Is $\mathbb{R}$ a subspace of the Euclidean Plane $\mathbb{R}^2$ 
It said that $\mathbb{R}$ a subspace of the Euclidean Plane $\mathbb{R}^2$. But $\mathbb{R}$ is not even a subset of $\mathbb{R}^2$. How can it be possible?  Thank you for helping.
 A: Identify $R$ with $R \times  \{0\}$, the Cartesian product of $R$ and $\{0\}$ by the isomorphism $f(x)=(x,0)$.
That is what they mean when they talk about the real line being a subspace of the Cartesian plane.  
A: I take as my definition of the vector space $(\mathbb{R}^{2},+,\cdot)$ the set $\mathbb{R}^{2} = \{(x,y) \, \mid \, x, y \in \mathbb{R}\}$ with the operations $(x,y) + (z,w) = (x + z, y + w)$ and $\lambda \cdot (x,y) = (\lambda x, \lambda y)$ with the addition and multiplication of the coordinates inherited from the corresponding operations in the complete ordered field $\mathbb{R}$.  I will prove that $\mathbb{R}$ is isomorphic to a linear subspace of $\mathbb{R}^{2}$.  In particular, I will prove that $\mathbb{R} \simeq \{(x,0)  \mid  x \in \mathbb{R}\}$.
Define a linear map $\Phi : \mathbb{R} \to \mathbb{R}^{2}$ by $$\Phi(x) = (x,0).$$  This is linear since 
\begin{align*}
\Phi(x + y) &= (x + y, 0) = (x,0) + (y,0) = \Phi(x) + \Phi(y) \\
\Phi(\lambda x) &= (\lambda x, 0) = (\lambda x, \lambda \cdot 0) = \lambda (x,0) = \lambda \Phi(x).
\end{align*}
Moreover, $\Phi$ is injective.  Indeed, if $\Phi(x) = \Phi(y)$, then 
\begin{align*}
\Phi(x) = (x,0) = (y,0) = \Phi(y)
\end{align*}
which implies $x = y$ by the definition of (the set) $\mathbb{R}^{2}$.  Thus, $\Phi$ is injective.  Recall that the image of a linear map is always a linear subspace of the target.  In particular, $\Phi(\mathbb{R})$ is a linear subspace of $\mathbb{R}^{2}$.  Since $\Phi$ is injective, $\Phi : \mathbb{R} \to \Phi(\mathbb{R})$ is a bijective linear map.  It is immediate that $\Phi(\mathbb{R}) = \{(x,0) \, \mid \, x \in \mathbb{R}\}$.  This completes the proof that $\mathbb{R} \simeq \{(x,0) \, \mid \, x \in \mathbb{R}\}$.
If you've read this far, I hope you will try to "draw" the line $\{(x,0) \, \mid \, x \in \mathbb{R}\}$ in the plane $\mathbb{R}^{2}$ (e.g. using Cartesian coordinates).  A picture is worth a thousand words.
A: Technically you are right and as was stated in the comments, maybe authors should not do this without explanation.
Consider the $\mathbb R^2 = $ the $x$-$y$ plane $= $ "the $x$-axis crossed with the $y$-axis" = $\{(x,y)\mid x\in \mathbb R, y\in \mathbb R\}$ or any other way we will introduce this concept to students for the first time.
Now, consider immediately introducing $\mathbb R^3 = $ the $x-y-z$ space $= $ "the $x$-axis crossed with the $y$-axis crossed with the $z$-axis" = $\{(x,y,z)\mid x,y,z\in \mathbb R\}$ 
Now suppose some well-meaning instructor or some bright elementary student claimed that "the" $x-y$ plane and "the" $x$-axis and $y$-axis we refered to in defining the 2-D plane were the "same" $x$-axis and $y$-axis as in 3D space but with $z$ being restricted to $0$.
Would the instructor/student be right or wrong?  Or would it be a case of "It depends on how you look at it".
Basically the author is viewing $\mathbb R \subset \mathbb R^2 \subset \mathbb R^3 \subset \cdots$ via $\mathbb R^n = \mathbb R^n \times \{0\} \subset \mathbb R^n \times \mathbb R = \mathbb R^{n+1}$.  In other words, as thought there is one universal abstract set of axes, and $\mathbb R^n$ and $\mathbb R^{n+1}$ have the same axes but $\mathbb R^n$ is restricted to $x_{n+1} = 0$.
Is this legitimate?  Is this fair?
Well, sort of.  But.... not really.
It's "clear" that $\mathbb R^n \times \{0\} = \{(x_1, x_2, \ldots , x_n, 0)\in R^{n+1}\}$ is isomorphic and in every possible sense equivalent. And it "doesn't matter" that $\mathbb R^n \times \{0\}$ are not the same thing; in every important sense they might as well be.
But then on the other hand they are clear not the same thing at all.
Except on the third hand; Aren't they?  How was the concept of $A \times B = \{(a,b)\mid a \in A; b\in B\}$ every introduced in the first place?  Does "putting two elements from two sets next to each other" make sense?  Is it well defined?
If so what is $A\times \emptyset = \{(a,b)\mid a \in A; b\in \emptyset\}$?  Wouldn't it mean $A\times \emptyset = \{(a,b)\mid a \in A; b\in \emptyset\} = \{(a,*) \mid a \in A\} = \{(a)\mid a\in A\} = \{a\mid a\in A\} = A$?  If not, why not?  Admittedly the "$1$-tuple" $(a)$ doesn't look like the element $a$ but if they are different what exactly is the difference?  Can we or can we not define $(a,b)$ where $b$ simply does not exist at all? i.e. if $b \in \emptyset$?
By all logic, $A \times \emptyset = A$ and $\emptyset \subset B$, so ..... logically $A = A\times \emptyset \subset A\times B$.  Thats logical isn't it?
Okay, It feels as though someone is pulling a fast one on us but...  are they?
I'll let you think this out.
