# A Cantor-Bernstein style theorem for finite posets.

WARNING: This only applies for FINITE posets. For arbitrary posets it's false (all answers of that question serve as counterexamples), (even for arbitrary chains as in this Asaf Karagila answer). I couldn't find proof of my theorem below for finite poset in that question, and neither on (1) or (2).

I want to prove this:

Theorem (Cantor-Bernstein for finite posets):

Let $P1$, $P2$ be finite posets, let $f1:P1 \mapsto P2$ , $f2:P2 \mapsto P1$ be monotone increasing (aka "order preserving", or "increasing") injective functions from $P1$ to $P2$. Then $P1$ and $P2$ are order-isomorphic.

Proof: Since $f1$ is injective, it follows that $|P1|$<=$|P2|$, and analogously exchanging 1 for 2, it follows $|P2|$<=$|P1|$, thus $|P1|$=$|P2|$. Since $P2$ is finite and $f1$ injective, it means $f1$ is bijective! And analogously $f2$. (This is called "argument 1").

Now the more muddy part. The set of elements of the order binary relation of $P1$, NOT the elements, but the ARROWS, let's call them $A1$ (for "arrows 1"). There is a finite number of arrows, since $P1$ is finite, and analogously that last remark applies for $P2$.

We define $F1$:$A1 \mapsto A2$ to map the arrrow $(a,b)$ (wich corresponds to $a \leq b$, to $(f1(a),f1(b))$. (This maps arrows to arrows, using the monotonicity!). It's easy to see $F1$ is injective: because $f1$ is injective. We define $F2$ analogously, and it's injective!

Lastly: Mimicking "argument (1)" exactly, Since $F1$ and $F2$ injective between finite sets of arrows $A1$ and $A2$, it follows they are bijective. With this I prove $f1$ is order reflecting: because for every arrow in $A2$, there is a preimage in $A1$ (since $F1$ suryective, because it's bijective) thus for with $x,y \in P2$ $x \leq y$ , it means there is $(a,b)$ in $A1$, but that means $f(a)=x, f(b)=y$, and $a<=b$ since that's what arrows are! So $f1$, it's order reflecting and thus since it was order preserving by hypothesis, and it was proved bijective, it's a order isomorphism, and thus $P1$ and $P2$ isomorphic.

The question is: is this theorem and proof correct? How to repair the hand waveness of "arrows"? Is sufficient to use the definition of binary relation, as pairs of the cartesian product?

• This looks correct. And it should work for arbitrary finite digraphs, not just posets. In fact, it should work for arbitrary finite relational structures. Nice argument, by the way. Apr 6, 2017 at 16:32
• @goblin Thanks a lot for checking! First, in my mind: digraph=binary relation. Second, Since I don't use any poset "axiom", it seems it works for [binary endo-relations ]() over finite sets, you seem right to me! Is your "relational structure" the same as this? Apr 6, 2017 at 16:52
• @Santropedro I suppose goblin means that or perhaps even more general, since in a relational structure we may have several relations, and in that link there's only one. Of course you would need to have the same type of structure in both sides (same number of relations and same arities). Apr 6, 2017 at 20:25
• Another interesting idea that occurred to me, but I don't have the time now to check it: the way you define those correspondences between arrows, they sound like functors between categories. Given that a poset can always be thought of as a category, perhaps you're just proving an isomorphism (or equivalence?) between categories? It seems so... Apr 6, 2017 at 20:30
• @amrsa First thanks a lot for the feedback! I'm interested on the "relational structure" however googling it, returns nothing interesting except for that link given by me in the earlier comment. If you could give a link to that concept, I would be grateful! Also, I would like to respond to the category theory part, although in a couple of months will be, since I still haven't learnt category theory. Apr 6, 2017 at 20:54