Chris Nolan#19

If all the comparable elements of two sets are equal, then the sets are equal. But posets also include an order, and the order of two sets could be different, even though the elements are the same.

Now, if by poset, you mean posets whose order is subset inclusion, if the corresponding sets are equal, then they would be equal as posets (because they are equal as sets, and have the same \\( \leq \\) ).

This is similar to saying that {1, 4, 6} and {1, 4, 6} are equal as sets with the usual order for integers.

Does that answer your question?