[Math] Totally Ordered Set successor and predecessor unique

Question

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.

Answer 1

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).