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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 next stage after 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 at most 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 $\kappa^\omega$.

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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 $E_\lambda=\bigcup_{\alpha\lt\lambda} E_\alpha$. 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. Thus, the union of the sets appears at the next stage after 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 at most 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 $\kappa^\omega$.