Equivalence relations: showing $[b] \subseteq [a]$ while knowing $[a] \subseteq [b]$

Question

Let $R$ be an equivalence relation on $A$ for two elements $a,b \in A$. Show that $[b]\subseteq [a]$.

In a previous section it was already proven that $[a]\subseteq [b]$. Can I just argue that by symmetry it must be true that $[b]\subseteq [a]$?

EDIT: Here's my attempt.

Assume $aRb$. Take any $t\in [a]$. Then $aRt$. By symmetry, $tRa$. Given $[a]\subseteq [b]$ then $t\in [b]$, and so $bRt$. Since $tRa$ and $bRt$, then $[a]=[b]$, therefore it follows $[b]\subseteq [a]$.

Answer 1

The proof is mostly superfluous, since it does effectively follow by a symmetric argument. You have that $[a] \subseteq [b]$ for all pairs $(a,b) \in R$. Taking the ordered pair $(a',b') := (b,a)$ will then have $[a'] = [b] \subseteq [a] = [b']$. But we can always go from a more axiomatic perspective, as you and your text did. (And there's no harm in fully justifying things.)

This would be a more basic set-inclusion proof though, and I think you overcomplicate matters just a bit. Here's how I would do it... Assume $aRb$; thus $bRa$ by symmetry. We want to show $[b] \subseteq [a]$, i.e. $\forall x \in [b]$, it follows $x \in [a]$. Observe:

\begin{align*} x \in [b] &\iff xRb \tag{def.} \\ &\implies xRa \tag{transitivity} \\ &\iff x \in [a]\tag{def.} \end{align*}

• The first is just definition of membership of an equivalence class represented by $b$
• The second is transitivity applied to the facts that $xRb$ and $bRa$
• The third is again definition of membership of equivalence classes

Thus, $[b] \subseteq [a]$.

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...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.