Do arbitrary union and arbitrary intersection operations commute? Let $X$ be a set, and $\mathcal{S}$ a set of subsets of $X$. Let $\sigma(\mathcal{S})$ be the set of all arbitrary unions of the elements of $\mathcal{S}$, and $\tau(\sigma(\mathcal{S}))$ the set of all arbitrary intersections (empty intersection being $X$) of the elements of $\sigma(\mathcal{S})$. Similarly, let $\tau(\mathcal{S})$ be the set of all arbitrary intersections of the elements of $\mathcal{S}$, and $\sigma(\tau(\mathcal{S}))$ the set of all arbitrary unions of the elements of $\tau(\mathcal{S})$. Then do we have
$$\tau(\sigma(\mathcal{S}))=\sigma(\tau(\mathcal{S})) \ ?$$
A couple of observations:


*

*If $\sigma$ and $\tau$ are finite union and finite intersection operations, respectively, then the equality seems true.

*If only $\tau$ is finite intersection operation, then the equality fails.


Consider that for subsets $A_{i,j} \subseteq X$ and index set $J$ (and for each $j \in J$ and index set $I_j$), we can rewrite a member of $\tau(\sigma(\mathcal{S}))$ as:
$$\bigcap_{j \in J} \bigcup_{i \in I_j} A_{i,j} = \bigcup_{f \in F} \bigcap_{j \in J} A_{j, f(j)}$$ 
where $F$ is the set of all functions $f$ from $J \to \bigcup_{j} I_j$ such that $(j) \in I_j$ for all $j$ (choice functions); this equality requires AC. This shows that $\tau(\sigma(\mathcal{S})) \subseteq \sigma(\tau(\mathcal{S}))$
