[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?

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.