$\mathbb{R}$ and $\mathbb{R}\setminus \{0\}$ are not isomorphic as linear orders.

Informal argument
So, we know that $\mathbb{R}$ with its usual topology is connected, and of course $\mathbb{R}\setminus\{0\}$ is not. Any $$supposed" isomorphism should grant us a homeomorphism from $\mathbb{R}$ with its usual topology to $\mathbb{R}\setminus\{0\}$ with the topology it inherits from $\mathbb{R}$. So we conclude that they can't be isomorphic.

More Formal Argument
Suppose a bijection $i: \mathbb{R} \to \mathbb{R}\setminus \{0\}$ is such that $\forall x \forall y: x<y \longrightarrow f(x)<f(y)$, and additionally, $\forall a \forall b: a<b \longrightarrow f^{-1}(a)<f^{-1}(b)$. Extract a subsets $\Psi$ and $\xi$ of $\mathbb{R}\setminus\{0\}$ such that: $$\Psi = \{x\in \mathbb{R}\setminus \{0\} \mid x<-\frac{1}{n} \forall n\in \mathbb{N} \}$$ $$\xi = \{x\in \mathbb{R}\setminus \{0\} \mid x>\frac{1}{n} \forall n \in \mathbb{N}\}$$ It is clear that $\Psi \cup \xi = \mathbb{R}\setminus \{0\}$. It is also the case that $\Psi$, $\xi$ are open in the topology that $\mathbb{R}\setminus \{0\}$. Since $i^{-1}$ is an order preserving bijection, $i^{-1}$ grants us that $\mathbb{R}$ is the union of two disjoint sets: $i^{-1}(\Psi) \cup i^{-1}(\xi)$. It follows that the preimage of $\Psi$, $\xi$ are open in $\mathbb{R}$ and $\mathbb{R}$ is the union of two disjoint open sets, a contradiction.

My question
Is there a way to avoid talking about topological notions of connectedness, or at least phrase connectedness in the language of linear orders? This is an extracurricular problem from my course in set theory and logic and it feels as if I'm cheating.

For example, we have the notion elementarily equivalence. If two linear orders are isomorphic then they are elementarily equivalent. The contrapositive gives us if two linear orders are not elementarily equivalent, then they are not isomorphic. So I guess, my real question is: does there exist a sentence $\phi$ in the language of linear orders that says something about connectedness? I mean surely $(\mathbb{R},<)$ satisfies the axioms of dense linear orders, does $(\mathbb{R}\setminus \{0\}, <)$ $\models \forall x \forall y \exists z (x<z<y)$? (I believe it does). What sort of sentence should I use?

Edit
We desire to use the fact that $\mathbb{R}$ is a complete linear order, in that every subset that is bounded above, has a least upper bound. $\mathbb{R}\setminus\{0\}$ does not have this property, as Goos's answer shows. Thus, they are not isomorphic as linear orders.

• You shouldn't need to bring topology in to it. I think the notion you are looking for is completeness - in $\mathbb R$, every bounded-below set has a greatest lower bound. May 5 '14 at 6:31
• @Alex: Your edit overran mine; and I think [order-theory] is better than [logic]+[elementary-set-theory]. May 5 '14 at 6:39
• @AsafKaragila Makes sense, I agree. May 5 '14 at 6:40
• @AnthonyCarapetis Psh! damn. >_< (face palms) May 5 '14 at 6:40
• As ordered sets, $\mathbb R$ and $\mathbb R\setminus\{0\}$ are elementarily equivalent to each other and to $\mathbb Q$. Every countable densely ordered set without greatest or least element is isomorphic to $\mathbb Q$ (Cantor), and every densely ordered set without greatest or least element is elementarily equivalent to $\mathbb Q$.
– bof
May 5 '14 at 7:10

In $\mathbb{R} \setminus \{0\}$, the set $[-1, 0)$ has an upper bound, but no least upper bound. No such set exists in the linearly ordered $\mathbb{R}$.