Isomorphism equivalence relation

Question

I am reading through Real Analysis by Fomin and Kolmogorov, and the book makes the statement that: "Isomorphism between partially ordered sets is an equivalence relation as defined in Sec. 1.4, being obviously reflexive, symmetric, and transitive".

So I have a couple questions regarding this:

1. When equivalence relations are expounded upon earlier in the book, it is between members of a given set, but here we are speaking of a mapping between two distinct sets. I don't understand how to translate this into an equivalence relation when it is a mapping between two sets.
2. How then does the isomorphism end up being symmetric? I understand the reflexivity and transitivity, as they are included in the definition of a set having a partial ordering, but how does the symmetry come into play?

Any help would be appreciated! Thanks!

Answer 1

They’re talking about a relation on the class of all partial orders, specifically, the relation of being isomorphic as partial orders. I’ll write $\langle P,\le\rangle\equiv\langle Q,\preceq\rangle$ to mean that the partial orders $\langle P,\le\rangle$ and $\langle Q,\preceq\rangle$ are order-isomorphic.

All that they’re saying is that this relation is reflexive, symmetric, and transitive.

• Any partial order $\langle P,\le\rangle$ is isomorphic to itself: $\langle P,\le\rangle\equiv\langle P,\leq\rangle$.
• A partial order $\langle P,\le\rangle$ is isomorphic to a partial order $\langle Q,\preceq\rangle$ iff $\langle Q,\preceq\rangle$ is isomorphic to $\langle P,\le\rangle$: $\langle P,\le\rangle\equiv\langle Q,\preceq\rangle$ iff $\langle Q,\preceq\rangle\equiv\langle P,\leq\rangle$.
• If a partial order $\langle P,\le\rangle$ is isomorphic to a partial order $\langle Q,\preceq\rangle$, and $\langle Q,\preceq\rangle$ is isomorphic to a partial order $\langle S,\sqsubseteq\rangle$, then $\langle P,\le\rangle$ is isomorphic to $\langle S,\sqsubseteq\rangle$: if $\langle P,\le\rangle\equiv\langle Q,\preceq\rangle$, and $\langle Q,\preceq\rangle\equiv\langle S,\sqsubseteq\rangle$, then $\langle P,\le\rangle\equiv\langle S,\sqsubseteq\rangle$