Let $X,Y$ be well-ordered sets with $X \subseteq Y$ and where $X$ inherits the order of $Y$.
I want to show:
$$Ord(X) \leq Ord(Y)$$
I know the basiscs about well-ordered sets. I know that there is at most one isomorphism between well-ordered sets and that $Ord(X) = Ord(Y) \iff X\cong Y$ for well-ordered sets $X,Y$.
I tried to take an isomorphism $$f: Y \to Ord(Y)$$
and then we have $X \cong f(X)$ and I want to prove that $f(X) = Ord(X)$, which holds iff $f(X)$ is an ordinal. Clearly $f(X)$ is well-ordered, and
$$f(X)_{f(x)} = f(X) \cap Ord(Y)_{f(x)}= f(X) \cap f(x)$$
Am I on the right track?
How would I show this?
First let us solve it for the special case where $X\subseteq Y$ and $Y$ is an ordinal.
There is a unique function $g$ having $X$ as domain and satisfying:$$g(\beta)=\{g(\gamma)\mid \gamma\in X\text{ and }\gamma<\beta\}$$
This with $g(X)=\mathsf{Ord}(X)$.
Further by induction we find that $g(\beta)\leq\beta<Y$ for every $\beta\in X$ so that $g(X)\subseteq Y$ or equivalently $\mathsf{Ord}(X)\leq Y$.
Now the more general case.
If $<$ denotes the well order on $Y$ then there is a unique function having $Y$ as domain and satisfying:$$f(y)=\{f(z)\mid z<y\}$$
Then $f(X)\subseteq f(Y)=\mathsf{Ord}(Y)$ and $f:Y\to f(Y)$ is an isomorphism.
Above it has been proved that $\mathsf{Ord}(f(X))\leq\mathsf{Ord}(Y)$
Here $X$ and $f(X)$ are isomorphic so also $\mathsf{Ord}(X)\leq\mathsf{Ord}(Y)$.