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What is the best way and easiest way to approach this problem?

My first relation is going to be defined as such

R = {(a,a), (b,b), (c,c)} which is reflexive, symmetric, and vacuously transitive.

My second relation will be defined as such

R = {(a,a), (b,b), (c,c), (a,c), (c,a)}

Am I on the right track here?

It seems to me that I can make more than 5 equivalence relations.. ie.

R = {(a,a), (b,b), (c,c)}

R = {(a,a), (b,b), (c,c), (a,c), (c,a)}

R = {(a, a), (b,b), (c, c), (a,b), (b,a)}

R = {(a, a), (b,b), (c, c), (b,c), (c,b)}

R = {(a, a), (b,b), (c, c), (a,c), (c,a), (b,a), (a,b)}

R = {(a, a), (b,b), (c, c), (a,c), (c,a), (b,a), (a,b) (b,c), (c,b)}

Hare Krishan
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2 Answers2

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The fifth relation in your list is not transitive: $(c, a), (a, b) \in R,$ but $(c, b)\not\in R$. Also note that $(b, a), (a, c) \in R$ but $(b, c)\notin R$. In order to make that relation transitive (i.e, finding the transitive closure of the relation) yields your last relation, which is indeed an equivalence relation.

So, after deleting the fifth relation in your list, you'll have found the five equivalence relations on $\{a, b, c\}$.

Jordan Glen
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If you know that there is a one-to-one correspondence between equivalence relations on a set and partitions of the set, it may be easier to find partitions. (Note: For each partition, the subsets making up the partition are the equivalence classes of the corresponding equivalence relation).

In your case, the five partitions are

$\{a\}, \{b\}, \{c\}$;

$\{a\}, \{b,c\}$;

$\{b\}, \{a,c\}$;

$\{c\}, \{a,b\}$; and

$\{a,b,c\}$

paw88789
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