Ordered set with a smallest element with element having a successor and and predecessor, not similar to the natural numbers I'm looking to give an ordered set that is not similar to $\mathbb N$ with a smallest element in which every element has a successor and a predecessor, which wouldn't apply to the least (that is the predecessor).
$\displaystyle \left\{\frac{-1}{n}, \frac{1}{n}, 1-\frac{1}{n} : n=1,2,3,...\right\}$ 
In this set, the least would be infinitesimally small as $n \rightarrow \infty$. Is this logic correct?
 A: Such an order exists: Let $X = \{0\} \times \mathbb N_0 \cup \{1\} \times \mathbb Z$ and for $(a,b),(c,d) \in X$ let $(a,b) \prec (c,d)$ iff $a < c$ or $a = c$ and $b < d$ (where $<$ denotes the natural order on $\mathbb Z$ and we consider $\mathbb N_0$ as a subset of $\mathbb Z$).
$(X, \prec)$ looks like $\mathbb N_0$ with $\mathbb Z$ "put on top", i.e. the first few elements of $(X, \prec)$ are $(0,0) \prec (0,1) \prec \ldots \prec (0,n)$ which may be thought of as the natural numbers $0 < 1 < \ldots < n$ and then we have $(0,n) \prec (1,z)$ for every $n \in \mathbb N_0$ and every $z \in \mathbb Z$.
It's easy to check that every element of $X$ has a unique successor and (if the element is not $(0,0)$) a unique predecessor. Since $(X, \prec)$ contains an infinite decreasing sequence (namely $(1,0) \succ (1,-1) \succ (1,-2) \succ \ldots$) it is not similar to $(\mathbb N_0, < )$.

edit: Note that André came up with the same ordering in his comment: His "red" copy of natural numbers relates to those $(0,n)$ for $n \in \mathbb N_0$ and his "blue" copy of integers are represented as $(1,z)$ for $z \in \mathbb Z$. By definition $\prec$ restricted to either "red" or "blue" elements is just the usual order on the "red natural numbers" or "blue integers" and every "red natural" is below every "blue integer".
