Union of $\sigma$-algebras is a $\pi$-system...? I thought that the  the union of $\sigma$-algebras is a $\pi$-system (which I believe is what is said at the beginning of this this wikipedia article). 
When I went to prove it myself, I had some trouble. 
I keep thinking of the following example: let $\Omega$ be the sample space, and let $A \neq B$ be two subsets of $\Omega$ such that $A\cap B\neq \emptyset$. Now
\begin{align*}
\mathcal{F}_{1} &= \{ \emptyset, A, A^{c}, \Omega\}\\
\mathcal{F}_{2} &= \{ \emptyset, B, B^{c}, \Omega\}\\
\\
\implies \mathcal{F}_{1} \cup \mathcal{F}_{2} &=  \{ \emptyset,A,A^{c}, B, B^{c}, \Omega\}\\
\end{align*}
where clearly $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ are $\sigma$-algebras, but $\mathcal{F}_{1}\cup\mathcal{F}_{2}$ is not a $\pi$-system. 
Can someone end my misery and either (i) point out the problem with my counterexample and provide a proof that the union of $\sigma$-algebras is a $\pi$-system, or (ii) clarify what the intro of this wikipedia articleis saying and elaborate on the conditions under which the union of $\sigma$-algebras is a $\pi$-system? 
 A: As you found, an arbitrary union of $\sigma$-algebras is not necessarily a $\pi$-system.
In the introduction of the wikipedia article, they look at an increasing sequence of $\sigma$-algebras,
$$\mathcal{A}_n = \sigma(E_1,\dotsc, E_n),$$
and note that the union of this sequence is a $\pi$-system. That follows because for any two sets $A,B$ in the union, one finds an $n$ such that $A,B \in \mathcal{A}_n$, whence $A\cap B \in \mathcal{A}_n$, and thus $A\cap B \in \bigcup \mathcal{A}_n$.
This generalises beyond an increasing sequence of $\sigma$-algebras, if $\{\mathcal{A}_i : i \in I\}$ is a family of $\sigma$-algebras such that for all $i,j\in I$ there is a $k\in I$ with $\mathcal{A}_i \cup \mathcal{A}_j \subset \mathcal{A}_k$, then
$$\Pi := \bigcup_{i\in I}\mathcal{A}_i$$
is a $\pi$-system. For if $A,B \in \Pi$, there are $i,j\in I$ such that $A\in \mathcal{A}_i$ and $B\in \mathcal{A}_j$. By the condition, there is a $k\in I$ such that $A, B \in \mathcal{A}_k$, and hence $A\cap B \in \mathcal{A}_k \subset \Pi$.
