# Does $\alpha + \beta = \alpha$ imply $\beta \le \aleph_0$

Just like in title, my question is : Does $\alpha + \beta = \alpha$ imply $\beta \le \aleph_0$ where, $\alpha$ and $\beta$ are cardinals?

P.S. I actualy have to prove $\alpha + \beta = \alpha$ $\iff$ $\alpha + \aleph_0 \cdot \beta = \alpha$. Any ideas?

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No. In general, $\alpha +\alpha = \alpha$ if $\alpha$ is an infinite cardinal. – Thomas Andrews Jun 19 '13 at 20:07
Sorry about my previous comment, I thought you were talking about ordinal arithmetic!! – Edoardo Lanari Jun 19 '13 at 20:08

If you mean $\alpha+\beta= \alpha$ as in addition of cardinals, then no. Consider $\alpha=\aleph_2, \beta=\aleph_1$.
Well $\aleph_0 \cdot \beta=\max(\beta, \aleph_0)$. Thus, there are two cases. If $\aleph_0 \cdot \beta=\beta$, then the two statements are equivalent. If $\aleph_0 \cdot \beta= \aleph_0$, then $\beta \leq \aleph_0$ and so $\alpha+ \beta \leq \alpha+\aleph_0 \cdot \beta= \alpha$ which gives one direction. Also, if $\alpha+ \beta= \alpha$, then $\alpha \geq \aleph_0$ and so $\alpha+ \aleph_0 \cdot \beta=\alpha+\aleph_0= \alpha$ – Alexander Jun 19 '13 at 20:31
$\alpha + \beta = max(\alpha, \beta)$ if one of them is an infinite cardinal, so it's easy to find counterexample.