Given: $(A,<)$ is a strictly ordered set and $b \notin A$.
Define: a relation $\prec$ in $ B = A \bigcup \{b\} $ as: $x\prec y $ if and only if $(x,y \in A \text{ and } x<y)$ or $(x \in A \text{ and } y=b)$
Show that $\prec $ is a strict ordering of $B \text{ and } \prec \bigcap A^{2} =<$.
I know that I need to show the relation is asymmetric and transitive.
Here is what I have :
Asymmetric: let $x \prec y $ then $ x < y \text{ OR } x \in A \text{ and } y = b$
if $x<y$ then $y \not< x$ since $<$ is a strict order.
if $x \in A \text{ and } y=B$ then $y \not\prec x \text{ since } y \notin A \text{ and } x\not= b$
therefore $\prec$ is asymmetric.
Transitive: I know how to show it is transitive if $x, y, z \in A$ as I would just fall back on the given that $<$ is a strict order.
However I don't know how to show it if $y=b$.
Hint: Note that there is no element $z$ such that $b\prec z$.