Is there any Dedekind-infinite set can be split to two smaller Dedekind-infinite sets? 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?
 A: 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
$$\omega\to B:n\mapsto f(2n)$$
and
$$\omega\to C:n\mapsto f(2n+1)\;.$$
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 $|C|=|\omega\uplus E|<|A|$.
A: 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.)
