Is the infinite integer set order isomorphic to its infinite proper subset

Question

It is quite obvious that the statement is true, but is there any theorem to show that the infinite integer set is order isomorphic to its infinite proper subset? Thanks!

Answer 1

"Obvious" or not, it's not true for subsets of the integers. For example $\Bbb N$ is an infinite subset of $\Bbb Z$ which is not order-isomorphic to $\Bbb Z$.

It is true for sets of natural numbers, which is perhaps what you meant to say. This is so trivial it's not going to be a Theorem with a name:

Unnamed Theorem. If $S\subset\Bbb N$ is infinite then $S$ is order-isomorphic to $\Bbb N$.

Proof. Define $f:\Bbb N\to S$ recursively, by $f(0)=$ the smallest element of $S$ and $f(n+1)=$ the smallest element of $S$ larger than $f(n)$. Then it's not hard to show that $f$ is an order isomorphism. (One detail: for each $n\in\Bbb N$ you can show by induction on $m$ that $n<m$ implies $f(n)<f(m)$,)