> **Puzzle 193.** We know that for any set \\(X\\) the power set \\(P(X)\\) becomes a monoidal poset with \\(S \le T\\) meaning \\(S \subseteq T\\), with product \\(S \otimes T = S \cap T\\), and with the identity element \\(I = X\\). Is \\(P(X)\\) closed? If so, what is \\(S \multimap T\\)?

In the case of \\(\otimes = \cap\\), then \\(A \multimap B = (X \backslash A) \cup B\\). Here \\(\backslash\\) denotes [set difference](https://en.wikipedia.org/wiki/Complement_(set_theory)).

**Proof.** In [Lecture 6](https://forum.azimuthproject.org/discussion/1901/lecture-6-chapter-1-computing-adjoints/p1) we saw that \\((A \otimes -) \dashv (A \multimap -)\\) if and only if

\[ A \multimap B = \bigcup \\{Y \in X : \; A \cap Y \subseteq B \\} \]

Since \\(A \cap ((X \backslash A) \cup B) = A \cap B\\) it must be \\((X \backslash A) \cup B \subseteq \bigcup \\{Y \in X : \; A \cap Y \subseteq B \\}\\).

All that is left to show is \\((X \backslash A) \cup B \supseteq \bigcup \\{Y \in X : \; A \cap Y \subseteq B \\}\\). It suffices to show the *contrapositive*: if \\(x \not\in (X \backslash A) \cup B\\) then \\(x \not\in \bigcup \\{Y \in X : \; A \cap Y \subseteq B \\}\\). Assume \\(x \not\in (X \backslash A) \cup B\\). It must be that \\(x \in A\\) and \\(x \not\in B.\\) This means if \\(x \in Y\\) then \\(A \cap Y \not\subseteq B,\\) hence \\(x \not\in\bigcup \\{Y \in X : \; A \cap Y \subseteq B \\}\\).\\(\qquad \blacksquare \\)