I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha<h(A)$, in which $h(A)$ is the Hartogs number of $A$.

If there is $\alpha$ s.t. $\omega\le\alpha<h(A)$, we can get $\alpha \le A$ and $A$ is Dedekind-infinite. So if $|A| \ne |A|+|\alpha|$ then both $|\alpha|$ and $|A|$ $<|A|+|\alpha|$. Note that they are all Dedekind-infinite, so if it can be proven that every Dedekind-infinite set cannot be split into two smaller Dedekind-infinite sets then $|A| = |A|+|\alpha|$ holds.

So my question: Is there any Dedekind-infinite set can be split into two smaller Dedekind-infinite sets?


The question is incorrect in this context.

If $\alpha<h(A)$ then you can split $A$ into one part of size $\alpha$ and another of size $|A|$. Your question is different, is whether we can always split some set into two strictly smaller-yet-infinite sets.

Indeed we can. Let $A$ be any Dedekind-infinite set which cannot be well-ordered and consider $A\cup h(A)$. This is a Dedekind-infinite set and it can be split into $A$ and $h(A)$. Both are Dedekind-infinite.

As I wrote in one of my previous answers to your question, assuming that there is no Dedekind-infinite set which can be split into two strictly smaller Dedekind-infinite sets is equivalent to the axiom of choice.

(The proof, let me remind you is that if this principle fails then for every infinite $A$ we cannot split $A+h(A)^+$ into two Dedekind-infinite sets which are strictly smaller, therefore $h(A)^+=A+h(A)^+$, and therefore $A<h(A)^+$ and can be well-ordered.)

  • $\begingroup$ It is another equivalent form of axiom of choice. Sounds good to me. $\endgroup$ – Popopo Dec 2 '12 at 13:13

If $A$ is Dedekind-infinite, there is an injection $f:\omega\to A$. Let $B=\{f(2n):n\in\omega\}$ and $C=A\setminus B$; then $B$ and $C$ are both Dedekind-infinite, since each clearly admits an injection from $\omega$, namely

$$\omega\to B:n\mapsto f(2n)$$


$$\omega\to C:n\mapsto f(2n+1)\;.$$

Added: Suppose further that $\omega<|A|$. Then certainly $|B|=\omega<|A|$ and $\omega\le|C|$. However, it’s clear that the map

$$A\to C:a\mapsto\begin{cases} a,&\text{if }a\in A\setminus\operatorname{ran}f\\ f(4n+1),&\text{if }a=f(2n)\\ f(4n+3),&\text{if }a=f(2n+1) \end{cases}$$

is a bijection, so $|C|=|A|$.

We can also say that if $A=\omega\uplus D$, where $D$ is amorphous, and $\{B,C\}$ is a partition of $A$ into Dedekind-infinite sets, then one of $B$ and $C$ is countably infinite, and the other has the same cardinality as $A$. On the other hand, if $A=\omega\uplus D\uplus E$, where $D$ and $E$ are amorphous, then $A$ has a partition $\{B,C\}$ such that $|B|=|\omega\uplus D|<|A|$ and $|C|=|\omega\uplus E|<|A|$.

  • $\begingroup$ Well, any Dedekind-infinite set indeed can be split into two D-infinite parts. But honestly my question is can a Dedekind-infinite set be split into two strict smaller D-infinite parts? $\endgroup$ – Popopo Dec 2 '12 at 9:32
  • $\begingroup$ That's all right. It is a nice example. Many thanks! $\endgroup$ – Popopo Dec 2 '12 at 10:05
  • $\begingroup$ @Popopo: You’re welcome. $\endgroup$ – Brian M. Scott Dec 2 '12 at 10:14
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    $\begingroup$ @Popopo: Do note that almost all this (including my answer) was covered in my answer and our discussion in the comments, to one of your previous questions. Also note that the disjoint union of an amorphous set and a countable set can be split into two Dedekind-infinite sets, but can also be split into a Dedekind-finite and a Dedekind-infinite set. $\endgroup$ – Asaf Karagila Dec 2 '12 at 11:31

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