# Ordering Cycle implies Equality

## Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $x_1$, $x_2$, and $x_3$ be elements of $S$.

Suppose that

 $\ds x_1$ $\preceq$ $\ds x_2$ $\ds x_2$ $\preceq$ $\ds x_3$ $\ds x_3$ $\preceq$ $\ds x_1$

Then $x_1 = x_2 = x_3$.

### General Case

Let $\struct {S,\preceq}$ be an ordered set.

Let $x_0, x_1, \dots, x_n \in S$.

Suppose that for $k = 0, 1, \dots, n - 1: x_k \preceq x_{k + 1}$.

Suppose also that $x_n \preceq x_0$.

Then $x_0 = x_1 = \dots = x_n$.

## Proof

Because $\preceq$ is an ordering, it is transitive and antisymmetric.

By transitivity, $x_1 \preceq x_3$.

Because $x_1 \preceq x_3$ and $x_3 \preceq x_1$, antisymmetry allows us to conclude that $x_1 = x_3$.

Because $x_1 = x_3$ and $x_1 \preceq x_2$, we must have $x_3 \preceq x_2$.

Because $x_3 \preceq x_2$ and $x_2 \preceq x_3$, antisymmetry allows us to conclude that $x_2 = x_3$.

Thus $x_1 = x_2 = x_3$.

$\blacksquare$