The Cantor-Schröder-Bernstein Theorem: Parts and Equipotence in Set Theory

Cantor-Schröder-Bernstein. However, the reader who has meditated on what we have written so far can also omit continuing the reading, because, in the final part of this work, we will only show how the presence of parts, the possibility of performing operations between them and combining them with the existence of functions clarifies in a truly very general sense, how the essence of the indefinite is that of being equipotent to one of its parts; in short, that it is controllable through a portion strictly contained within it.

The part for the whole and the Cantor-Schröder-Bernstein Theorem.

The Cantor-Schröder-Bernstein theorem shows how given two sets $A$ and $I$ such that there exist two injective functions $P: A \to I$ and $Q: I \to A$, then it is possible to construct a bijective function $f: A \to I$, that is, that the two sets $A$ and $I$ are equipotent.

A first step in the proof of the Cantor-Schröder-Bernstein Theorem consists in considering two arbitrary series of numbers $A$ and $I$ such that there exist two functions: $P: A \to I$ and $Q: I \to A$.

Since $A$ has parts, let us concentrate on what the procedures $P: A \to I$ and $Q: I \to A$ allow us to see if applied starting from $A$.

Let $B$ be a part of $A$. Naturally we obtain that the image $P(B)$ of $B$ is a part of $I$. Now let us consider the part of $I$ that is the complement of $P(B)$; that is $I \setminus P(B)$. Since $I \setminus P(B)$ is a part of $I$ constructed starting from the part $B$ of $A$, then the function $Q: I \to A$ allows us to return to $A$; that is, we look at this new part $Q(I \setminus P(B))$ of $A$.

We now have a new part $Q(I \setminus P(B))$ of $A$ obtained starting from the part $B$ and from the following procedures: $P$, passage to the complement and $Q$.

The mind sees clearly that there are therefore two types of parts of $A$. There are those parts $B$ of $A$ such that $B$ and $Q(I \setminus P(B))$ have elements in common and those parts $B$ of $A$ such that $B$ and $Q(I \setminus P(B))$ have no elements in common.

We can therefore construct a new part $B_0$, that is, that part of $A$ obtained by uniting all the parts $B$ of $A$ such that $B$ and $Q(I \setminus P(B))$ have no elements in common.

The mind observes the role of a double negation; the first is hidden in the complementation $I \setminus P(B)$, the second in there being no elements in common between $B$ and $Q(I \setminus P(B))$.

Finally, with meditation we expand this situation to its maximum and we see before our eyes the part $B_0$ of $A$.

We now apply to this particular part $B_0$ of $A$ the preceding procedures and it is not difficult to verify not only that $B_0$ and $Q(I \setminus P(B_0))$ have no elements in common, as is easy to see, but also that $B_0$ united with $Q(I \setminus P(B_0))$ coincides with all of $A$: $$A = B_0 \cup Q(I \setminus P(B_0))$$

It is important to dwell on this point. Through the passage to the opposite, impossible in the true Infinite, and for the sole existence of the two procedures $P: A \to I$ and $Q: I \to A$, we have been able to split $A$ into two disjoint parts, $B_0$ and $Q(I \setminus P(B_0))$, nothing of what is in $B_0$ can come from $I$ if it did not already come from $B_0$ and, finally, $B_0$ is maximal with respect to this property.

Let us now demonstrate the theorem assuming for simplicity that $Q: I \to A$ is the simple inclusion, that is, that $I$ is a part of $A$ and $P: A \to I$ is an injective function from $A$ into its part $I$.

We see, in the mind, that $A$ is an indefinite.

Now let us consider the part $B_0$ of $A$. Let $I_0 := I \setminus P(B_0)$.

We still see before our eyes that $B_0$ and $Q(I_0)$ have no elements in common and that $A = B_0 \cup Q(I_0)$.

We slowly move the mind to $I$. We observe that $I_0$ and $P(A)$ have no elements in common and that $I = I_0 \cup P(B_0)$.

The image $P(B_0)$ is such that for the injectivity of $P$ we have that $P: B_0 \to P(B_0)$ is a bijective application. The image $P(B_0)$ is disjoint from $I_0$. Similarly $Q: I_0 \to Q(I_0)$ is bijective.

Therefore there exists an inverse procedure to $Q$, if we limit ourselves to defining it on the part $Q(I_0)$ of $A$; that is exactly above the complement of $B_0$.

We call $R: Q(I_0) \to I_0$ that unique procedure such that $Q(i) = a \Leftrightarrow R(a) = i$, if and only if $i \in Q^{-1}(a)$.

In synchronic meditation we see that the procedure $P$ carries $B_0$ into $P(B_0)$ and the inverse of the procedure $Q$ carries the complement of $B_0$ into the complement of $P(B_0)$.

We therefore unify the two procedures as follows: $$A \ni a \mapsto f(a) := \begin{cases} P(a) & \text{if } a \in B_0 \\ R(a) & \text{if } a \in Q(I_0) \end{cases}$$ to obtain a bijective function $f: A \to I$.

In the case of an indefinite $A$, the presence of at least one of its parts $I$ equipotent to $A$ is clear manifestation that the indefinite is not a notion comparable to that of the Infinite. Even thanks to the excluded third $A$ and the possibility of decomposing $A$ into parts, it is even possible to put $A$ in bijection with one of its parts $I$.

Symbolism of divergent series - 1st part (Radice and Tabit)

There can be truths that surpass common sense, but there is none that contradicts mathematics - E.Levi