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Let $H=\{1,2,3,4,5,6,7\}$ and $K=\{1,2,3,4\}$ and let $P$ be the power set of $H$ and let $R$ be the relation on $P$ defined by: For all $X,Y$ in $P$ : $(X,Y)$ are in $R$ if and only if $|X \cap K| = |Y \cap K|$.

how many elements does the equivalence class of $\{1,2\}$ has? I don't know how to solve this one. Can you please help me?

Thanks

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For #2, there are $2^3$ total sets (power set of $\{5,6,7\}$) that could work, and you have correctly listed 4 of them.

For #3, a subset in that equivalence class must consist of some subset of $\{1,2,3,4\}$ of size two, combined with some elements outside of $K$. (Effectively, add each subset found in #2 to any subset of $K$ of size $2$.) There are $\binom{4}{2}$ subsets of $K$ of size $2$, and $2^3$ subsets from #2, so in total there are $\binom{4}{2} 2^3= 48$ subsets as you have correctly computed.

Apologies for my earlier mistake.

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  • $\begingroup$ 2) thank you! I only wrote those because it asked for 4 elements only. $\endgroup$
    – Just a girl
    Apr 4, 2017 at 3:56
  • $\begingroup$ 3) I still don't understand 3 though $\endgroup$
    – Just a girl
    Apr 4, 2017 at 3:56
  • $\begingroup$ 3 is asking how many sets have an intersection with $K$ that's as big as $\{1,2\}$'s intersection with $K$; which in this case is the same as asking how many sets in $P$ contain two elements of $K$. It's not asking you to know anything especially different than you know to answer 2, it's just asking for a count instead of for specific elements. $\endgroup$
    – Malice Vidrine
    Apr 4, 2017 at 5:46
  • $\begingroup$ @Justagirl Sorry for my mistakes. $\endgroup$
    – angryavian
    Apr 4, 2017 at 6:03
  • $\begingroup$ No worries, Thank you! I get it now! $\endgroup$
    – Just a girl
    Apr 4, 2017 at 6:05

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