Cantor-Bernstein-Schröder Theorem/Proof 4

== Theorem ==
== 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.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
Category:Cantor-Bernstein-Schröder Theorem