> **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\\)?

If \\(P\multimap Q\\) needs to be the coarsest partition that distinguishes everything distinguished in Q that is not distinguished in P, I think that there is no such thing in general, which means this monoidal poset is not closed. For instance, if \\(P=((1,2),(3,4))\\) and \\(Q=((1),(2),(3),(4))\\), then \\(P\wedge((1,4),(2,3))=P\wedge((1,3),(2,4)) = Q\\), but \\(((1,4),(2,3))\vee((1,3),(2,4))=I\\).

If \\(P\multimap Q\\) needs to be the coarsest partition that distinguishes everything distinguished in Q that is not distinguished in P, I think that there is no such thing in general, which means this monoidal poset is not closed. For instance, if \\(P=((1,2),(3,4))\\) and \\(Q=((1),(2),(3),(4))\\), then \\(P\wedge((1,4),(2,3))=P\wedge((1,3),(2,4)) = Q\\), but \\(((1,4),(2,3))\vee((1,3),(2,4))=I\\).