Totally Ordered Set successor and predecessor unique I'm trying to prove that, in a totally ordered set, an element can have at most one successor and at most one predecessor. 
I know that if $x < y$ and there is no $z\in X$ with $x < z < y$ then $x$ is a predecessor of $y$ and $y$ is a successor of $x$.
I know that the successor and predecessor are unique but don't know how to establish it in a proof. Any advice would be greatly appreciated. 
 A: Note that the characterizations that you gave of successor and predecessor are definitions. In particular this means:


*

*$x$ is a predecessor of $y$ if and only if $x<y$ and there is no $z$ such that $x<z<y.$

*$y$ is a successor of $x$ if and only if $x<y$ and there is no $z$ such that $x<z<y.$


So, suppose (for example) that $y_1,y_2$ are successors of $x.$ Since $y_1$ is a successor of $x,$ then $x<y_1,$ and there is no $z$ such that $x<z<y_1.$ In particular, we cannot have $x<y_2<y_1.$ Likewise, $x<y_2,$ and there is no $z$ such that $x<z<y_2.$ In particular, we cannot have $x<y_1<y_2.$ Since $x<y_2,$ but we can't have $x<y_2<y_1,$ it follows that we can't have $y_2<y_1.$ Similarly, we can't have $y_1<y_2.$ From this, we can conclude (why?) that $y_1=y_2,$ which shows that $x$ has at most one successor (since if it has any successors, then they are all equal).
As similar proof approach works for uniqueness of predecessors (if they exist).
A: Hint: Suppose $y$ and $y'$ are two successors for $x$. Now which is the case: $y<y'$ or $y'<y$?
A: It is impossible to have two distinct successors to a point $x \in X$.
Assume, to the contrary, that we have two distinct successors $y_1$ and $y_2$ for a point $x \in X$. We know that the subset $\{x,y_1,y_2\}$ of $X$ can be naturally viewed as a totally ordered structure and it is easy to see that both $y_1$ and $y_2$ must continue being successors to $x$ in this 3 point space. Since $x$ is first, there are only two ways of totally ordering the finite subset:
$\tag 1 x \lt  y_1 \lt  y_2 $
$\tag 2 x \lt  y_2 \lt y_1 $
But if (1) is true then $y_2$ can't be a successor, and if (2) is true then $y_1$ can't be a successor. This is a contradiction.
In a similar manner it can be shown that predecessors must be unique.
