MHB Definition of $S_n$ for Cantor-Schröder-Bernstein Theorem

[evinda]

Hi! (Smile)

I am looking at the proof of the theorem of Cantor- Schröder-Bernstein, that states the following:

Let $A,B$ be sets. If $A$ is equinumerous with a subset of $B$ and $B$ is equinumerous with a subset of $A$ then $A, B$ are equinumerous. Or equivalently, if $f: A \overset{1-1}{B}$ and $g: B \overset{1-1}{A}$ then there is a $h: A \overset{\text{surjective}}{\to}B$.

Proof:

Let $f: A \overset{1-1}{\to}B$ and $g: B \overset{1-1}{\to}A$.
We define recursively the following sequence of sets:

$$S_0:=A-g(B)\\S_{n+1}=g[f[S_n]]\\ \text{ for all } n \in \omega \\ \dots$$

We define $S:=\bigcup_{n \in \omega} S_n$ and the function $h: A \to B$ as follows:

$$f(x)=x \text{ if } x \in S \\ h(x)=g^{-1}(x) \text{ if } x \in A-S$$
Could you explain me the definition of $S_n$?

 

[caffeinemachine]

Hi! (Smile)

I am looking at the proof of the theorem of Cantor- Schröder-Bernstein, that states the following:

Let $A,B$ be sets. If $A$ is equinumerous with a subset of $B$ and $B$ is equinumerous with a subset of $A$ then $A, B$ are equinumerous. Or equivalently, if $f: A \overset{1-1}{B}$ and $g: B \overset{1-1}{A}$ then there is a $h: A \overset{\text{surjective}}{\to}B$.

Proof:

Let $f: A \overset{1-1}{\to}B$ and $g: B \overset{1-1}{\to}A$.
We define recursively the following sequence of sets:

$$S_0:=A-g(B)\\S_{n+1}=g[f[S_n]]\\ \text{ for all } n \in \omega \\ \dots$$

We define $S:=\bigcup_{n \in \omega} S_n$ and the function $h: A \to B$ as follows:

$$f(x)=x \text{ if } x \in S \\ h(x)=g^{-1}(x) \text{ if } x \in A-S$$
Could you explain me the definition of $S_n$?

Hi evinda.

You have defined $S_n$ inductively by declaring:

$$S_0:=A-g(B)\\S_{n+1}=g[f[S_n]]\\ \text{ for all } n \in \omega \\ \dots$$

Where exactly are you facing a problem?