Powerset

Given a set, there is a second set standing behind it: the set of all its subsets. The things that count as points of the powerset are whole subsets of the original, so a choice of subsets of the original is a single subset of the powerset.

Membership in \(\mathcal{P}(S)\) and membership in \(S\) are different relations, and reading \(\in\) carelessly is the usual source of trouble. For \(S = \{1,2,3\}\): \(1 \in S\) but \(1 \notin \mathcal{P}(S)\), while \(\{1\} \in \mathcal{P}(S)\) but \(\{1\} \notin S\). The element and the one-element set holding it are different objects.

Both \(\emptyset\) and \(S\) are subsets of \(S\), so both are elements of \(\mathcal{P}(S)\). In particular \(\mathcal{P}(\emptyset) = \{\emptyset\}\), a one-element set whose single element is the empty set.

Building a subset of a finite \(S\) means walking its elements and deciding independently, for each, whether to keep it. With \(n\) elements that is \(n\) independent binary choices:

\[ |S| = n \quad\Longrightarrow\quad |\mathcal{P}(S)| = 2^n. \]

The gap is not an artifact of finiteness. For any set at all there is no surjection \(S \to \mathcal{P}(S)\), so \(\mathcal{P}(S)\) is strictly larger than \(S\) whatever \(S\) is.

The powerset is rarely interesting in itself; its use is as the space a structure gets selected from. A topology is the standard case, the chosen subsets being the ones called the open sets. Writing that collection as \(\mathcal{O}\), a condition on which subsets belong to it becomes a condition on the single object \(\mathcal{O} \subseteq \mathcal{P}(S)\).