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Iosif Pinelis
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$T$ is the set (say $S^\ast$) of all intersections of finite unions of elements of $S$.

Indeed, clearly $T\supseteq S^\ast$. It is also clear that $S^\ast$ is closed wrt to intersections. Finally, $S^\ast$ is closed wrt to finite unions, because $$\bigcup_{j=1}^n\bigcap_{i_j\in I_j}A_{i_j,j} =\bigcap_{(i_1,\dots,i_n)\in I_1\times\cdots\times I_n}\,\bigcup_{j=1}^n A_{i_j,j}$$ for any sets $A_{i,j}$ and any index sets $I_j$ -- use now the displayed identity for the case when each of the $A_{i,j}$'s is the union of finitely many elements of $S$, and the fact that finite unions of finite unions of sets are finite unions of sets.


Cf. conversion to the conjunctive normal form and the disjunctive normal form.

Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229