What is $\mathbb R^{\mathbb R}$ as a vector space? In Sheldon Axler's Linear Algebra Done Right third edition the following is given as an example of a subspace:
The set of differentiable real-valued functions on $\mathbb R$ is a subspace of $\mathbb R^{\mathbb R}$
I'm looking for an intuitive explanation of the statement? Letting $S$ be the set of all differentiable real-valued functions, in order for the statement to be true, $S$ must be a subset of $\mathbb R^{\mathbb R}$(a subspace needs to be a subset).   
-How can $S \subset \mathbb R^{\mathbb R}$ when $S$ is a set containing functions and  $\mathbb R^{\mathbb R}$ is a set containing real numbers?
-What are the elements in $\mathbb R^{\mathbb R}$? How can we think of $  \mathbb R^{\mathbb R}$ as a tuple?
 A: $\Bbb R^2$ is a two dimensional vector space, a point in $\Bbb R^2$ is a pair $(x,y)$.  Now $\Bbb R^3$ is three dimensional, a point in $\Bbb R^3$ is a triple $(x,y,z)$.  Now  $\Bbb R^{\Bbb N}$ is countably many dimensions, indexed by the index set $\Bbb N$.  So a point in $\Bbb R^{\Bbb N}$ can be thought of as an infinite sequence $(x_1,x_2,x_3,\dots)$.  You can also think of that as a function from $\Bbb N$ to $\Bbb R$. Next $\Bbb R^{\Bbb R}$ is a vector space with uncountably many dimensions, indexed by $\Bbb R$.  So a point is an association of a value in $\Bbb R$ to every element of (the index set) $\Bbb R$.  You can think of a number of ways to denote it, but if you think about it that's exactly just the same thing as a function from $\Bbb R$ to $\Bbb R$.  That way an entire function from $\Bbb R$ to $\Bbb R$ is captured as a single point in $\Bbb R^{\Bbb R}$. Do you see how that works now?
A: If you go back to page 14 in chapter 1 of the text, at the bottom left corner of the page there is a footnote which mentions how $\mathbb{F}^{\infty}$ and $\mathbb{F}^n$ are special cases of $\mathbb{F}^S$, which Axler defines as the set of functions $g:S\to \mathbb{F}$.
Why is this so?
We can think of the $n$-tuples in $\mathbb{F}^n$ as being the assignment of elements of the set $\{1,2,3,....,n\}$ to elements of $\mathbb{F}$ by $g \in \mathbb{F}^{\{1, 2, 3,...,n\}}$ as $g(1)=x_1$, $g(2)=x_2$,...., $g(n)=x_n$ for the $n$-tuple $(x_1, x_2,..., x_n)$. We can also think of this as indexing a subset of $\mathbb{F}$ with $\{1,2,3,...,n\}$.
So similarly for $\mathbb{F}^{\infty}$, we have that the natural numbers are indexing our set (as Axler defined it), so $\mathbb{F}^{\infty}$ is really just $\mathbb{F}^{\mathbb{N}}$ i.e. functions $g: \mathbb{N} \to \mathbb{F}$ defined as $g(1)=x_1$, $g(2)=x_2$,...... which assign all of the natural numbers to elements of our field $\mathbb{F}$ in the form of countably infinite tuples $(x_1, x_2, ......)$. 
Now to your question. From the perspective I just explained, we can see that $\mathbb{R}^{\mathbb{R}}$ are just functions from $\mathbb{R} \to \mathbb{R}$, and that we are indexing elements $\mathbb{R}$ with elements of $\mathbb{R}$. Hence each real valued function is just a single ordering of a subset of $\mathbb{R}$ into an uncountably long tuple!(since $\mathbb{R}$ is uncountable) And so these real valued functions correspond to single points in $\mathbb{R}^{\mathbb{R}}$.  Amazing isn't it? 
From this perspective we can see how the set of differentiable real valued functions is a subset of $\mathbb{R}^{\mathbb{R}}$, since these functions are just tuples which order the elements of $\mathbb{R}$ in a way such that the graph of the function $f(i)=x_i$ for $i,x\in \mathbb{R}$ is differentiable. 
To start you off, note that our additive identity is just $g:\mathbb{R} \to \{0\}$ (since constant functions are differentiable). This is just an uncountably long tuple consisting only of zeroes, and this tuple is the origin of our $\mathbb{R}$-dimensional space! Since $0$ is the additive identity for $\mathbb{R}$, we can see how each component of any tuple we add with our uncountably long tuple of $0$'s will remain unchanged under addition of a $0$, hence the zero function is our additive identity.
Hope that helps :)
A: $\mathbb R^{\mathbb R}$ denotes the set of all maps from $\mathbb R$ to $\mathbb R$.
So $S \subset \mathbb R^{\mathbb R}$ is given.
To check if $S$ is a linear subspace, see if the zero element is in $S$, see if the sum of two members of $S$ is in $S$, and similarly for multiplication by a scalar.
A: Actually one usually uses the notation $A^B$ to mean the set of all functions from $B$ to $A$, so in your case we have
$$\Bbb{R}^{\Bbb{R}} = \{f\colon\Bbb{R}\to\Bbb{R}\ |\ f\text{ is a function}\}.$$
This is why it makes sense to consider $S$ as a subset of $\Bbb{R}^{\Bbb{R}}$.
