I need to prove that $[-4,0)\sim(0,12)$ using the Cantor-Bernstein theorem and also by constructing a bijection.
From what I understood, in order to prove it with the Cantor theorem, I need to construct an injection between the two. From looking at one example I found, I saw that they did it by using a function.
Example was, from $(0,1)$ to $(0,1]$, $f(x) = x$, $x \in (0,1)$, because $(0,1) \subset (0,1]$
I tried to apply the same logic to my question, but got stuck..
I'm very new to this topic and I don't have a clue about where to start.
Let's firstly consider $y=f(x)=3x+12$ which maps $(-4,0)$ to $(0,12)$. We can take some countable set from $(-4,0)$, for example $\left\{-\frac{1}{n}\right\}=\left\{x_n\right\}$ for $n\in \mathbb{N}$. By function $f$ these set goes to set $\left\{-\frac{3}{n}+12\right\}=\left\{y_n\right\}$ as $x_n \to y_n$. If now we consider set $\left\{4, x_n\right\}$, then is possible to define bijection $4 \to y_1, x_n \to y_{n+1}$.