Suppose that I have a set $\displaystyle \Omega$ over which I define a collection of $\displaystyle \sigma$-algebra: $\displaystyle \{\beta_i\}$; that is, each $\displaystyle \beta_1,\beta_2,...$ is a $\displaystyle \sigma$-algebra of $\displaystyle \Omega$. Is the countable union of those also a $\displaystyle \sigma$-algebra? That is, is $\displaystyle \bigcup_{i \in N} \beta_i = \beta^*$ a $\displaystyle \sigma$-algebra over $\displaystyle \Omega$? How about countable intersection?

1 - It seems pretty clear that $\displaystyle \Omega$ and $\displaystyle \emptyset$ are contained in $\displaystyle \beta^*$.

2 - It is also clear that if $\displaystyle A \in \beta^*$, then $\displaystyle A^C \in \beta^*$

So the first two conditions of a $\displaystyle \sigma$-algebra have been fulfilled: $\displaystyle \beta^*$ contains the full sample space / empty set and it is closed under complements.

3 - But if $\displaystyle A_1,A_2,... \in \beta^*$, is it true that $\displaystyle A^* = \bigcup_{i \in N} A_i \in \beta^*$? I am unable to prove this fact. Yet I cannot find a counterexample.

A counterexample or hint of proof would be very appreciated - thanks.