> **Puzzle 194.** From [Lecture 11](https://forum.azimuthproject.org/discussion/1991/lecture-11-chapter-1-the-poset-of-partitions/p1) we know that for any set \$$X\$$ the set of partitions of \$$X\$$, \$$\mathcal{E}(X)\$$, becomes a poset with \$$P \le Q\$$ meaning that \$$P\$$ is finer than \$$Q\$$. It's a monoidal poset with product given by the meet \$$P \wedge Q\$$. Is this monoidal poset closed? How about if we use the join \$$P \vee Q\$$?

Just as \$$(P(X), \subseteq, \cup, \emptyset)\$$ is not a monoidal poset, we have \$$(\mathcal{E}(X), \le, \vee, D)\$$ is not a monoidal poset for the same reasons.

On the other hand, I agree with Yoav: \$$P \multimap Q\$$ does not exist in general for partitions.

Recall from the [adjoint functor theorem for posets](https://forum.azimuthproject.org/discussion/2031/lecture-16-chapter-1-the-adjoint-functor-theorem-for-posets/p1) that \$$A \multimap -\$$ exists and \$$(A \wedge -) \dashv (A \multimap -)\$$ if and only if \$$(A\wedge -) \$$ preserves all joins.

This means, in particular, that if \$$(A \multimap -)\$$ exists then \$$A \wedge (B \vee C) = (A \wedge B) \vee (A \wedge C)\$$.

But if we let

\begin{align} A & = \lbrace \lbrace 1,4 \rbrace, \lbrace 2,3 \rbrace \rbrace \\\\ B & = \lbrace \lbrace 1,2 \rbrace, \lbrace 3,4 \rbrace \rbrace \\\\ C & = \lbrace \lbrace 1\rbrace, \lbrace 2,3 \rbrace, \lbrace 4 \rbrace \rbrace \\\\ \end{align}

then we have \$$A \wedge (B \vee C) = A\$$ but \$$(A \wedge B) \vee (A \wedge C) = C\$$ .

Hence \$$(A \multimap -)\$$ does not exist in general for partitions.