What is this $\mathbb{E}^{3}$ that Loomis and Sternberg talk about in their book "Advanced Calculus"? The authors introduce the "three dimensional Euclidean space", $\mathbb{E}^{3}$, without defining it, but explains that there is a correspondence between $\mathbb{E}^{3}$ and $\mathbf{R}^{3}$ by picking an arbitrary point $O$ in $\mathbb{E}^{3}$ as the origin, and three unit points $Q_{1}, Q_{2}, Q_{3}$ such that not all four points lie in a plane. Every point $\mathbf{X}$ in $\mathbb{E}^{3}$ thus determines a triple $\mathbf{x} = (x_{1},x_{2},x_{3})$ living in $\mathbf{R}^{3}$.
The correspondence isn't explained in detail, but they list four "theorems" about the correspondence which they claim to be quite tricky to prove, and assume them instead. They are:


*

*$T: \mathbb{E}^{3} \rightarrow \mathbf{R}^{3}$ as explained above, is a bijection.

*Two line segments $AB$ and $XY$ in $\mathbb{E}^{3}$ are equivalent (same length, parallel, equally directed) if and only if the vectors  $\mathbf{b} - \mathbf{a}$ and  $\mathbf{y} - \mathbf{x}$ in $\mathbf{R}^{3}$ are equal.

*If $X \neq 0$, then $Y$ is on the line through $O$ and $X$ if and only if $\mathbf{y} = t \mathbf{x}$ for some $t \in \mathbf{R}$.

*If the axes in $\mathbb{E}^{3}$ are orthonormal, then then the length $|OX|$ equals $(\sum_{i=1}^{3} x_{i})^{2}$.
How in the world would someone go about proving these theorems without knowing what $\mathbb{E}^{3}$ is? Intuitively, the authors seem to imply that it's the familiar $3$-dimensional space that contains points, lines, and planes that we're all familiar with. But it's hard to say without a definition of Euclidean space. Furthermore, is $\mathbb{E}^{3}$ a vector space? If it is, then I'm assuming that its base field is $\mathbf{R}$, but if that's the case, why do we even bother with $\mathbb{E}^{3}$ in the first place, when we can work with the vector space $\mathbf{R}^{3}$?
 A: $\mathbb{R}^n$ is a vector space but due to the fact that $\mathbb{R}^n = \mathbb{R} \times \cdots \times \mathbb{R}$ and $\textbf{e}^j = (0,....,1=x_j,0,...,0)$ forms a basis for $j = 1,...,n$ then $\textbf{v} \in \mathbb{R}^n \Rightarrow \textbf{v} = \sum_j v_j \textbf{e}^J$ i.e the coordinate representation of $\textbf{v}$ is the $n$-tuple $(v_1,...,v_n)$. We define $\mathbb{E}^n = (\mathbb{R}^n, \| \cdot \|)$ i.e $\mathbb{R}^n$ equipped with the standard euclidean topology induced by the distance metric (hence the $\mathbb{E}$). 
These two spaces are usually identified since $\mathbb{R}^n$ with the coordinate chart $\varphi = (x^1,...,x^n)$ where $x^j(\textbf{v}) := v_j$ turns $\mathbb{R}^n$ into a manifold trivially, how? Let $\mathcal{B} = \{\textbf{e}^j: j = 1,...,n\}$ then we see that $\Phi: \mathbb{R}^n \to \mathbb{R}^n_{\mathcal{B}}$ defined by $\textbf{v} \mapsto \textbf{v}_{\mathcal{B}}$ is a linear isomorphism i.e we identify $\mathbb{R}^n$ with $\mathbb{R}^n_{\mathcal{B}}$ and with this $\varphi: \mathbb{R}^n_{\mathcal{B}} \to \mathbb{E}^n$ becomes a global chart which is just the identity i.e a smooth diffeomorphism. 
Really we should define $\psi =  \varphi \circ \Phi: \mathbb{R}^n \to \mathbb{E}^n$ but since we know that we can transfer the necessary information about $\mathbb{R}^n_{\mathcal{B}}$ by a linear isomorphism (i.e a structure preserving map), we can suppress this map for now and reference it when need be. The same identification is done is calculus when you learn about arc-length parameterizations and defining the Frenet-Frame for an arbitrary curve with constant speed i.e $\textbf{T}(s):= \textbf{T}(t)$. But really $s,t$ are related by a diffeomorphism i.e a structure preserving map (in this case, manifold structure) i.e $\textbf{T}(s) = \textbf{T}(\beta(t))$.  In sum, $\mathbb{E}^n$ is the model space of $\mathbb{R}^n$. I hope this helps. 
