 Power-set product cardinality inequality

For arbitrary sets $A,B$ each containing at least two elements, the displayed inequality follows by explicitly encoding two subsets in disjoint parts of a Cartesian product. Choose $a_0\ne a_1$ and $b_0\ne b_1$, put $S$ along the row $A\times \{b_0\}$, put $T\setminus \{b_0\}$ along the column $\{a_0\}\times (B\setminus \{b_0\})$, and add the extra point $(a_1,b_1)$ precisely when $b_0\in T$. These regions are disjoint, so the encoding recovers both $S$ and $T$ and is an injection from $\mathcal{P}(A)\times \mathcal{P}(B)$ into $\mathcal{P}(A\times B)$. This proves the cardinality inequality even for infinite sets. For finite sizes $a,b\ge 2$, the same result is $2^{ab}\ge 2^{a+b}$, since $ab-a-b=(a-1)(b-1)-1\ge 0$. Mapping $(S,T)$ to $S\times T$ is insufficient because all pairs with an empty factor collapse.