- Let $X$ and $Y$ be two well-ordered sets. Let
$$S=(X\times\left\{0\right\})\cup( Y\left\{1\right\})$$
For all $x,x_1,x_2\in X$ and for all $y,y_1,y_2\in Y$
$(x_1,0)<(x_2,0)$ iff $x_1<x_2,$
$(y_1,1)<(y_2,1)$ iff $y_1<y_2,$
$(x,0)<(y,1)$.
Show the above relation well-orders $S$
My try. Let $U\subseteq (X\times\left\{0\right\})\cup( Y\left\{1\right\})$. If $U\cap (X\times\left\{0\right\})\neq\emptyset$, then define
$$W=\left\{x\in X: (x,0)\in U\right\}$$
$W$ does have a minimal element, say $a$.
If $U\cap (X\times\left\{0\right\})=\emptyset$, then if $U\cap (Y\times\left\{1\right\})\neq\emptyset$, so define
$$V=\left\{y\in Y: (y,1)\in U\right\},$$
$V$ does have a minimal element, say $b$.
I can't continue. Can you check my try, can you help to continue it? Thanks ...
Let $\langle A,<_R\rangle,\;\langle B,<_S\rangle$ be two well ordered sets such that $A\cap B=\varnothing$. Let $C$ be a nonempty subset of $A\cup B$. We shall prove that it has a minimal element in the sense of the order $<_{R\oplus S}\;=\;<_R\cup<_S\cup\;(A\times B)$
If $C\cap A\not=\varnothing$, then the minimal element of $C\cap A$ in the sense of $<_R$ is the minimal element of $C$ in the sense of $<_{R\oplus S}$.
If $C\cap A=\varnothing$, then $C\subseteq B$ and the minimal element of $C$ in the sense of $<_S$ is the minimal element of $C$ in the sense of $<_{R\oplus S}$
Now to your argument. Note how the minimal element of $W$, and $V$ are not the actual minimal elements of $U$ in any of the considered cases, in the sense of the well ordered sum, but $(a,0)$ and $(b,1)$ are, respectively.