Cardinality of a Quotient Set

Question

Let $X = \{1, 2, 3, 4, 5, 6, 7, 8, 9,10\}$ and $P=\{2,3,5,7\}$. In $P(X)$ define the equivalence $A\mathcal{R}B$ if $A\setminus P = B \setminus P$ . Then what is the cardinal of the quotient set? I noted that the equivalence class of $\{\}$ is $P(\{2,3,5,7\})$ - the sets containing no elements not in $P$ - and I conjecture that the cardinality of the quotient set is $2^6$, but am unsure of how to prove it.

Answer 1

You don't care anymore about 4 elements, so only 6 remaining. That's why the cardinality of your quotient should be $$2^6=64.$$

If you want to prove it formally show that $\overline{A}\mapsto A\setminus P$ is a bijection between the quotient $Q$ and $P(X\setminus P)$. Hence we have $$|Q|=|P(X\setminus P)|=2^{|X\setminus P|}=2^6=64.$$