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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?

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    $\begingroup$ "and then symmetrically: 12 and 17 < 10" Absolutely quite deadly sure about that? $\endgroup$ – Did Oct 6 '13 at 7:49
  • $\begingroup$ What about $x=3$, $y=11$? $\endgroup$ – Michael Hoppe Oct 6 '13 at 7:53
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You're not understanding the property of symmetry held by all equivalence relations.

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.

Also, a relation on the set is an equivalence relation if and only if it partitions that set, the integers in this case, into equivalence classes, (1) the union of which is the entire set, such that (2) the intersection of any two equivalence classes is empty.

The classes you've identified to not exhaust all possible ordered pairs of integers: What are we to do about: $$(x, y): \quad (4, 12),\;\text{ or }\; (12, 4),\; \text{ or } \; (10, 10)\quad?$$

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  • $\begingroup$ this needs a TU! +1 $\endgroup$ – Amzoti Oct 6 '13 at 17:15

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