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.

 A: 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 $\tau(\sigma(\mathcal{S})) \subseteq \sigma(\tau(\mathcal{S}))$
A: The following is just an elaboration of Henno Brandsma's idea:
Proposition.
Let $I$ be a set. For each $\alpha\in I$, let $f_\alpha$ be a function on a set $J_\alpha$.
Then
$$\bigcap_{\alpha\in I}\left(\bigcup_{j\in J_\alpha}f_\alpha(j)\right)=\bigcup_{g\in\prod_{\alpha\in I} J_\alpha}\left(\bigcap_{\alpha\in I} f_\alpha(g(\alpha))\right).$$
Proof.
Suppose that $x\in\bigcap_{\alpha\in I}\left(\bigcup_{j\in J_\alpha}f_\alpha(j)\right)$. Then for each $\alpha\in I$, there exists $j_\alpha\in J_\alpha$ such that $x\in f_\alpha(j_\alpha)$. Assuming the axiom of choice, choose such $j_\alpha$ for each $\alpha\in I$. Let $g\in\prod_{\alpha\in I}J_\alpha$ be the function defined by $g(\alpha)=j_\alpha$. Then
$x\in\bigcap_{\alpha\in I}f_\alpha(g(\alpha))$, so
$$\bigcap_{\alpha\in I}\left(\bigcup_{j\in J_\alpha}f_\alpha(j)\right)\subseteq\bigcup_{g\in\prod_{\alpha\in I} J_\alpha}\left(\bigcap_{\alpha\in I} f_\alpha(g(\alpha))\right).$$
Next, let $g\in\prod_{\alpha\in I}J_\alpha$, and $x\in\bigcap_{\alpha\in I}f_\alpha(g(\alpha))$. Since $g(\alpha)\in J_\alpha$,
$$x\in f_\alpha(g(\alpha))\in\bigcup_{j\in J_\alpha}f_\alpha(j).$$
It follows that $x\in\bigcap_{\alpha\in I}\left(\bigcup_{j\in J_\alpha}f_\alpha(j)\right)$. Hence,
$$\bigcap_{\alpha\in I}\left(\bigcup_{j\in J_\alpha}f_\alpha(j)\right)\supseteq\bigcup_{g\in\prod_{\alpha\in I} J_\alpha}\left(\bigcap_{\alpha\in I} f_\alpha(g(\alpha))\right).$$
Proposition.
Let $I$ be a set. For each $\alpha\in I$, let $f_\alpha$ be a function on a set $J_\alpha$.
Then
$$\bigcup_{\alpha\in I}\left(\bigcap_{j\in J_\alpha}f_\alpha(j)\right)=\bigcap_{g\in\prod_{\alpha\in I} J_\alpha}\left(\bigcup_{\alpha\in I} f_\alpha(g(\alpha))\right).$$
Proof.
Suppose that $x\in\bigcup_{\alpha\in I}\left(\bigcap_{j\in J_\alpha}f_\alpha(j)\right)$. Then $x\in \bigcap_{j\in J_\alpha}f_\alpha(j)$ for some $\alpha\in I$. Hence, if $g\in\prod_{\alpha\in I}J_\alpha$,
then
$$x\in f_\alpha(g(\alpha))\in\bigcup_{\alpha\in I}f_\alpha(g(\alpha)).$$
Hence,
$$\bigcup_{\alpha\in I}\left(\bigcap_{j\in J_\alpha}f_\alpha(j)\right)\subseteq\bigcap_{g\in\prod_{\alpha\in I} J_\alpha}\left(\bigcup_{\alpha\in I} f_\alpha(g(\alpha))\right).$$
Next, suppose that $x\not\in\bigcup_{\alpha\in I}\left(\bigcap_{j\in J_\alpha}f_\alpha(j)\right)$.
Then for each $\alpha\in I$, there exists $j_\alpha\in J_\alpha$ such that $x\not\in f_\alpha(j_\alpha)$. Assuming the axiom of choice, choose such $j_\alpha$ for each $\alpha\in I$. Let $g\in\prod_{\alpha\in I}J_\alpha$ be the function defined by $g(\alpha)=j_\alpha$. Then $x\not\in f_\alpha(g(\alpha))$ for all $\alpha\in I$, so $x\not\in\bigcup_{\alpha\in I}f_\alpha(g(\alpha))$. Hence, $x\not\in\bigcap_{g\in\prod_{\alpha\in I} J_\alpha}\left(\bigcup_{\alpha\in I} f_\alpha(g(\alpha))\right).$
We therefore have
$$\bigcup_{\alpha\in I}\left(\bigcap_{j\in J_\alpha}f_\alpha(j)\right)\supseteq\bigcap_{g\in\prod_{\alpha\in I} J_\alpha}\left(\bigcup_{\alpha\in I} f_\alpha(g(\alpha))\right).$$
Corollary.
Let $X$ be a set, and $\mathcal{S}$ a set of subsets of $X$. Then
$$\sigma(\tau(\mathcal{S}))=\tau(\sigma(\mathcal{S})).$$
