Please check my below proof. Thank you for your help!
Definition:
$I_n=\{m \in \mathbb{N} \mid m \leq n\}$
Theorem:
$I \subseteq I_n$ and $I \neq \varnothing \implies I$ has the greatest element.
Proof:
Let $M$ be $\{m \in \mathbb{N} \mid \forall i \in I, i \leq m\}$. Then $n \in M \implies M \neq \varnothing$. By well-ordering theorem, $M$ has the least element $m_0$. Now we prove $m_0 \in I$. There are two possible cases.
- $m_0=0$
$\implies I=\{0\} \implies I$ has only one element $0$ and $0$ is also the greatest element of $I$.
- $m_0>0$
$\implies m_0=k+1$. If $m_0 \not \in I$, then $\forall i \in I, i < m_0$. Thus $\forall i \in I, i \leq k \implies k \in M \implies k+1$ is not the least element of $M \implies m_0$ is not the least element of $M$. (CONTRADICTION).
As a result, $m_0 \in I$. So we have that $I$ has the greatest element.
