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.

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 $\kappa^\omega$. 


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. 

