Cantor-Bernstein-Schröder Theorem

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:

$$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \Longrightarrow S \sim T$$

Alternatively, from Dominates is Equivalent to Subset, this can be expressed:

$$\forall S, T: T \le S \land S \le T \Longrightarrow S \sim T$$

(Also known as the Cantor-Bernstein Theorem or the Schroeder-Bernstein Theorem.)

Proof 1
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 $$k \in \mathbb{N}$$.

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:

$$S = \left({S - S_1}\right) \cup \left({S_1 - S_2}\right) \cup \ldots \cup \left({S_k - S_{k+1}}\right) \cup \ldots \cup D$$ (1)

Similarly, we can represent $$S_1$$ as:

$$S_1 = \left({S_1 - S_2}\right) \cup \left({S_2 - S_3}\right) \cup \ldots \cup \left({S_k - S_{k+1}}\right) \cup \ldots \cup D$$ (2)

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 $$S \sim T$$.

Proof 2
By Hypothesis, 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:


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

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:


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

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 $$\lnot \exists t \in T: g \left({t}\right) = s_n$$. This is possible because it was not specified that $$g$$ is a surjection.


 * We reach some $$t_n \in T$$ such that $$\lnot \exists s \in S: f \left({s}\right) = t_n$$. This is possible because it was not specified that $$f$$ is 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) $$s \in S_A \Longrightarrow f \left ({s}\right) \in T_A$$;
 * 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 $$s \in S_A \Longrightarrow f \left ({s}\right) \in T_A$$.

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 disjoint;
 * 2) $$T_A$$, $$T_B$$ and $$T_C$$ are disjoint;
 * 3) $$f$$, $$f$$ and $$g^{-1}$$ are the bijections which respectively do the mappings between them.

Comments

 * This theorem states in set theoretical concepts the "intuitively obvious" fact that if $$a \le b$$ and $$b \le a$$ then $$a = b$$.

Care needs to be taken to make well sure of this, because when considering infinite sets, intuition is frequently misleading.


 * In order to prove equivalence, a bijection needs to be demonstrated. It can be significantly simpler to demonstrate an injection than a surjection, so proving that there is an injection from $$S$$ to $$T$$ and also one from $$T$$ to $$S$$ may be a lot less work than proving that there is both an injection and a surjection from $$S$$ to $$T$$.