Define exotic operations so that $\mathbb R^2$ has a $1$ dimensional basis I am working on an exercise from Linear Algebra with Applications 5ed by Otto Bretscher. The problem is: Can we define "exotic" operations on $\mathbb R^2$ so that its dimension is $1$. This seems impossible but the solution manual says it it possible. They say to let $T$ be an invertible function from $\mathbb R^2$ to $\mathbb R$. Then they use the existence of such a transformation to define exotic operations based off of that function and its inverse. Before I even try to understand the second part of their solution, I do not understand why we can just say that such an invertible transformation $T$ exists between $\mathbb R^2$ and $\mathbb R$. I can't even think of an example of such a transformation.
 A: Bijections $T : \mathbb{R} \to \mathbb{R}^2$ exist but are annoying to write down and poorly behaved. The argument that they exist is usually done more indirectly using the Cantor-Schroeder-Bernstein theorem. The CBS theorem says that if $A, B$ are two sets and you can find an injection $f : A \hookrightarrow B$ and an injection $g : B \hookrightarrow A$, then you can find a bijection $h : A \cong B$, so $A$ and $B$ have the same cardinality; this is done by weaving together $f$ and $g$ in a specific way.
The CBS can be used to prove that $\mathbb{R}$ has the same cardinality as the set $2^{\mathbb{N}}$ of infinite binary strings. This is slightly annoying but can be done as follows: given an infinite binary string $a_i \in \{ 0, 1 \}, i \in \mathbb{N}$ we can write down a real number with binary expansion $0.a_1 0 a_2 0 a_3 0 \dots $ (the $0$s here are to avoid an annoying issue with identities like $0.0111 \dots = 0.1$). This gives an injection $2^{\mathbb{N}} \hookrightarrow \mathbb{R}$. In the other direction, given $x \in \mathbb{R}$ we can first consider $\arctan x$, which lives in the interval $\left( - \frac{\pi}{2}, \frac{\pi}{2} \right)$ (this is a bijection and a very well-behaved one), then shift it around a bit to live in the open interval $(0, 1)$, for example via
$$\mathbb{R} \ni x \mapsto \frac{\arctan x}{\pi} + \frac{1}{2} \in (0, 1).$$
Next we can consider the binary expansion $0.d_1 d_2 d_3 \dots $  of this number, which produces an infinite binary string. This is an injection $\mathbb{R} \hookrightarrow 2^{\mathbb{N}}$ (almost but not quite a bijection, again because of identities like $0.0111 \dots = 0.1$). So by the CBS theorem we have a bijection $h : \mathbb{R} \cong 2^{\mathbb{N}}$.
(This is a very annoying argument as you can see but you only need to do it once and then you know forever that $\mathbb{R}$ and $2^{\mathbb{N}}$ have the same cardinality. A mathematician will immediately substitute one for the other in a cardinality argument without blinking.)
Now instead of finding a bijection $\mathbb{R} \to \mathbb{R}^2$ we'll find a bijection $2^{\mathbb{N}} \to (2^{\mathbb{N}})^2$. This is now easy: $(2^{\mathbb{N}})^2 \cong 2^{\mathbb{N} \sqcup \mathbb{N}}$ is the set of pairs of infinite binary strings, and so a bijection is given by interweaving:
$$2^{\mathbb{N}} \ni (a_1, a_2, a_3, \dots) \mapsto (a_1, a_3, a_5, \dots), (a_2, a_4, a_6, \dots) \in 2^{\mathbb{N}} \times 2^{\mathbb{N}}.$$
This gives us a bijection $\mathbb{R} \cong \mathbb{R}^2$ by first applying our CBS bijection $h : \mathbb{R} \cong 2^{\mathbb{N}}$, then interweaving, then applying two copies of the inverse bijection $h^{-1} : 2^{\mathbb{N}} \cong \mathbb{R}$. Like I said: annoying to write down! It's much faster to not even try to write down a bijection and just argue about the equivalence relation "has the same cardinality." Once you have some sense of how this works the argument can be blazed through like this:
$$\mathbb{R} \cong 2^{\mathbb{N}} \cong (2^{\mathbb{N} \sqcup \mathbb{N}}) \cong (2^{\mathbb{N}})^2 \cong \mathbb{R}^2.$$
