# $A+\alpha\sim A$ when $\omega\le\alpha<h(A)$

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.

However by a counterexample showed by Brian M. Scott in this question, this path is obstructed.

So could anyone give me a hint?

-

Note that $\alpha+\alpha\sim\alpha$ for any infinite ordinal.
Therefore if $A>\alpha\geq\omega$ we can write $A\sim\alpha\cup B$ for some $B\subseteq A$ disjoint from $\alpha$. Then $\alpha+|A|=\alpha+\alpha+|B|=\alpha+|B|=|A|$, as wanted.
The requirement that $\alpha\geq\omega$ is clear because finite ordinals do not have the property $n+n\sim n$, and so it is false.