Cantor-Bernstein-Schröder Theorem/Proof 4

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


 * $f \sqbrk S = T_1 \subseteq T$
 * $g \sqbrk T = S_1 \subseteq S$

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 $S$:
 * $\ldots, a_{-2 n}, \ldots, a_{-2}, a_0, a_2, \ldots, a_{2 n}, \ldots$

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

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 $\map {f^{-1} } x$ and $\map {g^{-1} } y$ 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 $\sqbrk a_S$

and
 * the elements of $T$ with an odd index be denoted $\sqbrk a_T$

that is:

We are given that $f$ and $g$ are injections.

So by the definition of $\sqbrk a_S$, for any two $a, b \in S$, $\sqbrk a_S$ and $\sqbrk b_S$ are either disjoint or equal.

The same applies to $\sqbrk a_T$ and $\sqbrk b_T$ for any $a, b \in T$.

It follows that:
 * $\AA_S = \set {\sqbrk a_S: a \in S}$ is a partition of $S$

and
 * $\AA_T = \set {\sqbrk a_T: a \in S}$ is a partition of $T$.

So:
 * every element of $S$ belongs to exactly one element of $\AA_S$

and:
 * every element of $T$ belongs to exactly one element of $\AA_T$.

So let $a \in S$ and $b \in T$ such that $\map f a = b$.

It follows that a bijection can be constructed from $\AA_S$ to $\AA_T$.

Now there are two different kinds of $\sqbrk a_S$ sets:


 * $(1):$ It is possible that no repetition occurs in the sequence $\sequence {a_{2 n} }$.

As a consequence, no repetition occurs in the sequence $\sequence {a_{2 n - 1} }$ either.

By the method of construction it can be seen that $\sqbrk a_S$ and $\sqbrk a_T$ are both countably infinite.

So a bijection can be constructed between $\sqbrk a_S$ and $\sqbrk a_T$.


 * $(2):$ There may be a repetition in $\sqbrk a_S$.

Suppose such a repetition is $a_{2 m} = a_{2 n}$ for some $m \ne n$.

Then $a_{2 m + 1} = a_{2 n + 1}$ and $a_{2 m + 2} = a_{2 n + 2}$ and so on.

In general $a_{2 m + k} = a_{2 n + k}$ for all $k \in \N$.

But because $f$ and $g$ are injections it follows that $a_{2 m - 1} = a_{2 n - 1}$ and $a_{2 m - 2} = a_{2 n - 2}$ and so on, where they exist.

So in general $a_{2 m - k} = a_{2 n - k}$ for all $k \in \N$, where they exist.

Given $m$, we can choose $n$ so that the elements $a_{2 m}, a_{2 m + 2}, \ldots, a_{2 n - 2}$ are distinct.

Then $a_{2 m + 1}, a_{2 m + 3}, \ldots, a_{2 n - 1}$ are likewise distinct elements of $T$.

Thus we can set up a bijection:
 * $a_{2 m} \mapsto a_{2 m + 1}, a_{2 m + 2} \mapsto a_{2 m + 3}, \ldots, a_{2 n - 2} \mapsto a_{2 n - 1}$

between the elements of $\sqbrk a_S$ and $\sqbrk a_T$.

It follows that a bijection can be constructed between any two elements of the partitions of $S$ and $T$.

These maps then automatically yield a bijection from $S$ to $T$.

Hence the result.