[Chris Nolan #19 wrote:](https://forum.azimuthproject.org/discussion/comment/16259/#Comment_16259):

> So, can we compare posets to other posets?

Yes. The best way is with monotone functions. There's a lot to say about this, but we can use monotone mappings to define when two posets are "isomorphic".

> So, if all corresponding comparable subsets of one poset are equal to the other then can we say the two posets are equal?

Alex has already answered your question quite nicely but I'll just say a bit more.

A mathematician wouldn't ask that question unless they had first told us that the two posets under discussion are equal _as sets_. They'd say something like this:

Suppose I have a set \\(S\\) with two different rrelations on it, \\(\le_1\\) and \\(\le_2\\), which give me two posets \\((S,\le_1)\\) and \\((S,\le_2)\\). Suppose \\(x \le_1 y\\) if and only if \\(x \le_2 y\\) for all \\(x,y \in S\\) . Does that mean these two posets are equal?

And the answer is yes: the posets \\((S,\le_1)\\) and \\((S,\le_2)\\) are equal if and only if

$$ x \le_1 y \textrm{ if and only if } x \le_2 y \textrm{ for all } x,y \in S. $$

However, equality of posets is ultimately less interesting than other ways of comparing posets, using monotone mappings.

Note: everything I just said would still be true if I replaced the word "poset" by "preorder" throughout this comment.