Let $\alpha,\beta,\gamma$ be cardinal numbers. We define $\alpha+\beta=|S|$, where $S=A\cup B$ with $|A|=\alpha$, $|B|=\beta$, and $A\cap B=\emptyset$.
If $\alpha\le\beta$ and $\aleph_0\le\beta$, then $\alpha+\beta=\beta$.
If $\alpha+\beta=\gamma$, $\aleph_0\le\gamma$, and $\alpha<\gamma$, then $\beta=\gamma$.
My attempt:
Lemma: $\aleph_0\le\beta\implies \beta+\beta=\beta$ (I presented a proof here)
We first prove claim 1:
We have $\beta\le\alpha+\beta\le\beta+\beta$ and $\beta+\beta=\beta$ by Lemma. Thus $\alpha+\beta=\beta$.
We next prove claim 2:
We have $\alpha+\beta=\gamma$, then $\beta\le\gamma$. Assume the contrary that $\beta\neq\gamma$, then $\beta<\gamma$.
If $\alpha\le\beta$, then $\alpha+\beta=\beta$ by Claim 1. It follows that $\alpha+\beta=\beta<\gamma$, which is a contradiction.
If $\beta<\alpha$, then $\alpha+\beta=\alpha$ by Claim 1. It follows that $\alpha+\beta=\alpha<\gamma$, which is a contradiction.
Hence $\beta=\gamma$.
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!
