# Proving the symmetry of an equivalence relation

When proving the symmetry of an equivalence relation, must each equivalence class be closed under symmetry.

for example:

the relation

both x and y > 10

or

both x and y < 10

across all integers.

this clearly has two equivalence classes, the first being all numbers greater than 10 and the second being all numbers less than 10.

take x = 12 y = 17

so:

12 and 17 > 10

and then symmetrically:

12 and 17 < 10

It has kept within the relation but has moved to the other equivalence class. is it still an equivalence relation?

• "and then symmetrically: 12 and 17 < 10" Absolutely quite deadly sure about that? – Did Oct 6 '13 at 7:49
• What about $x=3$, $y=11$? – Michael Hoppe Oct 6 '13 at 7:53

The relation here is $$x R y \iff (x > 10 \;\land \;y> 10)\;\text{ or } \;(x < 10 \;\land \;y < 10).$$
So, if $xRy$, we have that \begin{align} x R y &\iff (x > 10 \land y > 10) \lor (x \lt 10 \land y<10) \\ \\ & \iff (y > 10 \land x > 10) \lor (y\lt 10 \land x\lt 10)\\ \\ & \iff y R x\end{align}, so the relation is symmetric.
The classes you've identified to not exhaust all possible ordered pairs of integers: What are we to do about: $$(x, y): \quad (4, 12),\;\text{ or }\; (12, 4),\; \text{ or } \; (10, 10)\quad?$$