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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!

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