Is the union of partially ordered sets also partially ordered?

Question

I am in the process of doing a proof exercise for showing that a function $f : L_1 \to L_2$ between partially ordered sets $L_1 = (L_1, \sqsubseteq_1)$ and $L_1 = (L_1, \sqsubseteq_2)$ is monotone (or isotone or order-preserving) if $\forall l, l^\prime \in L_1 : l \sqsubseteq_1 l^\prime \Rightarrow f(l) \sqsubseteq_2 f(l^\prime)$. I want to use the idea that the union of partially ordered sets also partially ordered; however, I'm unsure if this is even true. Is it true that the union of partially ordered sets also partially ordered?

Answer 1

I assume you mean disjoint union. That is, $(L, \sqsubseteq) = (L_1, \sqsubseteq_1) \sqcup (L_2, \sqsubseteq_2)$ has as underlying set $L = L_1 \sqcup L_2$ and we set $x \sqsubseteq y$ iff $x, y \in L_i$ and $x \sqsubseteq_i y$ for $i \in \{1,2\}$ (so for example $x \not \sqsubseteq y$ if $x \in L_1$ and $y \in L_2$).

Then yes, this is a partial order. You can easily verify this by checking the axioms of a partial order. For example, we can check reflexivity. Let $x \in L$ we have $x \in L_i$ for some $i \in \{1, 2\}$. Since $(L_i, \sqsubseteq_i)$ is a partial order we have $x \sqsubseteq_i x$. So $x \sqsubset x$, as required. The other properties follow similarly, and I leave that to you to check.

Edit. Just as I posted this answer, another answer By Dániel G appeared, which shows very nicely why we need to consider the disjoint union.

Answer 2

It depends on what you mean by "a union of partially ordered sets". Naively, the answer is no: you can take $(\{a,b\}, \leq_1)$ and $(\{b,c\}, \leq_2)$, where $a \leq_1 b$ and $b \leq_2 c$ (and $a \neq b \neq c)$. Then in the union $(\{a,b,c\}, \leq)$ we have $a \leq b \leq c$, but $a \not \leq c$, contradicting transitivity.

In general, the union of the two order relations is reflexive and symmetric but not necessarily transitive (unless the posets are disjoint). You can rectify this by taking the transitive closure after taking union.