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Let $A$ be an arbitrary infinite set and $x\not\in A$. I want to find a bijection between $A$ and $A\cup\{x\}$.

Proof 1: A possible proof of that is to play with cardinal numbers as follows: Let $\kappa,\lambda$ be cardinal numbers such that $\kappa$ is infinite and $\lambda\leq \kappa$. Then

$$ \kappa\leq\kappa+\lambda\leq\kappa+\kappa=\kappa $$

and hence, Schroeder-Bernstein Theorem implies that $\kappa+\lambda=\kappa$, which for $\lambda=1$ implies the desired result.

Proof 2: Another way of proving it is the following: We know that since $A$ is infinite there exists a countably infinite proper subset of $B\subseteq A$. Then, it is easy to see that $B$ can be partitioned as $B=B_{1}\cup B_{2}$, where $B_{1},B_{2}$ are countably infinite subsets of $B$. In particular, there exists a bijection $g:B\to B_{1}$. Hence, by writing $A=B\cup(A\setminus B)$, we see that the map $f:A\to B_{1}\cup(A\setminus B)$ where

$$ f(x)=\left\{\begin{matrix} x, & x\in A\setminus B \\ g(x), & x\in B\end{matrix}\right. $$

is a bijection between $A$ and its proper subset $B_{1}\cup(A\setminus B)$. Then, using this fact we proceed of proving the desired result. Let $C\subseteq A$ be a proper subset of $A$ such that $|C|=|A|$. Let $h:A\to C$ be a bijection. Let also $y\in A\setminus C$. Obviously, we have that $|A|\leq|A\cup\{x\}|$. For the other direction observe that the map $f:A\cup\{x\}\to A$ where

$$ f(a)=\left\{\begin{matrix} h(a), & a\in A \\ y, & a=x\end{matrix}\right. $$

is 1-1 and so $|A\cup\{x\}|\leq |A|$. Then, again from Schroeder-Bernstein Theorem, we derive the equality.

Question: My question is whether I can prove this fact without relying in the axiom of choice. I know that Schroeder-Bernstein Theorem is independent from AC but the proofs of the facts:

  • if $\kappa$ is an infinite cardinal then $\kappa+\kappa=\kappa$, and
  • if $A$ is an infinite set then it contains a countably infinite proper subset

rely in AC (as far as those proofs that I know). However, it seems to me that the statement $|A\cup\{x\}|=|A|$ is very simple to rely in AC and can be proven without it, though I may be completely wrong. Any comments?

richarddedekind
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  • Cantor–Bernstein is not independent from AC. That would mean that it is neither provable nor refutable from ZFC. – Asaf Karagila May 17 '20 at 13:30
  • You didn't define infinite? Do you mean Dedekind infinite? Or did you mean Tarski infinite. If the latter then you cannot prove that since we can indeed have a set $A$ that is Tarski infinite but Dedekind finite, and so $|A+1| > |A|$. However if the former then I think its provable without choice. – Zuhair May 17 '20 at 16:01

1 Answers1

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If you have a bijection $f:A\cup\{x\}\to A$, then consider the sequence recursively defined by $x_0:=x$, $x_{n+1}:=f(x_n)$. Then $\{x_1,x_2,\ldots,\},$ is a countable infinite subset of $A$. So $|A\cup\{a\}|=|A|$ is equivalent to $A$ having a countable infinite subset.

I think the theorem that states that an infinite set has a countable infinite subset requires a weak form of Choice.

amWhy
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Angina Seng
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