Does there exist an injective function $f: \mathbb{R}^2 \mapsto I$

Question

Here, $I$ is an interval of real-numbers, e.g. $[1,5]$. Any citations would be greatly appreciated so I can look further into this.

If one such function does exist, can you please provide it?

Answer 1

I can break it up for you into steps. The map you want is bound to be very unwieldy, but it certainly exists, and, even stronger, it can be made a bijection.

Sequences of zeros and ones

The crux of the proof is that, if $2^\mathbb N$ is the set of all sequences $(a_n)_{n\in\mathbb N}$ indexed by natural numbers $\mathbb N$, where $a_n\in\{0,1\}$, then there is a bijection between $2^\mathbb N\times2^\mathbb N$ and $2^\mathbb N$ alone.

It is easy to prove this bijection. Namely, a pair of sequences $((a_0, a_1, a_2,\ldots), (b_0, b_1, b_2, \ldots))$ would map into a sequence $(a_0, b_0, a_1, b_1, a_2, b_2,\ldots)$.

Real numbers

Now, the link to the real numbers. You know about the decimal number system, but you may have heard of the binary number system (using only the digits $0$ and $1$), ternary number system (using the digits $0$, $1$ and $2$) etc.

So - you can imagine that every real number in $[0,1)$ can be represented as a binary number $0.a_0a_1a_2\ldots$ (with digits $a_n\in\{0, 1\}$), or as a ternary number $0.a_0a_1a_2\ldots$ (with digits $a_n\in\{0, 1, 2\}$) etc. Now this almost gives us a bijection between $[0,1)$ and the sequences of digits $0, 1$ of the kind we saw above. Sadly, this correspondence is not a bijection. First of all, the sequence $0.1111\ldots$ actually has a value $1$ (so the map would be to $[0,1]$, rather than $[0,1)$), but, worse, other sequences ending with "only ones" will map into the same number where the "tail of ones" is replaced with the "tail of zeros" and the last digit before the tail is incremented. For example, $0.1011111\ldots=0.1100000\ldots$, as binary decimal expansions.

The fix, and Cantor-Schroeder-Bernstein's theorem

How to "fix" that problem? One possible way is this:

• There is an injection from the set $2^\mathbb N$ to $[0,1)$, which you make by looking not at binary but at ternary expansion. In that case, the ambiguous representations would have the "tail of twos" (will end with $22222\ldots$), but we are not looking at those.
• There is an injection from $[0,1)$ to $2^\mathbb N$, which we get by taking the binary representation, and in case of ambiguity we refuse to take the one with the "tail of ones".

So, we have an injection of $2^\mathbb N$ to $[0,1)$ and an injection from $[0,1)$ to $2^\mathbb N$. Does that now guarantee that there is a bijection between the two sets? The answer is yes, but it is slightly non-trivial, and is contained in the famous Cantor-Schroeder-Bernstein theorem. The proof is nice and constructive (see the Wikipedia article) but it will end up with a fairly unwieldy map.

Open, semi-open and closed intervals

Okay, enough about semi-open intervals. Can we now construct a bijection between $[0,1)$ and one of $[0,1]$ and $(0,1)$? The answer is again "yes": there are obvious injections $(0,1)\to[0,1)\to[0,1]$, and you can explicitly construct a bijection $[0,1]\to(0,1)$ by mapping like the following:

• Pick a sequence from $(0,1)$, e.g. $a_n=\frac{1}{n+2}$, which is the sequence $(1/2, 1/3, 1/4, 1/5,\ldots)$.
• Map $0$ into the first element $1/2$
• Map $1$ into the second element $1/3$
• Map any element of the sequence into one that is "two places away", i.e. $a_n\mapsto a_{n+2}$
• All the other elements from $(0,1)$ map to themselves.

Excellent. Now to prove there are bijections between $[0,1)$ and any of $[0,1]$ and $(0,1)$, you can either invoke a very similar construction as above - or use Cantor-Schroeder-Bernstein theorem again. (The injection $[0,1)\to[0,1]$ composed by the bijection $[0,1]\to(0,1)$ gives you an injection $[0,1)\to(0,1)$.)

Closing in

Next step: there is a bijection between $(0,1)$ and $\mathbb R$: $f(x)=\tan\left(\pi\left(x-\frac{1}{2}\right)\right)$, for example. There is also a bijection between $[0,1]$ and $[1,5]$: $f(x)=4x+1$, for example.

The completion of the argument

First, notice we've made a bijection from $\mathbb R$ to $(0,1)$ to $[0,1)$ to $2^\mathbb N$, which gives us a bijection of $\mathbb R^2$ to $2^\mathbb N\times 2^\mathbb N$.

Then, notice that the first thing we proved above was that there was a bijection between $2^\mathbb N\times 2^\mathbb N$ and $2^\mathbb N$.

