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How do I show equivalence class of "="

let R={(a,a) such that $a\in$ A}.

I know I have to show reflexive, transitive, symmetry properties to show that this relation is an equivalence relation.

  1. Reflexive: For all a in A, a=a. proved.
  2. Symmetry: Suppose a=b, then a=a =>b=a.
  3. Transitive: Suppose a=b and b=c, then a=b=c. So, a=c.

But, I have trouble showing its equivalent class of "=". Any help is appreciated. Thank you.

3 Answers 3

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Since you never have any $(a,b) \in R$ with $a \not = b$, that means that every $a$ forms an equivalence class all by itself for any $a \in A$. Or, if you want: $[a] = \{ a \}$ for every $ a \in A$.

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    [a]=[a] is the equivalent class of "="? – 2017-02-14
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    No, $[a] = [a]$ will always be true, no metter what $[a]$ is. They are asking what the equivalence classes look like, i.e. What elements from $A$ are in $[a]$ for any $a$ .. And the answer is that for any $a$, only $a$ is in $[a]$ since for any $a$, only $a$ stands in the relation $R$ to $a$. – 2017-02-14
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    Therefore, equivalent class of a is a itself. – 2017-02-14
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    or I can write it as [a]=[a'] – 2017-02-14
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    The first one: the equivalence class of a is itself. Or: $[a] = \{ a \}$ – 2017-02-14
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    ok. This makes sense. Thank you. – 2017-02-14
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    @Lily I know you haven't been on this site for a long time ... Did you know you can accept answers by checking on the check mark next to them? – 2017-03-13
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Let $(a,b)\in A^2$.

  • If $(a,b)\in R$, then $a=b$.

  • If $a=b$, then $(a,b)\in R$.

Therefore, $(a,b)\in R$ if and only if $a=b$.

Does it help?

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    That is the definition of equivalent class. – 2017-02-14
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    I don't really get your problem then. – 2017-02-14
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By definition of $R$, $(a, a) \in R$. So, $R$ is reflexive.

Suppose $(a, b) \in R$. By definition, $(a, b)$ must be of the form $(c, c)$ for some $c$. But then $(c, a) \in R$ by the indiscernibility of identicals (substitutivity of equals). Similarly, $(b, a) \in R$ by a second application of the same principle. So, if $(a,b) \in R$, $(b, a) \in R$. That is, $R$ is symmetric.

Suppose $(a,b) \in R$ and $(b,c) \in R$. By definition, $(a,b)$ must be of the form $(d,d)$ for some $d$. But, then, $(b,c) \in R$ implies that $(d,c) \in R$ by the indiscernibility of identicals (substitutivity of equals). A second application of the same principle yields $(a,c) \in R$. So, $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$. That is, R is transitive.

Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation.

This is one of those nasty little bootstrap proofs that seem to make no sense unless one understands that one is relating two distinct notations to one another. The sign of equality "warrants" substitutions in the notation of the class elements as ordered pairs.

And, I just noticed -- proving that a relation is an equivalence relation is not "showing the equivalence class" of a relation. I think your wording is leading you to some confusion.