Example of a set and two $\sigma$ algebras such that union is not a $\sigma$-algebra What is an example of a set $X$ and two $\sigma$-algebras $\mathcal{A}_1$ and $\mathcal{A}_2$, each consisting of subsets of $X$, such that $\mathcal{A}_1 \cup \mathcal{A}_2$ is not a $\sigma$-algebra?
 A: Assume $X$ has more than $2$ elements, and let two of those elements by $a,b\in X$.  Then $\mathcal{A}_1 = \{\emptyset,\{a\},X\setminus\{a\},X\}$ and $\mathcal{A}_2= \{\emptyset,\{b\},X\setminus\{b\},X\}$ are both $\sigma$-algebras over $X$.  
However, $\mathcal{A}_1\cup \mathcal{A}_2 = \{\emptyset,\{a\},\{b\},X\setminus\{a\},X\setminus\{b\},X\}$ is not a $\sigma$-algebra, as $\{a\}\cup \{b\}=\{a,b\}\notin \mathcal{A}_1\cup \mathcal{A}_2$, for example.
A: For example, $X=\{1,2,3\}$ and
$$ \mathcal A_1 = \{\varnothing,\{1\},\{2,3\},X\} \\
\mathcal A_2 = \{\varnothing,\{2\},\{1,3\},X\} $$
A: For a more natural example, you could consider $\mathscr{B}([0,1])$, The $\sigma-$algebra generated by the open sets on $[0,1]$ (called the Borel sets) and $\mathscr{B}([-1,0])$. 
Note that
$$ (-1,0) \in \mathscr{B}([-1,0]) \qquad \text{and} \qquad (0,1) \in \mathscr{B}([0,1]).$$
And $\{0\}$ is contained in both.
Though each is a $\sigma-$algebra, Their union, 
$$\mathscr{B}([-1,0]) \cup \mathscr{B}([0,1])$$
Fails to contain $(-1,1)$, even though
$$ (-1,0) \cup \{0\} \cup (0,1)=(-1,1) $$
So it fails to even be a Ring, much less a $\sigma-$algebra.
A: Hint: Construct two sigma algebras $S_1, S_2$ on $X$  such that $S_1$ is not contained in $S_2$ and $S_2$ is not contained in $S_1$
