how to combine different partial orders (Poset)

Question

Given two posets $\prec_A$ and $\prec_B$ where $A\neq B$ and $A\cap B\neq \emptyset$, is there any way to combine them while preserving the exact information they exhibit - namely dominance relation ( $x\prec y$) and incomparability.
(i.e. if $x$ and $y$ are incomparable in $\prec_A$ then they are incomparable in $\prec_{AB}$)?

Answer 1

If $A\cap B=\varnothing$ you can define an order $<$ on $A\cup B$ by $x<y$ if, and only if:

1. $x,y\in A$ and $x\prec_A y$.
2. $x,y\in B$ and $x\prec_B y$.

This will preserve comparability. A diagram for this order can be obtained by placing the diagram for $A$ on the same plane as $B$.

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Again assuming $A\cap B=\varnothing$, you can define an order $\ll$ on $A\cup B$ by $x\ll y$ if, and only if:

1. $x,y\in A$ and $x\prec_A y$
2. $x,y\in B$ and $x\prec_B y$.
3. $x\in A$ and $y\in B$

This will also preserve comparability. A diagram for this order can be obtained by placing the diagram for $A$ below the diagram for $B$ and connecting $A$'s maximal elements to $B$'s minimal elements if they exist.

Answer 2

Example where the combination of 2 Posets is not a Poset:

• <A={{1},{2,3},{4,5}}=SA and <B={{2,3},{5,6},{7}}=SB
• <(SA||SB)={{1},{2,3},{2,3},{4,5},{5,6},{7}} is not equal to <(∪)

where :

• ∪ ={1,2,3,4,5,6,7}
• SA||SB being the concatenation of SA with SB
• SA||SB={{1},{2,3},{4,5},{2,3},{5,6},{7}}
• <(SA||SB) being the sorted concatenation of SA and SB