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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 $A$.)

FizzleDizzle
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    Why do you think that last set is nonempty? There could be elements beyond $a_0,a_1,a_2,....$ Please use a standard definition of infinite. – Somos Jun 08 '19 at 20:20
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    Writing $A={a_0,a_1,a_2,\dots}$ is misleading (especially after "it must be nonempty, i.e."). This suggests that $A$ is a countable set which is not what is assumed. That said, if $A\setminus{b,a_0,a_1,\dots}$ is empty that's not a problem. (Also, it makes no sense to include $b$ there since we already know $b\notin A$). – Derek Elkins left SE Jun 08 '19 at 20:26
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    That's a pretty horrible "proof". Look for a better book, if that is its standard. – hmakholm left over Monica Jun 09 '19 at 00:39
  • That was given in a book. I'd say the proof is out and out incorrect. – fleablood May 26 '20 at 21:04
  • The above statement can be stronger https://math.stackexchange.com/a/1967807/1059606. – An5Drama May 02 '24 at 02:18

2 Answers2

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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.

Mrcrg
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There is a little bit of confusion regarding your question. For instance, you are assuming that $A$ is infinite countable, and from the statement in the title of your question, that could simply not be the case.

Here is a proof of the right hand side implication. First, since $A$ is infinite, then there exists an infinite countable subset of $A$ (this requires at least assuming AC$_\omega$). Let this set be denoted by $B$, and choose an enumeration of $B$; $B=\{b_n|\;n\in\omega\}$.

On the one hand, we clearly have that $A\preccurlyeq A\cup\{b\}$, via the injective function $i:A\longrightarrow A$ given by: for all $a\in A$, $\;i(a)=a$, that is, $i$ is the inclusion of $A$ into $A\cup\{b\}$.

On the other hand, the function $f:A\cup\{b\}\longrightarrow A$ defined by: for all $x\in A\cup\{b\}$:

$$f(x)=\begin{cases} b_0 \qquad\text{if }x=b\\ b_{n+1}\quad\text{if }x=b_n\text{ for some }n\in\omega\\ x\qquad\text{ in any other case} \end{cases}$$

Is clearly injective, so $A\cup\{b\}\preccurlyeq A$. From the Cantor-Bernstein theorem, we can conclude that $A\approx A\cup\{b\}$.

To prove the other implication, take into account that $A$ is a proper subset of the set $A\cup\{b\}$ which is equinumerous to $A$, so we immediately get that $A\cup\{b\}$ is infinite, as well as $A$ (in some literature this result is known as Dedekind's theorem; if you are iterested about this, i.e., that a set is infinite if and only if it contains a proper subset to which it is equinumerous, you should check my answer to this question).

Akerbeltz
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