I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha
If there is $\alpha$ s.t. $\omega\le\alpha
However by a counterexample showed by Brian M. Scott in this question, this path is obstructed.
So could anyone give me a hint?
I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha
If there is $\alpha$ s.t. $\omega\le\alpha
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.