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$1$. True or false? If $R$ is an equivalence relation on a set $S$ and it has only finitely many equivalence classes altogether, then $S$ itself is a finite set.

I think the answer is true because if $R$ is an equivalence relation then it will be reflexive,symmetric and transitive.

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    $\begingroup$ That is the definition of an equivalence relation, but I don't see how it's relevant or how the given statement follows from it. $\endgroup$ – anomaly Sep 8 '14 at 4:42
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    $\begingroup$ False. Take the set $\mathbb{Z}$ of all integers, and consider the universal relation (that relates every number with every other number). Then there is only one equivalent class, but the set $\mathbb{Z}$ itself is infinite. $\endgroup$ – E W H Lee Sep 8 '14 at 4:45
  • $\begingroup$ thanks....can provides some more example ,so that i exactly understand on how to model this question $\endgroup$ – anish Sep 8 '14 at 5:00
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    $\begingroup$ What you think is false and the reason why you think that it is true, is also false. $\endgroup$ – Hirak Sep 8 '14 at 5:23
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On the set of integers $\mathbb Z$ define a relation $\rho$ by $a\rho b$ iff $3|a-b$ Then $\rho$ is equivalence relation and eqivalence classes are $[0],[1],[2]$ i.e the remainders when divided by $3$ But $\mathbb Z$ is not finite.

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Let $S$ be any set. Consider any partition of $S$ into subsets. That partition determines an equivalence relation on $S$, namely, two elements of $S$ are equivalent if and only if they are in the same subset. So there are no restrictions on the size of $S$, nor on the number of equivalence classes, except that the latter can't be any bigger than the former.

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