Bijective map $[0,1]$ to $\mathbb{R}$

Question

If I wanted a bijective map $[0,1]$ to $\mathbb{R}$ and I would define $f$ as: \begin{equation} f(x)=\begin{cases} 0&x=0\\{}\\ 1&x=1\;\\{}\\ -cot\pi x,&\text{otherwise}\\ \end{cases} \end{equation} Would it be a legitimate bijection?

Answer 1

First, let me clear up a misconception about the values of $\cot(={\sin\over \cos}$): $\cot(x)$ is $0$ if $\cos(x)=0$ (so e.g. $x={\pi\over 2}$), and $\cot(x)$ is $1$ if $\cos(x)=\sin(x)$ (so e.g. if $x={\pi\over 4}$). (I think you're mixing up domain and codomain here?)

Meanwhile, your construction fails for a more basic reason: the map $x\mapsto -\cot(\pi x)$ is already a bijection from $(0, 1)$ to $\mathbb{R}$, so you can't extend it to a bijection from $[0, 1]$ to $\mathbb{R}$. There's nowhere to send $0$ or $1$ to that isn't already covered.

In order to build a bijection $b: [0, 1]\rightarrow \mathbb{R}$ from a bijection $a: (0, 1)\rightarrow\mathbb{R}$ as you are trying to do, you need to "free up" some values from $a$: e.g. restricting $a$ to the map $a'$ with domain $(0, 1)\setminus\{{\pi\over 2}, {\pi\over 4}\}$ leaves the codomain values $0$ and $1$ free. So the map $$b': [0, 1]\setminus \{{\pi\over 2}, {\pi\over 4}\}\rightarrow\mathbb{R}$$ gotten by extending $a'$ to send $0$ to $0$ and $1$ to $1$ is a bijection; but now there's nowhere to send ${\pi\over 2}$ and ${\pi\over 4}$!

You need to free up infinitely many values, doing a trick like in Hilbert's hotel. For example, let $$r_1={1\over 2}, r_2={1\over 4}, ..., r_i={1\over 2^i}, . . .$$ and let $D=(0, 1)\setminus\{r_i: i\in\mathbb{N}\}$. Then define $b$ as follows:

• For $s\in D$, $b(s)=\cot(\pi s)$.

• For $s=0$, $b(s)=r_1$.

• For $s=1$, $b(s)=r_2$.

• For $s=r_i$, $b(s)=r_{i+2}$.

Do you understand why this is a bijection?

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Here's another neat bijection, based on the following observation: we can chop $[0, 1]$ and $(0, 1)$ into a bunch of half-open intervals, with one point left over.

For $[0, 1]$ this is easy: just let $$I_1=[0, {1\over 2}),\quad I_2=[{1\over 2}, {3\over 4}),\quad ... , \quad I_n=[{n-1\over n}, {n\over n+1}), \quad. . .$$ Clearly the $I_n$s are disjoint, and the point left over is $1$.

For $(0, 1)$ it's a bit trickier: we could try a bunch of $(\cdot, \cdot]$ intervals starting from the left ($0$) and moving right, but then we wouldn't have a point left over! Instead, we'll start from both ends and move towards the middle, so that ${1\over 2}$ is the point left over (I'll leave it as an exercise to define the intervals $J_n$ correctly).

Then pick bijections $f_i: I_n\cong J_n$, and let $f$ be the union of the $f_i$s together with $1\mapsto {1\over 2}$. Then $f$ is a bijection from $[0, 1]$ to $(0, 1)$. From here we just apply our favorite bijection $(0, 1)\cong \mathbb{R}$ - say, $\cot(\pi x)$.