> **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

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 \\\\

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.