Proof of: For all linearly ordered sets $X$, if there is a bijection from $n∈ℕ$ to $X$, then $n$ and $X$ are order-isomorphic

Question

The theorem is:

This is the full proof given by my textbook:

I don't understand the structure of this proof.
The proof goes:

1. Inductive hypothesis (the one i stated in the title of the question)
2. Define $f|ₙ:n→A\setminus\{f(n)\}$ is a bijection
3. Codomain of $f|ₙ$ is linearly ordered
4. Then by the inductive hypothesis there exist an order-isomorphism $g:n→A\setminus\{f(n)\}$
5. construct an order-isomorphism $h$ using $g$

and then i get lost.

Why do they define $h$ in two ways (the definition before the image and the definition after the image)? Another thing i don't understand is, a proof by induction is a proof of the type:

Where is the $p(n+1)$ in this proof?

Answer 1

It is a case study. First definition of $h$ is for the case $S\ne\emptyset$ and the second one is for the case $S=\emptyset$. In both cases, they show such an order-isomorphism exists.

For your second question, $p(n+1)$ is the construction of an order-isomorphism from $n+1$ (or $n^+, S(n))$ to $A$ when it is given that there is a bijection between $n+1$ and $A$.