Question on strict order of a disjoint union

Question

We consider two strict ordered set $(A,<_A)$ and $(B,<_B)$ and we define the set $(X,<_X)$ with $X=A\sqcup B$ and $u<_X v\Leftrightarrow (u,v\in A , u<_A v$) or ($u,v\in B, u<_B v$) or ($u\in A$ and $v\in B$).

So is $(X,<_X)$ a strict ordered set ?

The answer is trivial for the two first cases because $<_A$ and $<_B$ define strict order. But the last case I don't see why because we can't compare $u,v$. Same problem if the two sets are totally or well-ordered (strict).

Thanks in advance !

Answer 1

The order $<_X$ is strict:

• If $x\in X$ then $x\not <x$ because each of $(x\in A, x<_Ax)$, $(x\in B, x<_Bx)$, $(x\in A, x\in B)$ is false
• If $x<y$ and $y<z$ then $x<z$: If $x\in B$, then $y<_Xy$ implies $y\in B$ and $x<_By$. Then $y<_XZ$ implies $z\in B$ and $y<_Bz$, hence $x<_Bz$ and $x<_Xz$. Similarly, if $z\in A$, then $y\in A$ and $y<_Az$; then $x\in A$ and $x<_Ay$, hence $x<_Az$ and $x<_Xz$. Finally, if $x\in A$ and $z\in B$, we have $x<_Xz$ immediately.

If $<A,<_B$ are additionally total, then $<_X$ is also total: If $x,y\in X$ then either both are in $A$ and hence comparable, or both are in $B$ and hence comparable, or one is in $A$ and the other in B$ and hence they are comparable.

If $<_A,<_B$ are additionally well-orders, then so is $<_X$. If $S\subset X$ is non-empty, the either $S\subseteq B$, which makes $\min_X(S)=\min_B(S)$; or $S\cap A$ is non-empty and we find $\min_X(S)=\min_A(S\cap A)$.