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In the book of Richard Hammack, I come accross with the following question:

There are two different equivalence relations on the set $A = \{a,b\}$. Describe them.

OK, I found that the solution is,

$$R_1 = \{(a,a),(b,b),(a,b),(b,a)$$ and $$R_2 = \{(a,a),(b,b)\}$$

Then I thought two more equivalence classes $R_3 = \{(a,a)\}$, $R_4 = \{(b,b)\}$. But when I looked the answer, I saw that, $R_1$ and $R_2$ are true but others are false. Why is that?

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4 Answers 4

up vote 4 down vote accepted

Because they're not reflexive. An equivalence relation is reflexive, i.e. it contains all pairs of the form $(x,x)$.

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A relation is an equivalence relation if and only if it is reflexive, symmetric, and transitive. Your first two relations are indeed equivalence relations.

A relation $R$ is reflexive on a set $A$ if and only if for all $x \in A, (x, x) \in R$.

In $R_3$, we do not have that for $b \in A$, $(b, b) \in R_3$.

And in $R_4$, we do not have that for $a \in A$, $(a, a) \in R_4$.

So neither $R_3$ nor $R_4$ are reflexive, hence neither can be an equivalence relation.

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Read the reflexive property again.

Reflexive property is not conditional.

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$R_3 = \{(a, a)\}$ and $R_4 = \{(b, b)\}$ are not equivalence relations because they are not reflexive; $b$ must be related to itself.

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