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I want to prove that given a partition $\lbrace A_i|i\in I\rbrace$ of $A$(where $I$ is some indexing set), that there exists some equivalence relation $\sim$ such that the equivalence classes of $\sim$ are the sets $A_i$.

Here is my attempt at the proof:

Let $\lbrace A_i|i\in I\rbrace$ be a partition of $A$. Let there be a binary relation $\sim$ on $A$ such that $\forall a,b\in A$, $a\sim b$ if $a,b\in A_i$. If $a\in A_i$, then $a\sim a$ must be true. If $a,b\in A_i$, then $b,a\in A_i$, meaning $a\sim b$, and $b\sim a$. If $a,b\in A_i$, and $b,c\in A_i$, then $a,c\in A_i$, meaning $a\sim b$, $b\sim c\implies a\sim c$. Therefore, $\sim$ is an equivalence relation. For any element $a\in A$, $[a]=\lbrace x\in A|\space x\sim a\rbrace=\lbrace x\in A|\space x\in A_i\rbrace$. Therefore, each equivalence class of $\sim$ is a unique set $A_i$.

I want to know if

$1)$ There are any corrections to be made and

$2)$ Better wording for the proof.

Thank you all in advance.

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    $\begingroup$ Looks fine to me.$\ddot \smile$ $\endgroup$ – Donald Splutterwit Sep 30 '17 at 19:23
  • $\begingroup$ There are some ways to improve the wording: you want to say that $a\sim b$ if there is some $i\in I$ such that $a,b\in A_{i}.$ As it's worded now, it might sound like you're only referring to one set in the collection $A_{i},\,i\in I.$ Similarly, at the end, you want to say "for any $a\in A_{i}$, $[a]=\{x\in A:x\sim a\}=A_{i}$ (which equals the last set you wrote), since it's not true that for all $a\in A,$ $[a]=A_{i},$ as you've written. $\endgroup$ – RideTheWavelet Sep 30 '17 at 19:26
  • $\begingroup$ @RideTheWavelet Ah, I see thank you. $\endgroup$ – 高田航 Sep 30 '17 at 19:27

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