Cantor-Bernstein-Schröder Theorem/Proof 2

Theorem
Let $$S$$ and $$T$$ be sets, such that:
 * $$S \preccurlyeq T$$, that is: $$T$$ dominates $$S$$;
 * $$T \preccurlyeq S$$, that is: $$S$$ dominates $$T$$

Then:
 * $$S \sim T$$

that is, $$S$$ is equivalent to $$T$$.

Proof
Suppose $$S \preccurlyeq T$$ and $$T \preccurlyeq S$$.

By definition, we have that there exist injections $$f: S \to T$$ and $$g: T \to S$$.

We are going to try to build a sequence $$t_1, s_1, t_2, s_2, t_3 \ldots$$ as follows.

Consider any $$t_1 \in T$$.

By the Law of the Excluded Middle there are two choices:

$$ $$

Suppose $$\exists s_1 \in S: f \left({s_1}\right) = t_1$$.

Because $$f$$ is injective, such an $$s_1$$ is unique.

So we can choose $$s_1 = f^{-1} \left({t_1}\right)$$.

Again, by the Law of the Excluded Middle there are two further choices:

$$ $$

Suppose $$\exists t_2 \in T: g \left({t_2}\right) = s_1$$.

Because $$f$$ is injective, such an $$t_2$$ is unique.

Similarly, we choose $$s_2 = f^{-1} \left({t_2}\right)$$, if it exists.

This process goes on until:


 * We reach some $$s_n \in S$$ such that $$\neg \exists t \in T: g \left({t}\right) = s_n$$. This may be possible because $$g$$ may not be a surjection.


 * We reach some $$t_n \in T$$ such that $$\neg \exists s \in S: f \left({s}\right) = t_n$$.This may be possible because $$f$$ may not be a surjection.


 * The process goes on for ever.

For each $$t \in T$$, then, there is a well-defined process which turns out in one of the above three ways.

We partition $$T$$ up into three subsets that are mutually disjoint:


 * Let $$T_A = \{$$ all $$t \in T$$ such that the process ends with some $$s_n\}$$;
 * Let $$T_B = \{$$ all $$t \in T$$ such that the process ends with some $$t_n\}$$;
 * Let $$T_C = \{$$ all $$t \in T$$ such that the process goes on for ever$$\}$$.

We can do exactly the same thing with the elements of $$S$$:


 * Let $$S_A = \{$$ all $$s \in S$$ such that the process ends with some $$s_n\}$$;
 * Let $$S_B = \{$$ all $$s \in S$$ such that the process ends with some $$t_n\}$$;
 * Let $$S_C = \{$$ all $$s \in S$$ such that the process goes on for ever$$\}$$.

What we need to do is show that $$S \sim T$$.

We do this by showing that $$S_A \sim T_A$$, $$S_B \sim T_B$$ and $$S_C \sim T_C$$.


 * The restriction of $$f$$ to $$S_A$$ is a bijection from $$S_A$$ to $$T_A$$.

To do this we need to show that:
 * 1) $$s \in S_A \implies f \left ({s}\right) \in T_A$$;
 * 2) $$\forall t \in T_A: \exists s \in S_A: f \left ({s}\right) = t$$.

Let $$s \in S_A$$. Then the process applied to $$s$$ ends in $$S$$.

Now consider the process applied to $$f \left ({s}\right)$$. Its first step leads us back to $$s$$. Then it continues the process, applied to $$s$$, and ends up in $$S$$. Thus $$f \left ({s}\right) \in T_A$$.

Thus $$s \in S_A \implies f \left ({s}\right) \in T_A$$.

Now suppose $$t \in T_A$$. Then the process applied to $$t$$ ends in $$S$$.

In particular, it must have a first stage, otherwise it would end in $$T$$ with $$t$$ itself.

Hence $$t = f \left ({s}\right)$$ for some $$s$$.

But the process applied to this $$s$$ is the same as the continuation of the process applied to $$t$$, and therefore it ends in $$S$$.

Thus $$s \in S_A$$ as required.

Hence the restriction of $$f$$ to $$S_A$$ is a bijection from $$S_A$$ to $$T_A$$.

We can use the same argument to show that $$g: T_B \to S_B$$ is also a bijection. Hence $$g^{-1}: S_B \to T_B$$ is a bijection.

Finally, suppose $$t \in T_C$$.

Because $$f$$ is an injection, $$t = f \left({s}\right)$$ for some $$s$$, and the process applied to $$t$$ must start.

And this $$s$$ must belong to $$S_C$$, because the process starting from $$s$$ is the same as the process starting from $$t$$ after the first step. This never ends, as $$t \in T_C$$.

Now we can define a bijection $$h: S \to T$$ as follows:



h \left({x}\right) = \begin{cases} f \left({x}\right): x \in S_A \\ f \left({x}\right): x \in S_C \\ g^{-1} \left({x}\right): x \in S_B \\ \end{cases} $$

The fact that $$h$$ is a bijection follows from the facts that:
 * 1) $$S_A$$, $$S_B$$ and $$S_C$$ are mutually disjoint;
 * 2) $$T_A$$, $$T_B$$ and $$T_C$$ are mutually disjoint;
 * 3) $$f$$, $$f$$ and $$g^{-1}$$ are the bijections which respectively do the mappings between them.