Ordering Cycle implies Equality/General Case

Theorem
Let $\left({S,\le}\right)$ be an ordered set.

Let $x_1, x_2, \dots, x_n$ be elements of $S$.

Suppose that for $k=1,2,\dots,n-1: x_k \le k_{k+1}$.

Suppose also that $x_n \le x_1$.

That is, suppose that $x_1 \le x_2 \le \dots \le x_n \le x_1$.

Then $x_1 = x_2 = \dots = x_n$.

Proof
The proof proceeds by induction.

If $n=2$, then the conclusion follows immediately from the fact that $\le$ is antisymmetric.

Suppose the theorem holds for some $n$.

Let $x_1, x_2, \dots, x_n, x_{n+1}$ be elements of $S$.

Suppose that for $k=1, 2, \dots, n, n+1$, $x_k \le k_{k+1}$

Suppose also that $x_{n+1} \le x_1$.

Then since $\le$ is transitive, $x_n \le x_1$.

Inductively, $x_1 = x_2 = \dots = x_n$.

Since $x_n \le x_{n+1} \le x_1$, we see by antisymmetry that $x_{n+1}=x_n$, so
 * $x_1 = x_2 = \dots = x_{n+1}$

Also known as
refers to this property as anti-circularity.