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Suppose $R$ is an equivalence relation on $A$ such that there are only finitely many distinct equivalence classes $A_1,A_2,\ldots,A_k$ w.r.t $R$. Show that $$A=\bigcup_{i=1}^k A_i$$

Since $A_i\subset A$ for each $i$, $$\bigcup_{i=1}^{k}A_i \subseteq A$$. For the other side, Let $a \in A$. Then $a \sim a$. Hence $a \in Cl(a)$. Since there are only finitely many equivalence classes, $Cl(a)=A_i$. Then $a \in \bigcup_{i=1}^{k} A_i$. And we are done.

Another one on the same lines: Show that $$R=\bigcup_{i=1}^{k}A_i\times A_i$$ For this: Let $(x,y) \in R$. Then $x \sim y$. Hence $x,y \in A_i$ for some $i$. Then $(x,y) \in A_i \times A_i \implies (x,y)\in \bigcup_{i=1}^{k}A_i \implies R \subseteq \bigcup _{i=1}^{k}A_i $.

For the other side let $(a,b) \in \bigcup _{i=1}^{k}A_i \times A_i$ which gives $(a,b) \in A_j \times A_j$ for some $j$ . Hence $a \in A_j$ and $b \in A_j$ and hence $a \sim b$. So $(a,b) \in R$. So $\bigcup _{i=1}^{k}A_i \times A_i \subseteq R$ .

Is this alright??

Thanks for the help!!

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  • $\begingroup$ Normally one asserts in addition that the union is disjoint. You might want to double check if you need to do this. (To be clear: it might be that you are not asked to do this, but it also might be you overlooked some dot over the union symbol or something like that.) $\endgroup$ – quid Aug 5 '14 at 14:54
  • $\begingroup$ finiteness is not necessary as long as the family you are given is of distinct equivalences classes the result holds true. $\endgroup$ – Anurag A Aug 5 '14 at 14:54
  • $\begingroup$ @AnuragA the question mentions finite $\endgroup$ – tattwamasi amrutam Aug 5 '14 at 14:55
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    $\begingroup$ @quid. There is no dot over the union symbol $\endgroup$ – tattwamasi amrutam Aug 5 '14 at 14:56
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    $\begingroup$ @tattwamasiamrutam yes I know but I just wanted to point out the statement is true in general. All you need to make sure is that the family $\{A_i\}$ has all possible equivalence classes. $\endgroup$ – Anurag A Aug 5 '14 at 14:56
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Since $a \in [a]$, we must have $A = \cup_{a \in A} [a]$. This is true whether or not the number of equivalence classes is finite or not.

If we let $A/{\sim} $ denote the equivalence classes, we have $A = \cup_{B \in A /{\sim}} B$.

If it happens that $A /{\sim} = \{ A_1,...,A_k \}$, you can write $A = \cup_{i=1}^k A_i$.

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  • $\begingroup$ take a look now $\endgroup$ – tattwamasi amrutam Aug 5 '14 at 15:53
  • $\begingroup$ The addition looks good. In general, you should add a new question rather than piggy-backing on an existing one. $\endgroup$ – copper.hat Aug 5 '14 at 15:56

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