TITLE: What is $\mathbb R^{\mathbb R}$ as a vector space?
QUESTION [9 upvotes]: 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?

REPLY [6 votes]: 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 :)