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I have the sets $X = \{0,1,2,3,4,5,6,7,8,9\}$ and $Y = \{0,2,4,6,8,9\} $ .In $\mathcal P(X)$ I define the equivalence relation such that $A\mathrel RB$ iff $A - Y = B - Y$. How many equivalence classes are there?or, cardinal of the quotient set $ \mathcal{P} (X)/R$?

I know the cardinal value of the set $\mathcal{P} (X)/R$ is $16$ but I do not know how to get to this result.

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  • $\begingroup$ What does $A/Y$ mean? $\endgroup$ – Asaf Karagila Sep 13 '14 at 13:49
  • $\begingroup$ I read it as $A\setminus Y$ and $B\setminus Y$, where $\mathcal{P}(X)/R$ denotes the collection of equivalence classes. However, I'd expect there to be $32$ of those, currently. @Jose: are you sure $10\in X$? $\endgroup$ – HSN Sep 13 '14 at 13:51
  • $\begingroup$ @HSN: Yes, $\mathcal P(X)/R$ is the quotient set; I was asking specifically about the $A/Y$ part. As for the numerical answer, $16$ is correct (under the set difference interpretation). $\endgroup$ – Asaf Karagila Sep 13 '14 at 13:55
  • $\begingroup$ @AsafKaragila are you sure $16$ is correct? I would have agreed with HSN on $32$. $\endgroup$ – David Sep 13 '14 at 13:57
  • $\begingroup$ I agree with @David. It's actually $P(\{1,3,5,7,10\}) = 2^5 = 32$. $\endgroup$ – Aharon Sep 13 '14 at 13:58
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I am assuming that $A/Y$ means the set difference, $A$ with elements of $Y$ (if any) removed. Then $A\,R\,B$ means that $A$ and $B$ contain the same elements which are not in $Y$, that is the same out of the elements $\{1,3,5,7,10\}$. For example, $\{1,2,4\}$ is related to $\{1,4,6,8\}$. Can you finish the problem from here?

Comment: $A/Y$ is more usually written as $A-Y$ or $A\backslash Y$.

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Hint: There is a unique equivalence class for each subset of $\{1,3,5,7\}$ (this is the set of elements in $X$ but not in $Y$.

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