# Ordering Cycle implies Equality/General Case

## Theorem

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 Transitive Chaining, it follows from the first premise that for all $k$ with $0 \le k \le n$:

$x_0 \preceq x_k$

and also:

$x_k \preceq x_n$

The other premise states that $x_n \preceq x_0$.

By transitivity of $\preceq$, this combines with the above to:

$x_k \preceq x_0$

Since $\preceq$ is antisymmetric, this means that $x_0 = x_k$ for $0 \le k \le n$.

That is, $x_0 = x_1 = \dots = x_n$.

$\blacksquare$

## Also known as

1967: Garrett Birkhoff: Lattice Theory (3rd ed.) refers to this property as anti-circularity.