaf notes, let me point out that you should be saying $2^{\aleph_0}$ or equivalently the continuum $c$ or $\beth_1$, rather than $\aleph_1$ in your question. The cardinal $\aleph_1$ is the first uncountable cardinal, which is only the same as the continuum $c$ if the Continuum Hypothesis holds.
Now, to your question. One begins with a family $E=E_0$ of size at most continuum, and then successively adds complements and countable unions. This idea can be made precise by transfinite recursion, as explained in the notes you cite. For any countable ordinal $\alpha$, we define $E_{\alpha+1}$ to consist of all countable unions of elements of $E_\alpha$ or their complements. And for a limit ordinal $\lambda$, we let $E_\lambda=\bigcup_{\alpha\lt\lambda} E_\alpha$. (One can combine these into one case by saying: $E_\alpha$ for $\alpha\gt 0$ consists of all countable unions of elements appearing before $\alpha$ and their complements.) It follows now that $E_{\omega_1}$ is a $\sigma$algebra, since any countable subcollection of $E_{\omega_1}$ will consist of sets added at various countable stages $\alpha$, but the supremum of a countable number of countable ordinals is still countable and hence below $\omega_1$, the first uncountable ordinal. Thus, the union of the sets appears at the supremum stage, which is before $\omega_1$. And similarly for complements.
It is not difficult to see that $E_{\omega_1}$ is the smallest $\sigma$algebra containing $E_0$, since whenever $E_{\alpha}$ is contained in a $\sigma$algebra, then so is $E_{\alpha+1}$, and so by induction all the $E_{\alpha}$ for countable $\alpha$ will be contained in any $\sigma$algebra containing $E_0$.
Finally, one shows that if $E_0$ has continuum many elements, then so does every $E_\alpha$. This can be seen by transfinite induction. If $E_\alpha$ has size $2^{\aleph_0}$, then the number of countable subsets of $E_\alpha$ is $(2^{\aleph_0})^{\aleph_0}$, but this is the same as $2^{\aleph_0}$ by basic cardinal arithmetic (an $\omega$sequence of $\omega$sequences can be coded into a single $\omega$sequence). If every $E_\alpha$ has size continuum for $\alpha\lt\lambda$, then $E_\lambda$ has size at most $\lambda\cdot c$, which again is $c$ for $\lambda\leq c$ since $\kappa^2=\kappa$ for any infinite cardinal. In particular, $E_{\omega_1}$ has size continuum, as desired.
The main idea can be summed up as follows: at each stage, you add countable sequences over the previous stage, and there are only continuum many countable seqeunces from a continuum. Thus, every $E_\alpha$ has size $c$ for countable $\alpha$.
More generally, one can show that if $E_0$ has size $\kappa$, then the resulting $E_\alpha$ will have size at most $\kappa^\omega$, and again $E_{\omega_1}$ will be the smallest $\sigma$algebra containing $E_0$, and it will have size at most $\kappa^\omega$.

$\begingroup$ I used the settheorist notation $\omega_1$ for the first uncountable ordinal, which your text denotes $\Omega$. $\endgroup$ Sep 23, 2010 at 12:56

$\begingroup$ 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! $\endgroup$– zzzhhhSep 23, 2010 at 13:38

$\begingroup$ 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. $\endgroup$ Sep 23, 2010 at 13:43

$\begingroup$ 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. $\endgroup$ Sep 23, 2010 at 13:51
Joel's answer is excellent as usual, but he did not really emphasize the main point that is getting you confused. The $E_\alpha$, $\alpha<\omega_1$, are not necessarily of size $\leq\aleph_1$. They are of size at most $c=2^{\aleph_0}$, which is, as has been pointed out several times above, not necessarily the same as $\aleph_1$. We always have $\aleph_1\leq c$, however.

$\begingroup$ We always have $\aleph_1\leq c$ ... but of course that depends on AC and is not explicit $\endgroup$ Sep 23, 2010 at 14:30

3$\begingroup$ Yes, but we also consider the existence of bases in vector spaces, the HahnBanach theorem and the Tychonov theorem theorems, or don't we? $\endgroup$ Sep 23, 2010 at 15:09