Cardinality of sets regarding Consider the following sets of functions on $\mathbb{R}$.
$W=$The set of all constant functions on $\mathbb{R}$
$X=$The set of polynomial functions on $\mathbb{R}$
$Y=$ The set of continuous functions on $\mathbb{R}$
$Z=$ The set of all functions on $\mathbb{R}$
Which of the sets has the same cardinality as that of $ \mathbb{R} $?


*

*Only $ W $ 

*Only $ W $ and $X$ 

*Only $ W, X$ and $Y$ 

*All of $ W,X,Y $ and $Z$


I suppose $W$ is the correct answer as $X$ is equivalent to a subset of $\mathbb R\times \mathbb R$ and $X\subset  Y\subset Z$. Am I correct?
 A: Assuming that all functions are mapping to $\mathbb{R}$, I believe the correct answer is $W, X$, and $Y$ (as being the sets with the same cardinality as $\mathbb{R}$.)
To see that $Z$ has cardinality at least as great as the power set of $\mathbb{R}$ (and hence cardinality greater than $\mathbb{R}$), consider for each subset $S\subseteq\mathbb{R}$ the function $f_S(x)=\left\{ \begin{array}{cc} 1&\text{if } x\in S\\ 0&\text{otherwise}\end{array} \right.$
This gives a one-to-one mapping of subsets of $\mathbb{R}$ to functions $\mathbb{R}\to\mathbb{R}$.
To see that $Y$ has cardinality same as $\mathbb{R}$, consider any continuous function $\mathbb{R}\to\mathbb{R}$.  The graph of this function is a closed set in the plane.  Thus there is a one-to-one mapping from the set of continuous functions to the set of closed sets in $\mathbb{R}^2$. However, the set of closed sets in the plane has the same cardinality as $\mathbb{R}$. (This last statement follows because $\mathbb{R}^2$ has a countable basis, and so at most $2^{\aleph_0}$ open sets; and the same number of closed sets.)
