Bijection from [-1,1) to the Reals $Proposition. [-1,1)\approx\mathbb{R}.$
I know for this problem I need to find a bijection from $[-1,1)\rightarrow\mathbb{R}$. However, I am having trouble establishing a function that fits the criteria.
 A: How about
$$f(x)=\begin{cases}\tfrac1x-1&\text{if }x>0,\\
0&\text{if }x=0,\\
\tfrac1x&\text{if }x=-\tfrac1n\text{ for some }n\in\mathbb N,\\
\tfrac1x+1&\text{otherwise?}\end{cases} $$
A: Extend reals with a singleton $\{\psi\}$ (of course $\psi\notin\Bbb R$). Then define $f_1:[-1,1) \to \Bbb R\cup\{\psi\}$ as
$$
f_1:x\mapsto \begin{cases}
  \psi & \text{if } x=-1 \\
  \tan\frac{\pi x}2 & \text{otherwise}
\end{cases}
$$
Next eat the superfluous item with $f_2: \Bbb R\cup\{\psi\} \to \Bbb R$:
$$
f_2:x\mapsto \begin{cases}
   x+1 & \text{if } x \in \Bbb N\cup\{0\} \\
   0   & \text{if } x=\psi \\
   x   & \text{otherwise}
\end{cases}
$$
Then the composition $f_2\circ f_1$ is a desired bijection from $[-1,1)$ to $\Bbb R$.
A: $x\mapsto\tan\left(\frac{\pi}{2}x\right)$ gives you a bijection $(-1,1)\to\mathbb{R}$ so $(-1,1)\approx\mathbb{R}$. If you don't like to use a complicated fonction like $\tan$, $x\mapsto\frac{1}{1-x}+\frac{1}{1+x}$ works to. Then just adding one point to $(-1,1)$ does not change the result because we are dealing with infinite sets.
