Please check if my proof has any error. Thank you for your help!
Definition:
$A$ is an initial segment of $B \iff A \neq \varnothing$ and $\forall b \in B(a \in A \wedge b \leq a \implies b \in A)$.
$I_n=\{m \in \mathbb{N} \mid m \leq n\}$
Theorem:
$I$ is an initial segment of $\mathbb{N} \implies I=\mathbb{N}$ or $I=I_n$
Proof:
It is clear that $(m \not \in I \wedge m \leq n \implies n \not \in I)$.
Let $C = \mathbb{N} \setminus I$. There are two possible cases.
- $I=N$
The theorem is trivially true.
- $I \neq N$
$\implies C \neq \varnothing$. By well-ordering principle, $C$ has the least element $c_0 \implies c_0 > 0$ (If $c_0=0$, then $I=\varnothing$).
$c_0 > 0 \implies c_0=n+1$.
$p \in I \implies p < c_0 \implies p<n+1 \implies p \leq n \implies p \in I_n$. Thus $I \subseteq I_n$.
$p \in I_n \implies p \leq n \implies p < n+1 \implies p<c_0 \implies p \not \in C \implies p \not \in \mathbb{N} \setminus I \implies p \in I$. Thus $I_n \subseteq I$.
$I \subseteq I_n \wedge I_n \subseteq I \implies I=I_n$.
The are two cases.
If I has an upper bound, then I has a maximum m.
Thus I = $I_m$.
If I does not have an upper bound, then show I = N.