This is a GRE math question:
My thoughts: I guess as for the cardinality, (A)=(B) and (D)=(E),but I couldn't prove whether it is true or not. Also, how to find the cardinality of (C), can someone tell me how to analyze this problem? I still have no idea how to find the largest one.
A) The cardinality of $\Bbb{R}$ is just $|\Bbb{R}|$.
B) It is really easy to prove the set is uncountable. Assume it's countable, and use a diagonal argument.
D) D is at least as big as $\Bbb{R}$ since it contains all the singleton sets of elements in $\Bbb{R}$. This question was already solved here: The cardinality of the set of all finite subsets of an infinite set
E) For any infinite set $S$ of some cardinality, the set $S[x]$ has the same cardinality: A set and polynomials with coefficients in that set
C) has a higher cardinality: See this paper for a solution of that fact
Regarding the cardinality of all functions $\mathbb{R} \rightarrow \{0;1\}$, one can prove that the corresponding set $C$'s cardinality is strictly greater than that of $\mathbb{R}$ by assuming there is an injective map $f: C \rightarrow \mathbb{R}$, and looking at the map $g: x \mapsto 1 - f^{-1}(x)(x)$, where $f^{-1} \circ f = Id_C$.
$g \in C$ so $g(f(g))$ makes sense, but $g(f(g)) = 1 - f^{-1}(f(g))(f(g)) = 1 - f \circ f^{-1}(g)(f(g)) = 1 - g(f(g)) \neq g(f(g))$, which is impossible. So there can't be an injective map from $C$ to $\mathbb{R}$.
This can be generalized to any set instead of $\mathbb{R}$, though you would have to treat the case of the empty set separately.
