Ordering Cycle implies Equality

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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 $x_1 = x_2 = x_3$.

$\blacksquare$