> **Hint** Use the fact that, if $\alpha_{1},\alpha_{2},\cdots$is a sequence of
> ordinals satisfying $\alpha_{n}<\Omega$, then there exists an ordinal
> $\alpha$ such that $\alpha<\Omega$ and $\alpha_{n}<\alpha$ for all
> $n$.

**Proof** By the $\mathbb{R^{1}}$ ordered topology *([H&S] (4.42))*, the first uncountable $\Omega$ *([H&S] (4.49) proved its existence)* is often written as $[0,\Omega)$ Every increasing sequence ${\alpha_{i}}$ of elements of $[0,\Omega)$ converges to a limit in $[0,\Omega)$ since the supremum ($\cup_{i=1}^{\infty}\alpha_{i}$) of every countable set of countable ordinals is another countable ordinal.
This proves the hint above. The $\alpha$ in that hint is actually the limit of the sequence ${\alpha_{i}}$.#

Since for any class $H$ of sets in $\Omega$ the class $H^{∗}$ consist of sets in $H$, the complements of sets in $H$ and the finite and countable unions of sets in $H$, we have:

> **Prop 1** If the cardinality of  $H$ is countable, the cardinality of $H^{*}$
> cannot exceed countable cardinality.

**Proof** All the complements of this class is another class with the same cardinality, and the countable unions of countably many elements in $H$ is still countable.#


> **Prop 2** $|A_{\alpha}|\neq \omega$(countable cardinal)

**Proof** If there exists some $\alpha_{0}$ such that $|A_{\alpha_{0}}|=\omega$, then choose the least such $\alpha_{0}$ *([H&S] (4.47))*. $|A_{\alpha_{0}}|=|(\cup_{\beta<\alpha}A_{\beta})^{*}|=\omega$, hence by Prop 1 $|\cup_{\beta<\alpha}A_{\beta}|=\omega$ and hence $|A_{\beta}|=\omega,\forall \beta < \alpha_{0}$. Now each such $\beta$ violates the minimality of $\alpha_{0}$#

And since $c\geq |A_{\alpha}|\neq \omega$, it has to be $|A_{\alpha}|=c $ due to the continuum hypothesis.

>**Prop 3**$\cup_{\beta<\Omega}A_{\beta}=\sigma(A)$

**Proof** $\cup_{\beta<\Omega}A_{\beta}\supset \sigma(A)$ is clear since every element in this sigma field should be obtained via a sequence(countable or not) of these operations. $\cup_{\beta<\Omega}A_{\beta}\subset \sigma(A)$ is also clear by definition of $\sigma(A)$ that it is the smallest sigma field containing these operations' result.#

Then by Prop 2 and the formula $\cup_{\beta<\Omega}A_{\beta}=\sigma(A)$, $|\sigma(A)|=c^{\omega}=c$

I still feel a bit uncertain about Prop 3, can anyone make it clearer to me? I am deeply appreciate all your help.

Reference
[H&S] Hewitt & Stromberg, Real and Abstract Analysis.