## Ordered relations

a. Let (A,<) be a strictly ordered set and b not in A. Define a relation $\prec$ in $B=A \cup${b} as follows:
x $\prec$y iff (x,y in A and x<y) or (x in A and y=b). Show that $\prec$ is a strict ordering of B and $\prec \cap A^2 =<.$ (Intuitively, $\prec$ keeps A ordered in the same way as < makes b greater than every element of A.)

b. Generalize part (a): Let ( $A_1, <_1$) and ( $A_2, <_2$) be strict orderings, $A_1 \cap A_2 = \emptyset$. Define a relation $\prec$ on B = $A_1 \cup A_2$ as follows:
x $\prec$y iff x,y in $A_1$ and $x <_1 y$
or x,y in $A_2$ and $x <_2 y$
or x in $A_1$ and y in $A_2$

Show that $\prec$ is a strict ordering of B and $\prec \cap A^{2}_1 = <_1, \prec \cap A^{2}_2=<_2$. (Intuitively, $\prec$ puts every element of $A_1$ before every element of $A_2$ and coincides with the original orderings of $A_1$ and $A_2$

Does anyone have any suggestions on how to do this?