Cantor-Bernstein-Schröder Theorem/Proof 1

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


 * $f: S \to T_1$
 * $g: T \to S_1$

Let:
 * $S_2 = g \sqbrk {f \sqbrk S} = g \sqbrk {T_1} \subseteq S_1$

and:
 * $T_2 = f \sqbrk {g \sqbrk T} = f \sqbrk {S_1} \subseteq T_1$

So $S_2 \subseteq S_1$ and $S_2 \sim S$, while $T_2 \subseteq T_1$, and $T_2 \sim T$.

For each natural number $k$, let $S_{k + 2} \subseteq S$ be the image of $S_k$ under the mapping $g \circ f$.

Then:
 * $S \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_k \supseteq S_{k + 1} \ldots$

Let $\ds D = \bigcap_{k \mathop = 1}^\infty S_k$.

Now we can represent $S$ as:

where $S \setminus S_1$ denotes set difference.

Similarly, we can represent $S_1$ as:

Now let:

and rewrite $(1)$ and $(2)$ as:

Now:

and so on.

So $N \sim N_1$.

It follows from $(3)$ and $(4)$ that a bijection can be set up between $S$ and $S_1$.

But $S_1 \sim T$.

Therefore $S \sim T$.