Is $\mathbb{R}$ a subspace of the Euclidean Plane $\mathbb{R}^2$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.