# Preordering induces Equivalence Relation

## Theorem

Let $\struct {S, \precsim}$ be a preordered set.

Define a relation $\sim$ on $S$ by letting $x \sim y$ if and only if $x \precsim y$ and $y \precsim x$.

Then $\sim$ is an equivalence relation.

## Proof

To show that $\sim$ is an equivalence relation, we must show that it is reflexive, transitive, and symmetric.

By the definition of preordering, $\precsim$ is transitive and reflexive.

### Transitive

Let $p, q, r \in S$.

Suppose that $p \sim q$ and $q \sim r$.

Then $p \precsim q$, $q \precsim r$, $r \precsim q$, and $q \precsim p$.

Since $\precsim$ is transitive:

- $p \precsim r$ and $r \precsim p$.

Thus by the definition of $\sim$, $p \sim r$.

Since this holds for all $p$, $q$, and $r$, $\sim$ is transitive.

$\Box$

### Reflexive

Let $p \in S$.

Since $\precsim$ is reflexive:

- $p \precsim p$

Thus by the definition of $\sim$:

- $p \sim p$

As this holds for all $p$, $\sim$ is reflexive.

$\Box$

### Symmetric

Let $p, q \in S$ with $p \sim q$.

Then $p \precsim q$ and $q \precsim p$.

Thus $q \sim p$.

Since this holds for all $p$ and $q$, $\sim$ is symmetric.

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: Further exercises: $5$