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So , I have been revising my lecture notes on relations . When I read the definition of an equivalence class of an element a of a relation R, I came forward those bullets:

  1. reflective: $ a \in [a]_R$
  2. symmetric: if $b \in [a]_R $ then $a \in [b]_R$
  3. transitive: if $ b \in [a]_R$ and $c \in [a]_R$ then $(b,c) \in R$ ( I don't quite get that : If b is related to a and c is related to a , why it is sure that b is related to c?)

So this is supposed to show us that an equivalence class is an equivalence relationship itself? As far as I have undestood theory , a relation basically defines a set ( of elements that satisfy whatever this relation defines). An equivalence class, also defines a set

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  • $\begingroup$ The property of transitivity is wrong. The correct statement would be $-$ If $a \in [b]_R$ and $b \in [c]_R,$ then $a \in [c]_R$ (so $(a,c) \in R$.) $\endgroup$
    – Air Mike
    Commented Aug 31, 2020 at 10:33
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    $\begingroup$ It seems that you want to prove that belonging to the same equivalence class defines an equivalence relation. Actually, it is immediate, because by definition, "belonging to the same equivalence class" means "to be in $R-$relation". $\endgroup$ Commented Aug 31, 2020 at 10:33
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    $\begingroup$ @AirMike Actually, once you have proved symmetry, the transitivity can be written this way (it is equivalent to $b \in [a]$ and $a \in [c]$...) $\endgroup$ Commented Aug 31, 2020 at 10:35
  • $\begingroup$ @TheSilverDoe that is true! I was just stating in a way that it doesn’t depend on symmetry. But both work just fine as well :) $\endgroup$
    – Air Mike
    Commented Aug 31, 2020 at 10:42
  • $\begingroup$ @AirMike , probably not (check an answer below ) $b \in [a]_R \rightarrow (a,b) \in R$ and $c \in [a]_R \rightarrow (a,c) \in R$ and since we said that R is an equivalence relation ( hence symmetric, transitive) then also $(c,a) \in R \rightarrow (c,a) and (a,b) \in R \rightarrow (c,b) \in R$ $\endgroup$ Commented Aug 31, 2020 at 15:56

2 Answers 2

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Intuitively you can think of an equivalence class as a set of things you can transform between. So if $a \in [b]_{R}$ then you can transform from $a$ to $b$ (and vice versa due to the symmetric property) which I will denote by $a\rightarrow b$. Since you can perform the transform $a\rightarrow b$ and $c \rightarrow a$ (again symmetry property) then you can combine them to get:

$$ c\rightarrow a \rightarrow b$$

and therefore you can transform from $c$ to $b$ (and vice versa) and so $b$ is in the equivalence class of $c$. This means there is a relation between $b$ and $c$ and so $(b,c)\in R$.

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  • $\begingroup$ i think you are right, , $b \in [a]_R \rightarrow (a,b) \in R$ and $c \in [a]_R \rightarrow (a,c) \in R$ and since we said that R is an equivalence relation ( hence symmetric, transitive) then also $(c,a) \in R \rightarrow (c,a) and (a,b) \in R \rightarrow (c,b) \in R$ $\endgroup$ Commented Aug 31, 2020 at 15:55
  • $\begingroup$ For implies its best to use double lines (so capitalise \Rightarrow and \Leftarrow), single lines tend to imply a transformation (or morphism) rather then an implication. I'm also guessing you meant your last equation to read $(c,a)\in R \ \mathrm{and} \ (a,b)\in R \Rightarrow (c,b) \in R$ (use \mathrm{text} to get non-math text and you can add spaces in math mode by a backslash with a space after it, or \quad, or \qquad for various lengths) $\endgroup$ Commented Sep 1, 2020 at 9:57
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Given a set $A$ and a relation $R$ on $A$, $R$ does not define a set; rather $R$ is a set. More precisely, it is a subset of $A\times A$.

If it turns out that $R$ is an equivalence relation, then, for each $a\in A$, you define the equivalence classe of $a$ as:$$[a]=\{b\in A\mid a\mathrel Rb\}.$$Note that $[a]$ is a subset of $A$, whereas $R$ is a subset of $A\times A$. And the set of all equivalence classes is a subset of $\mathcal P(A)$.

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  • $\begingroup$ okay , so I can have in my mind that a relation is a set , right? $\endgroup$ Commented Aug 31, 2020 at 15:58
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    $\begingroup$ Since that is part of the definition of relation, yes, you can. $\endgroup$ Commented Aug 31, 2020 at 16:05

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