I've been struggling to understand the proof of the following theorem given in a book.
Let $A$ be a set and $b \notin A$. Then $A$ is infinite iff there is a
bijection between $A$ and $A \cup \{b\}$.
Proof: Since $A$ is infinite, it must be nonempty, i.e. $A=\{a_0,a_1,a_2,…\}$. A bijection $f: (A \cup \{b\}) \rightarrow A$ can be defined by:
$f(b):=a_0$
$f(a_n):=a_{n+1}$, for $n \in \mathbb{N}$
$f(a):=a$, for $a\in (A\setminus \{b, a_0,a_1,… \})$.
q.e.d.
In the last line, I don't understand why the set $A\setminus \{b, a_0,a_1,… \}$ is nonempty. (It is literally the set formed by taking all the elements of $A$ away from $A$.)
Best Answer
We know, that the set $A$ is infinite, then it can be countable or uncountable, but regardless of which, the set $A$ contains an infinite countable subset. Suppose $B$ is such a set. So, $B=\{a_1,a_2,...,a_n,...\} \subset A$.
Defining $a_0 = b$, so $B'=B \cup \{b\}=\{b, a_1,a_2,...,a_n,...\} \subset (A~\cup \{b\})$ and the bijection $f: A \cup \{b\} \rightarrow A $ by
$$f(x) = \begin{cases} x , &\text{ if $x \notin B'$} \\ a_{n+1} , &\text{ if $x \in B'$ } \end{cases}$$
Note that, $f(b)=f(a_0)=a_1$, $f(a_1)=a_2, ...$
For the demonstration of the statement "If the set $A$ is infinity, contains an infinite countable subset", se below this ProofWiki.