Nonisomorphism of real numbers and reals without 0 (the only symbol is the < symbol) Why are $(\mathbb{R}, <)$ and $(\{ x \in \mathbb{R} | x \neq 0\}, <)$ not isomorphic?
My approach is to assume that there is an isomorphism and attempt to derive a contradiction by showing that there is a formula which the first structure satisfies but the second does not. I'm having trouble finding one.
 A: Your idea is good. Indeed, two isomorphic structures satisfy the same formulas. But pay attention, the converse is not true. For example, if you covered the notion of back-and-forth, you may have shown that the theory of dense linear orders without extremities is complete. Both $(\mathbb R,<)$ and $(\mathbb R\backslash\{0\},<)$ are such orders, so they are elementarily equivalent, and hence satisfy the same formulas.
So how can you show that they are not isomorphic? By showing that they don't satisfy the same properties that are not necessarily expressed in first order logic. For example, for the same reasons as above, $(\mathbb Q,<)$ and $(\mathbb R,<)$ are both elementarily equivalent, yet they're clearly not isomorphic because $\mathbb Q$ is countable but not $\mathbb R$. You can't say in the first order logic with the language $\{<\}$ that a model is countable or not.
Same thing goes for $(\mathbb R,<)$ and $(\mathbb R\backslash\{0\}$,<). One of them satisfies the least upper bound property, but not the other. This shows that they are not isomorphic.
Leo163's comment gives another proof. The usual topology on $\mathbb R$ coincides with its order topology, and same goes for $\mathbb R\backslash\{0\}$. The way this topology is defined, you can see that if two totally ordered sets are isomorphic (as ordered sets) then they're homeomorphic (as topological spaces, where the topology of each one is the order topology). So, if $(\mathbb R,<)$ and $(\mathbb R\backslash\{0\},<)$ were isomorphic, $\mathbb R$ would be homeomorphic to $\mathbb R\backslash\{0\}$, but one is connected and the other is not.
