Ordering Cycle implies Equality
Theorem
Let $\left({S,\preceq}\right)$ be an ordered set.
Let $x_1$, $x_2$, and $x_3$ be elements of $S$.
Suppose that
\(\displaystyle x_1\) | \(\preceq\) | \(\displaystyle x_2\) | |||||||||||
\(\displaystyle x_2\) | \(\preceq\) | \(\displaystyle x_3\) | |||||||||||
\(\displaystyle x_3\) | \(\preceq\) | \(\displaystyle x_1\) |
Then $x_1 = x_2 = x_3$.
General Case
Let $\left({S,\preceq}\right)$ 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
Since $\preceq$ is an ordering, it is transitive and antisymmetric.
By transitivity, $x_1 \preceq x_3$.
Since $x_1 \preceq x_3$ and $x_3 \preceq x_1$, antisymmetry allows us to conclude that $x_1 = x_3$.
Since $x_1 = x_3$ and $x_1 \preceq x_2$, we must have $x_3 \preceq x_2$.
Since $x_3 \preceq x_2$ and $x_2 \preceq x_3$, antisymmetry allows us to conclude that $x_2 = x_3$.
Thus $x_1 = x_2 = x_3$.
$\blacksquare$