Cantor-Bernstein-Schröder Theorem/Proof 4

Theorem
If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
 * $\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$

Proof
From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two injections:


 * $f \left({S}\right) = T_1 \subseteq T$
 * $g \left({T}\right) = S_1 \subseteq S$

Using $f$ and $g$, $S$ and $T$ are divided into disjoint subsets such that there exists a bijection between the subsets of $S$ and those of $T$, as follows.

Let $a \in S$.

Then we define the elements of $S$:
 * $\ldots, a_{-2n}, \ldots, a_{-2}, a_0, a_2, \ldots, a_{2n}, \ldots$

and the elements of $T$:
 * $\ldots, a_{-2n+1}, \ldots, a_{-1}, a_1, \ldots, a_{2n-1}, \ldots$

recursively as follows:

Let:

This construction is valid for all $n \ge 1$, but note that some of the $a_n$'s may coincide with others.

We set up a similar construction for negative integers:

As $f$ and $g$ are injections, it follows that if $f^{-1} \left({x}\right)$ and $g^{-1} \left({y}\right)$ exist for any $x \in T$, $y \in S$, then those elements are unique.

Let:
 * the elements of $S$ with an even index be denoted $\left[{a}\right]_S$

and
 * the elements of $T$ with an odd index be denoted $\left[{a}\right]_T$

that is:
 * $\left[{a}\right]_S = \left\{{\ldots, a_{-2n}, \ldots, a_{-2}, a_0, a_2, \ldots, a_{2n}, \ldots}\right\}$
 * $\left[{a}\right]_T = \left\{{\ldots, a_{-2n+1}, \ldots, a_{-1}, a_1, \ldots, a_{2n-1}, \ldots}\right\}$