# How to show that symmetry holds for equivalence classes of an equivalence relation $R$ on a set $A$?

Given that equivalence classes of an equivalence relation form a partition of a set $$A$$, how does one show that symmetric holds? If two equivalence class $$[a]$$ and $$[b]$$ of an equivalence relation are disjoint and since equivalence relations are reflexive, $$a\in[a]$$ and $$b\in[b]$$, which means that $$a$$ cannot be in $$[b]$$ and vice versa. However, since equivalence relations are also symmetric, $$aRb$$ implies $$bRa$$, which to me implies that $$b$$ should be in $$[a]$$ and vice versa? Is there a way (by example) to show that symmetry is satisfied despite the equivalence classes being pairwise disjoint?

• If equivalence classes of $a$ and $b$ are disjoint, then it simply means that $a$ is not related to $b$. So no need to look for symmetry in this case. – PJK Jul 1 at 4:46
• Symmetry is one of the defining properties of equivalence relations. If you have an equivalence relation, then by definition, you have symmetry, and the relation partitions the set into equivalence classes. Symmetry only applies to relations, not partitions! I'm guessing that you're looking at a proof that, given a partition, you can find an equivalence relation, the set of whose equivalence classes is the partition. Can you confirm this is the case? – Theo Bendit Jul 1 at 4:49
• Yes, it does mean $[c] = [d]$. You can prove this using a classic subset argument; show that if $x \in [c]$, then $x \in [d]$, i.e. if $x R c$, then $x R d$ (using $c R d$). Then show if $x \in [d]$, then $x \in [c]$. – Theo Bendit Jul 1 at 5:11
• The whole point to equivalence classes is this: $aRb$ if and only if $[a]=[b]$. So symmetry of $R$ comes down to "$[a]=[b]$ if and only if $[b]=[a]$". – David C. Ullrich Jul 1 at 16:37
• Consider an example: On $\Bbb Z$ let $aRb \iff (a-b)/2\in \Bbb Z.$ There are just two equivalence classes: $[0]= \{2n: n\in \Bbb Z\}$ and $[1]=\{2n+1: n\in \Bbb Z\}.$ – DanielWainfleet Jul 1 at 19:41

## 1 Answer

If $$C_i, i \in I$$ form a partition of a set $$X$$ (so $$\cup_{i \in I} C_i=X$$ and $$\forall_{i,j \in I}: i \neq j \implies C_i \cap C_j= \emptyset$$), defining $$xRy$$ iff $$\exists_{i \in I}: (x \in R_i \land y \in R_i)$$ then $$R$$ is symmetric because "and" is symmetric, essentially: $$x \in R_i \land y \in R_i \iff y \in R_i \land x \in R_i$$. It turns out that $$R$$ is indeed an equivalence relation and the class of $$x$$ is the unique $$C_i$$ it is in. So if you start with a set of classes the induced relation $$R$$ is very easily symmetric.