Recipe

Check a relation and find its classes

Use when: you're handed a specific relation on a set and need to know whether it's an equivalence relation, and if so, what its classes are

Goal: either a confirmed equivalence relation with its full list of classes, or the specific property that fails

Taught in Equivalence relations and partitions, revisited

1. 1

   Confirm a \sim a holds for every element a in the set. Don't just try a few examples and stop there.

   Example

   Let A = \{0,1,\ldots,8\} and x \sim y mean x and y leave the same remainder mod 4. Every x obviously relates to itself, so x \sim x holds for every x \in A.

   Why: an element missing from its own class right away breaks the whole idea of a class containing it.

2. 2

   Confirm a \sim b implies b \sim a for every related pair. Check both directions, not just one.

   Example

   If x \sim y, then y \sim x too: the relation doesn't care which one is named first.

   Why: without this check, an element could count as related to a while a doesn't count as related back. That leaves "the class containing a" unclear.

3. 3

   Confirm a \sim b and b \sim c together force a \sim c. Check every such chain in the relation, not just the pairs that come to mind first.

   • all three checks pass — continue to the next step and build the classes.
   • any single check fails — stop here. The relation is not an equivalence relation, and no class list can be built from it.

   Why: the transitive check is the one most often skipped. Skipping it is exactly what lets "classes" overlap or leave elements out.

4. 4

   For each element a, collect [a] = \{x \in A : x \sim a\}. Repeat until every element of A sits in some class.

   Example

   Continuing with A=\{0,\ldots,8\}: sort by remainder mod 4 to get [0]=\{0,4,8\},\quad [1]=\{1,5\},\quad [2]=\{2,6\},\quad [3]=\{3,7\}

   Why: once all three checks pass, the recalled theorem guarantees these classes partition A automatically. No element is left out or counted twice.