There are more numbers in $\mathbb{R}$ than in $\mathbb{N}$. There are as many vectors in $\mathbb{R}^n, n \in \mathbb{N}$ as numbers in $\mathbb{R}$.
How many real functions are there? If I denote $\{f,f: \mathbb{R} \rightarrow \mathbb{R}\}=\mathbb{R}^\mathbb{R}$ I get another quite.. very big, infinite, uncountable set, right?
Is $|\mathbb{R}^\mathbb{R}| > |\mathbb{R}|$?.. another uncountable infinite beyond continuity?
What about $\left|(\mathbb{R}^\mathbb{R})^{(\mathbb{R}^\mathbb{R})}\right|$ then? How far can we build huge sets in this direction?
Is the number of various infinites itself countable? $\left\{|\mathbb{N}|,|\mathbb{R}|,|\mathbb{R}^\mathbb{R}|,\left|(\mathbb{R}^\mathbb{R})^{(\mathbb{R}^\mathbb{R})}\right|,\dots\right\}$ would be, right?
Or maybe $|\mathbb{R}^\mathbb{R}| = |\mathbb{R}|$ and we fall back on our feet?
The set $S$ of functions $\mathbb{R}\to\mathbb{R}$ has the same cardinality as the power set of $\mathbb{R}$.
Step 1: Associate to each function in $S$ its graph in $\mathbb{R}^2$. This graph can be considered as a subset of $\mathbb{R}^2$. And no two functions in $S$ have the same graph. So the cardinality of $S$ is not greater than the cardinality of the power set of $\mathbb{R}^2$ which has the same cardinality as the power set of $\mathbb{R}$. So the cardinality of $S$ is not greater than the cardinality of the power set of $\mathbb{R}$.
Step 2: Let $A$ be the set of subsets of $\mathbb{R}$ that include the element $0$. So $A$ has the same cardinality as the power set of $\mathbb{R}$. For each element $K$ of $A$, we can find a function in $S$ whose range is $K$ (namely take the function from $\mathbb{R}\to\mathbb{R}$ that sends each element of $K$ to itself and each nonelement of $K$ to $0$). This gives a mapping from $S$ to $A$ that is onto. So the cardinality of $S$ is at least as great as the cardinality of $A$ which is equal to the cardinality of the power set of $\mathbb{R}$.