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Show that for equivalence relation $\sim$ on the set $A$ and $a,b\in A,[a]=[b] \iff a\sim b$.

This is an advanced practice problem.

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Can you at least prove the $\implies$ direction? It’s absolutely immediate from the definition of $[a]$. –  Brian M. Scott Feb 6 '13 at 0:10
    
What is the definition of equivalence class for you? –  Sigur Feb 6 '13 at 0:10
    
Yes, I have an idea of $\implies$ involving the reflexivity of $\sim$. I'm having more trouble trying to assemble the argument in the opposite direction. –  beethree Feb 6 '13 at 0:18
    
My definition of equivalence class is $[a]:=\{a_2\in A : a_1\sim a_2\}$ –  beethree Feb 6 '13 at 0:19
    
This ended up being much simpler than I had expected. –  beethree Feb 6 '13 at 0:58

2 Answers 2

up vote 2 down vote accepted

Recall that an equivalence relation is reflexive, symmetric and transitive.

By reflexivity $a\in[a]$ and $b\in[b]$. If $[a]=[b]$ then $b\in[a]$ as well, and by definition of $[a]$ we have $a\sim b$.

If $a\sim b$, we need to show a double-sided inclusion, let $c\in[b]$ then $b\sim c$, and by transitivity $a\sim c$ and therefore $c\in[a]$, so $[b]\subseteq[a]$.

I am leaving you the final case of showing $[a]\subseteq[b]$ and deducing the equality.

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Thank you for your assistance. –  beethree Feb 6 '13 at 0:58

Note also that equivalence classes partition $A$, and so for $a,b\in A,[a] \neq[b] \iff [a] \cap [b] = \emptyset$, or equivalently $ [a] = [b] \iff [a]\cap [b] \neq \emptyset$.

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How can you prove that the equivalence classes form a partition before you proved that every two are either equal or disjoint? This answer is exactly what there is to be proved here. –  Asaf Karagila Feb 6 '13 at 0:25

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