Finally, notice we have a bijection from $2^\mathbb N$ to $[0,1)$ to $[0,1]$ to $[1,5]$.

This means we have a bijection from $\mathbb R^2$ to $2^\mathbb N\times 2^\mathbb N$ to $2^\mathbb N$ to $[1,5]$, which is what you wanted all along.

Answer 2

The map $$\begin{align}\alpha\colon \Bbb R&\to (0,\infty)\\x&\mapsto e^x\end{align}$$ is injective. The map $$ \begin{align}\beta\colon (0,\infty)&\to (0,1)\\x&\mapsto \frac1{1+x}\end{align}$$ is injective. The map $$ \begin{align}\gamma\colon (0,1)&\to \mathcal P(\Bbb N)\\x&\mapsto \{\,n\in\Bbb N\mid \lfloor 2^nx\rfloor\text{ is odd}\,\}\end{align}$$ is injective(!). The map $$ \begin{align}\delta\colon \mathcal P(\Bbb N)\times \mathcal P(\Bbb N)&\to \mathcal P(\Bbb N)\\(X,Y)&\mapsto \{\,2n\mid n\in X\,\}\cup \{\,2n+1\mid n\in Y\,\}\end{align}$$ is injective, The map $$ \begin{align}\epsilon\colon \mathcal P(\Bbb N)&\to [0,1]\\ X&\mapsto \sum_{n\in X}3^{-n}\end{align}$$ is injective(!), and finally, for any closed intgerval $[a,b]$, the map $$ \begin{align}\zeta\colon [0,1]&\to [a,b]\\ X&\mapsto a+(b-a)x\end{align}$$ is injective.

All in all, we obtain an injective map $$ \begin{align} \Bbb R^2&\to [a,b]\\ (x,y)&\mapsto \zeta(\epsilon(\delta(\gamma(\beta(\alpha(x))),\gamma(\beta(\alpha(y))))))\end{align}$$

Answer 3

Here's an outline for how a function could be explicitly constructed.

The injective function $f_1 : \mathbb{R} \to (0,1)$ is given by

$$f_1(x) = \frac{1}{2} + \frac{1}{2 \pi} \tan^{-1}(x)$$

The injective function $f_2 : [0,1) \to \{0,1\}^\mathbb{N}$ maps a real number to the sequence of digits after the radix point in its base-$2$ representation. That is, $f_2(x)$ is the sequence $\{a_n\}_{n \geq 1}$ with values $0$ or $1$ given by

$$ f_2(x) = \{a_n\}_{n \geq 1} = \left\{ \lfloor 2^n x \rfloor - 2 \lfloor 2^{n-1} x \rfloor \right\}_{n \geq 1} $$

The inverse of $f_2$ is this infinite series, which always converges:

$$ f_2^{-1}\left(\{a_n\}_{n \geq 1}\right) = \sum_{n=1}^\infty a_n 2^{-n} $$

This formula for $f_2^{-1}$ is not injective, since sequences $\{a_1, \ldots, a_N, 1, 0, 0, \ldots\}$ and $\{a_1, \ldots, a_N, 0, 1, 1, \ldots \}$ give the same sum. (For example $1/4 = 0.01\bar{0}_2 = 0.00\bar{1}_2)$ But a sequence with an infinite run of $1$ digits is not possible in the definition of $f_2$, so those sequences are not in the image of $f_2$, and $f_2^{-1}$ is in fact an inverse of $f_2$.

The injective function $f_3 : \{0,1\}^\mathbb{N} \times \{0,1\}^\mathbb{N} \to \{0,1\}^\mathbb{N}$ maps two sequences of binary digits into a single such sequence, by alternating their digits:

$$ \begin{align*} f_3(\{a_n\}, \{b_n\}) &= \{c_n\} \mathrm{,~where} \\ c_{2n-1} &= a_n \\ c_{2n} &= b_n \end{align*} $$

Note that if sequences $\{a_n\}$ and $\{b_n\}$ do not contain an infinite run of $1$ digits, then neither does $\{c_n\}$.

To get from the "standard" interval $[0,1)$ to a general $[a,b]$, we can just use a linear transformation. $f_4 : [0,1) \to [a,b]$ is given by

$$ f_4(x) = a + (b-a) x $$

So finally, the desired function $f : \mathbb{R}^2 \to [a,b]$ can be pieced together:

$$ f(x,y) = f_4 \bigg(f_2^{-1} \Big( f_3 \big( f_2(f_1(x)), f_2(f_1(y)) \big) \Big) \bigg) $$

Since the $f_2$ values do not have infinite runs of $1$ digits, neither does the $f_3$ value, and so the $f_2^{-1}$ value will not repeat for different $(x,y)$ points. So the total function $f$ is injective.