Cantor-Bernstein-Schröder Theorem/Proof 2

Theorem
Let $S$ and $T$ be sets, such that:
 * $S \preccurlyeq T$, that is: $T$ dominates $S$
 * $T \preccurlyeq S$, that is: $S$ dominates $T$.

Then:
 * $S \sim T$

that is, $S$ is equivalent to $T$.

Proof
Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$.

By definition, we have that there exist injections $f: S \to T$ and $g: T \to S$.

We are going to try to build a sequence $t_1, s_1, t_2, s_2, t_3 \ldots$ as follows.

Consider any $t_1 \in T$.

By Law of Excluded Middle there are two choices:

Suppose $\exists s_1 \in S: f \left({s_1}\right) = t_1$.

Because $f$ is injective, such an $s_1$ is unique.

So we can choose $s_1 = f^{-1} \left({t_1}\right)$.

Again, by Law of Excluded Middle there are two further choices:

Suppose $\exists t_2 \in T: g \left({t_2}\right) = s_1$.

Because $f$ is injective, such an $t_2$ is unique.

Similarly, we choose $s_2 = f^{-1} \left({t_2}\right)$, if it exists.

This process goes on until:


 * We reach some $s_n \in S$ such that $\neg \exists t \in T: g \left({t}\right) = s_n$. This may be possible because $g$ may not be a surjection.


 * We reach some $t_n \in T$ such that $\neg \exists s \in S: f \left({s}\right) = t_n$.This may be possible because $f$ may not be a surjection.


 * The process goes on for ever.

For each $t \in T$, then, there is a well-defined process which turns out in one of the above three ways.

We partition $T$ up into three subsets that are mutually disjoint:


 * Let $T_A = \{$ all $t \in T$ such that the process ends with some $s_n\}$
 * Let $T_B = \{$ all $t \in T$ such that the process ends with some $t_n\}$
 * Let $T_C = \{$ all $t \in T$ such that the process goes on for ever$\}$.

We can do exactly the same thing with the elements of $S$:


 * Let $S_A = \{$ all $s \in S$ such that the process ends with some $s_n\}$
 * Let $S_B = \{$ all $s \in S$ such that the process ends with some $t_n\}$
 * Let $S_C = \{$ all $s \in S$ such that the process goes on for ever$\}$.

What we need to do is show that $S \sim T$.

We do this by showing that $S_A \sim T_A$, $S_B \sim T_B$ and $S_C \sim T_C$.


 * The restriction of $f$ to $S_A$ is a bijection from $S_A$ to $T_A$.

To do this we need to show that:
 * 1) $s \in S_A \implies f \left ({s}\right) \in T_A$;
 * 2) $\forall t \in T_A: \exists s \in S_A: f \left ({s}\right) = t$.

Let $s \in S_A$. Then the process applied to $s$ ends in $S$.

Now consider the process applied to $f \left ({s}\right)$. Its first step leads us back to $s$. Then it continues the process, applied to $s$, and ends up in $S$. Thus $f \left ({s}\right) \in T_A$.

Thus $s \in S_A \implies f \left ({s}\right) \in T_A$.

Now suppose $t \in T_A$. Then the process applied to $t$ ends in $S$.

In particular, it must have a first stage, otherwise it would end in $T$ with $t$ itself.

Hence $t = f \left ({s}\right)$ for some $s$.

But the process applied to this $s$ is the same as the continuation of the process applied to $t$, and therefore it ends in $S$.

Thus $s \in S_A$ as required.

Hence the restriction of $f$ to $S_A$ is a bijection from $S_A$ to $T_A$.

We can use the same argument to show that $g: T_B \to S_B$ is also a bijection. Hence $g^{-1}: S_B \to T_B$ is a bijection.

Finally, suppose $t \in T_C$.

Because $f$ is an injection, $t = f \left({s}\right)$ for some $s$, and the process applied to $t$ must start.

And this $s$ must belong to $S_C$, because the process starting from $s$ is the same as the process starting from $t$ after the first step. This never ends, as $t \in T_C$.

Now we can define a bijection $h: S \to T$ as follows:


 * $\displaystyle h \left({x}\right) = \begin{cases}

f \left({x}\right): x \in S_A \\ f \left({x}\right): x \in S_C \\ g^{-1} \left({x}\right): x \in S_B \\ \end{cases}$

The fact that $h$ is a bijection follows from the facts that:
 * $(1): \quad S_A$, $S_B$ and $S_C$ are mutually disjoint
 * $(2): \quad T_A$, $T_B$ and $T_C$ are mutually disjoint
 * $(3): \quad f$, $f$ and $g^{-1}$ are the bijections which respectively do the mappings between them.