Given a ordered set $(X,\preceq)$ for any $\xi\in X$ we call the set $$ I_\xi:=\{x\in X:x⪱\xi\} $$ the initial segment of $\xi$.
Now if $(X,\preceq)$ and $(Y,\precsim)$ are two ordered sets then it is a well know result that the proposition $$ (x_1,y_1)\precapprox(x_2,y_2)\Leftrightarrow(y_1\precnsim y_2)\vee(x_1\preceq x_2\wedge y_1=y_2) $$ define a order on $X\times Y$ called horizontal lexicographic order.
So provided that $X$ and $Y$ are well ordered in my text is written that if $W$ is an initial segment of $(X\times Y,\precapprox)$ then there exists an initial segment $V$ in $Y$ such that $X\times V$ is an initial segment in $X\times Y$ containing $W$ however I was not able to prove this so that I thought to put a specific question where I ask to prove it.
In particular I tried to use the following result
A subset $W$ of $X\times Y$ is an initial segment if and only if there exist an initial segment $U$ of $X$ and an initial segment $V$ of $Y$ such that $$ W=X\times V\cup U\times\{u\} $$ where $u$ is the minimum of $Y\setminus V$.
So clearly $$ W\subseteq X\times V\cup X\times\{u\}=X\times (V\cup\{u\}) $$ but unfortunately $V\cup\{u\}$ is not an inital segment. Anyway if $Y$ has not a maximum element then the set $$ Y:=\{y\in Y:u\precnsim y\} $$ is not empty so that it has an element $\eta$ and any initial segment $E$ of $\eta$ contains $V\cup\{u\}$ so that $X\times E$ would be an initial segment of $X\times Y$ such that $$ W\subseteq X\times E $$ However if $Y$ has a maximum element then it seem to me that the result does not holds.
So how prove the result? could it be not generally true? could someone help me, please?