Cantor-Bernstein-Schröder Theorem/Proof 1

Theorem
If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
 * $\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$

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:
 * $S_2 = g \left({f \left({S}\right)}\right) = g \left({T_1}\right) \subseteq S_1$

and:
 * $T_2 = f \left({g \left({T}\right)}\right) = f \left({S_1}\right) \subseteq T_1$

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 $\displaystyle D = \bigcap_{k=1}^\infty S_k$.

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 $S \sim T$.