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:
 * $$\forall S, T: T \sim S_1 \subseteq S \and S \sim T_1 \subseteq T \implies S \sim T$$

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


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

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

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 $$T_2 \sim T$$.

Let $$S_3 \subseteq S$$ be the image of $$S_1$$ under the mapping $$g \circ f$$,

and let $$S_4 \subseteq S$$ be the image of $$S_2$$ under the mapping $$g \circ f$$.

We can generalise this to:

Let $$S_{k+2} \subseteq S$$ be the image of $$S_k$$ under the mapping $$g \circ f$$, where $$k \in \N$$.

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

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

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 $$S \sim T$$.