The following extract is taken from Terence Tao's Analysis 1 (Third Edition)
Exercise 8.5.19. Let $X$ be set, and let $\Omega$ be the space of all pairs $(Y,≤)$ where $Y$ is a subset of $X$ and $\leq$ is a well-ordering of $Y$. If $(Y,≤)$ and $(Y',≤')$ are elements of $\Omega$, we say that $(Y,≤)$ is an initial segment of $(Y',≤')$ if there exists an $x \in Y'$ such that $Y = \{ y \in Y': y <' x\}$ (so in particular $Y \subsetneq Y'$), and for any $y, y' \in Y, y \le y'$ if and only if $y \le 'y'$ ...
The problem and the one I am currently struggling with is
Let $(Y,≤)$ be an initial segment of $(Y',≤')$. For all $y,y' \in Y$, we have $y ≤ y'$ if and only if $y ≤' y'$.
What I have I tried is to use
Proposition 8.5.10 (Principle of strong induction). Let $X$ be a well-ordered set with an ordering relation $\le$, and let $P(n)$ be a property pertaining to an element $n\in X$ (i.e., for each $n \in X, P(n)$ is either a true statement or a false statement). Suppose that for every $n \in X$, we have the following implication: if $P(m)$ is true for all $m \in X$ with $m \lt \ n$, then $P(n)$ is also true. Prove that $P(n)$ is true for all $n \in X$.
to prove the above statement, and arguing by contradiction, but I am not making any progress. I would appreciate any help/hints that anyone can offer.
EDIT: I may have parsed the definition incorrectly. Am I correct in saying that the condition I am trying to prove is actually just part of the definition of an initial segment?