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I might be interpreting this wrong but in my book it says: If ~ defines an equivalence relation on $A$ then the set of equivalence classes of ~ form a partition of $A$. To me, this means that the set of equivalence classes of the elements of $A$ forms a partition $A_i$ for some $i \in I$ (I is an indexing set). ex: let $A = \{a,b,c\}$ and let ~$ \subseteq A \times A$ denote the equivalence relation ~, where the set of equivalence classes are $\{\langle a,c\rangle,\langle c,a\rangle \}$ since the equivalence of $a$ is $c$ and the equivalence of $c$ is $a$. (is this a correct use of these terms?).

So the set $\{\langle a,c\rangle,\langle c,a\rangle \}$ is a partition of $A$? I don't get why this becomes a partition, assuming what I did was correct.

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    $\begingroup$ What you did is not correct. $a=c$ does not denote an equivalence relation. You should define an equivalence relation $\sim \subset A \times A$. $\endgroup$
    – Crostul
    Apr 6, 2016 at 20:45
  • $\begingroup$ Oh damn I forgot that an equivalence relation is a set of coordinate pairs I have to edit this one second. $\endgroup$
    – Obliv
    Apr 6, 2016 at 20:49

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I'm not entirely sure what you mean by "$a=c$" defining a relation, but if you mean "$a\sim c$", then it is not an equivalence relation for several reasons (e.g. $b$ is not related to itself, making it non-reflexive).

But maybe you meant to define the following equivalence relation on $A=\{a,b,c\}$:

$a\sim a$, $b\sim b$, $c\sim c$, $a\sim c$, $c\sim a$.

The equivalence classes of this relation are $\{a,c\}$ and $\{b\}$, and $\{\{a,c\},\{b\}\}$ is a partition of $\{a,b,c\}$.

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  • $\begingroup$ Oh the equivalence relation must apply to ALL elements of $A$. Okay I understand why it is a partition now. Because they do not intersect and the union of all sets equal $A$. Thank you! $\endgroup$
    – Obliv
    Apr 6, 2016 at 20:55
  • $\begingroup$ Last question: Can a set of ordered pairs like $A \times A$ be united to form the set $A$? Like, $\cup$ is not restricted to just sets right? $\endgroup$
    – Obliv
    Apr 6, 2016 at 20:57
  • $\begingroup$ @Obliv I'm glad it makes sense. Let me also point something out from your post. The set $\{\langle a,c\rangle,\langle c,a\rangle\}$ is not a partition of $A$. It is also not an equivalence class, since any equivalence class is a subset of $A$, not of $A\times A$. As for your second question, maybe this comment answers it. A union of subsets of $A\times A$ cannot be equal to $A$. The elements of $A\times A$ are of a different type than those of $A$, e.g. $(a,b)\in A\times A$, where $a, b\in A$. Only a union of subsets of $A$ can be equal to $A$. I hope this answers your question. $\endgroup$
    – Mankind
    Apr 6, 2016 at 21:09

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