proof that sets [4, 7) and (2,12) are equivalent

Question

prove that [4, 7) and (2,12) are equivalent:

1. Cantor-Bernstein theorem
2. constructing bijection between these sets.

1. A=[4, 7) and B=(2,12) injection from A to B, $f:A \to B$ y=x, $x \in [4,7). $ $f([4,7))=(4,7) \subset (2,12)$

injection from B to A, $g:B \to A$

$y=4+ \frac{1}{4}x$

$g((2,12))=(4.5,7) \subset [4,7)$

2. but what about the second proof?

Answer 1

We seek to construct a bijection between the sets $[4,7)$ and $(2,12)$. We will do so by first creating a bijection $h:(2,12) \to (4,7)$, and then creating a bijection $g:(4,7) \to [4,7)$. Since the composition of bijections is a bijection, the bijection $g \circ h:(2,12) \to [4,7)$ will be a desired one.

For the bijection $h:(2,12) \to (4,7)$, define $h$ as the function of the line passing through $(2,4)$ and $(12,7)$. (This trick works to define a bijection between any two open intervals.) This line is

$$ h(x)= \frac{3}{10}(x-2)+4$$

Where $2<x<12$. You may graph $h(x)$ to convince yourself that this is a correct bijection.

For the bijection $g:(4,7) \to [4,7)$, I will use an approach very similar to Ben's answer to the post linked by Eric Towers in the comments.

Let $A_1= \{7-\frac{3}{2},7-\frac{3}{3},7-\frac{3}{4},\ldots\}$. Note that $A_1\subseteq (4,7)$.

Let $A_2= \{4\}\cup A_1 = \{7-\frac{3}{1}, 7-\frac{3}{2},7-\frac{3}{3},\ldots\}$.

Define the function $f: A_1 \to A_2$ by $f(7-\frac{3}{n})=7-\frac{3}{n-1}$. $f$ is a bijection.

Now define $g:(4,7) \to [4,7)$ as $g(y)=y$ if $y\notin A_1$, otherwise $g(y)=f(y)$. $g$ is a bijection.

We are now ready to create a composition $g \circ h:(2,12) \to [4,7)$.

$(g \circ h)(x) = g(h(x))=h(x)$ if $h(x)\notin A_1$, otherwise $g(h(x))=f(h(x))$.

$g \circ h$ is the desired bijection. My apologies for inaccuracies and/or incorrect etiquette; this is one of my first answers on this site. I would appreciate any feedback. I hope this is helpful!