$(X, \le) $ and $(Y, \le')$ be two well ordered sets.
$f:X\to Y$ with
i) $f$ is bijective.
ii) $a\le b \implies f(a) \le' f(b) $ $(\forall a, b\in X) $
is said to be an order isomorphism between $X$ and $Y$.
$X=\mathbb{N}$ and $Y=\mathbb{N}\cup\{\infty\}$
$\le$ : usual ordering in $\mathbb{N}$.
$n< {\infty}$ ,$\forall n\in {\mathbb{N}}$
Both are well ordered sets.
Question : Is there any order isomorphism between $X$ and $Y$?
(Proof by contradiction) : Assume $\exists f:X \to Y $ an order Isomorphism.
Then, there exists an unique $u \in X$ s.t. ${f(u) =\infty}\in Y$
Since, $X$ is not bounded above, we can find an element $v$ s.t. $u<v$ (in that case we can choose $u+1$)
$u+1 \in X \implies f(u+1) \in Y \setminus \{\infty\}$
And, $f(u+1) < \infty =f(u) $
So, $\exists u+1 \in X$ s.t $u<u+1$ but $f(u+1)<f(u) $.
Hence, the contradiction.
Is my proof correct? Or, is there an other way to proof this? Thanks.