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!

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    $\begingroup$ The equivalence relation is on the space of all partially ordered sets (POS). Two POSs are equivalent if there is an isomorphism between them. For symmetric just use the inverse bijection. $\endgroup$ – jaRedDRedmp Jul 27 '20 at 19:09
  • $\begingroup$ Ah so then the equivalence relation that is the isomorphism accepts the sets themselves as arguments? $\endgroup$ – user132849 Jul 27 '20 at 19:10
  • $\begingroup$ Is question 2 conflating the isomorphism (as the equivalence relation) with the partial order (as attempting to and failing to be an equivalence relation)? Or is it just asking why an isomorphism should go in both directions? $\endgroup$ – Adina Goldberg Jul 27 '20 at 19:11
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    $\begingroup$ @AdinaGoldberg I actually get it now thanks to jaRedDRedmp. Thanks! $\endgroup$ – user132849 Jul 27 '20 at 19:12
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    $\begingroup$ @user132849 Yeah, the whole set is the set of POS and the equivalence relation (being isomorphic) partitions this whole space into classes of isomorphic POSs. See here en.wikipedia.org/wiki/Order_isomorphism#Order_types $\endgroup$ – jaRedDRedmp Jul 27 '20 at 19:13

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 $\langle P,\le\rangle\equiv\langle S,\sqsubseteq\rangle$
  • $\begingroup$ Yeah actually another user commented that, and I get it now. Thanks for the help! $\endgroup$ – user132849 Jul 27 '20 at 19:18
  • $\begingroup$ @user132849: You’re welcome! $\endgroup$ – Brian M. Scott Jul 27 '20 at 19:27

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