Proving that the two structures $(\mathbb{R},<)$ and $(\mathbb{R}\setminus\{0\},<)$ are not isomorphic For an exercise on model theory I have to prove that the structure $(\mathbb{R},<)$ is not isomorphic to $(\mathbb{R}\setminus\{0\},<)$. 
I found the function $f(x)=\begin{cases}
x & x\notin \mathbb{N}\\
x+1 & x\in\mathbb{N}
\end{cases}$, which is a bijection between $\mathbb{R}$ and $\mathbb{R}\setminus\{0\}$. For this function it is clear that the relation $<$ won't be the same under $f$, for instance $3<\pi$ but $f(3)=4\not < \pi=f(\pi)$.
How can I prove that every bijection from the one to the other will give the same problem  with the relation $<$
 A: In the first structure, every increasing sequence which is bounded above has a least upper bound. In the second, there is a bounded increasing sequence that does not have a least upper bound. 
The property that every bounded increasing  sequence has a least upper bound is preserved by any order isomorphism. 
A: Note that if $(A,<_A)$ and $(B,<_B)$ are two isomorphic ordered sets with $f\colon A\to B$ such isomorphism, and $X\subseteq A$ is bounded, then $f[X]$ is bounded in $B$. This is because for some $a\in A$, every $x\in X$ satisfies $x<_A a$, and therefore $f(x)<_B f(a)$. Similarly the property of being the least upper bound is also preserved under isomorphisms.
It is important to point that these properties need not be expressible as a first-order property of some element or set, or whatever the above $X$ or $a$ need not be definable. But isomorphisms preserve more than just the first-order theory of the structure, and we can talk about these properties as we would, externally to the structure.
Now notice that $\Bbb R\setminus\{0\}$ satisfies that $\{x\mid x<0\}$ is bounded but does least upper bound; whereas in $\Bbb R$ every bounded set has a least upper bound.
