> **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 \$$