# Let $\alpha,\beta,\gamma$ be cardinal numbers where $\aleph_0\le\gamma$, $\alpha+\beta=\gamma$, and $\alpha<\gamma$. Then $\beta=\gamma$

Let $$\alpha,\beta,\gamma$$ be cardinal numbers. We define $$\alpha+\beta=|S|$$, where $$S=A\cup B$$ with $$|A|=\alpha$$, $$|B|=\beta$$, and $$A\cap B=\emptyset$$.

1. If $$\alpha\le\beta$$ and $$\aleph_0\le\beta$$, then $$\alpha+\beta=\beta$$.

2. 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!