How do I show there isn't an order isomorphism b/w the two sets $\{1, 2, 3,...\}$ and $\{1, 2, 3, ..., \omega \}$

Question

That is, how can I prove there isn't a bijection $f$ from one set to the other such that $f(x) < f(y)$ iff $x < y$?

Answer 1

Suppose that $f:\{1,2,3,\dots\}\to\{1,2,3,\dots,\omega\}$ is an order-preserving bijection. There must be some $n\in\{1,2,3,\dots,\}$ such that $f(n)=\omega$. What can $f(n+1)$ be?

Answer 2

Note that an order isomorphism preserves maximum properties, namely if $a$ is a maximum then $f(a)$ is a maximum of the image.

In particular a linearly ordered set with a maximum is never isomorphic to one without.