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: {{begin-eqn}} {{eqn | l = a_0       | r = a }} {{eqn | l = a_1       | r = \map f {a_0} }} {{eqn | l = a_2       | r = \map g {a_1} }} {{eqn | o = \vdots }} {{eqn | l = a_{2 n - 1}       | r = \map f {a_{2 n - 2} } }} {{eqn | l = a_{2 n}       | r = \map g {a_{2 n - 1} } }} {{eqn | o = \vdots }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = a_{-1}       | r = \map {g^{-1} } {a_0}       | c = if such an element exists in $T$ }} {{eqn | l = a_{-2}       | r = \map {f^{-1} } {a_{-1} }       | c = if such an element exists in $S$ }} {{eqn | o = \vdots }} {{eqn | l = a_{-2 n + 1}       | r = \map {g^{-1} } {a_{-2 n + 2} }       | c = if such an element exists in $T$ }} {{eqn | l = a_{2 n}       | r = \map {f^{-1} } {a_{-2 n + 1} }       | c = if such an element exists in $S$ }} {{eqn | o = \vdots }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \sqbrk a_S       | r = \set {\ldots, a_{-2 n}, \ldots, a_{-2}, a_0, a_2, \ldots, a_{2 n}, \ldots} \subseteq S }} {{eqn | l = \sqbrk a_T       | r = \set {\ldots, a_{-2 n + 1}, \ldots, a_{-1}, a_1, \ldots, a_{2 n - 1}, \ldots} \subseteq T }} {{end-eqn}} 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. {{qed}}