prove that $|\aleph_{\beta}+\aleph_{\beta}|=\aleph_{\beta}$

Question

Assume that I have already proved for $ \beta\in On$ that $ |\aleph_{\beta}\times\aleph_{\beta}|=\aleph_{\beta}$. Is it correct to conclude that:

$$ |\aleph_{\beta}+\aleph_{\beta}|=|\aleph_{\beta}\times2|\leq|\aleph_{\beta}\times\aleph_{\beta}|=\aleph_{\beta} \ \text{?} $$

Also, if the answer for the first question was yes, is it correct to conclude that if $A, B$ are two sets such that $ |A|\leq\aleph_{\beta}$ and $|B|\leq\aleph_{\beta} $, then

$$ |A\cup B|\leq |A\ \dot{\cup}\ B| \leq|\aleph_{\beta}+\aleph_{\beta}| \ \text{?} $$ (Here, $\dot{\cup}$ is disjoint union.)

Thanks in advance.

Answer 1

Yes, both are correct.

One thing to notice, though, is that $\aleph_\beta$ is already a cardinal, and $+$ between two cardinals is always assumed to be cardinal addition. Therefore the correct notation is $\aleph_\beta+\aleph_\beta$, rather than $|\aleph_\beta+\aleph_\beta|$. Similarly for $\times$.

You can also prove this directly by noting that $\omega_\beta$, the initial ordinal of size $\aleph_\beta$ satisfy that $\{\alpha+1\mid\alpha<\omega_\beta\}$ and $\{\alpha+\omega\mid\alpha<\omega_\beta\}$ are disjoint and both have cardinality $\aleph_\beta$, which gives us a quick and direct proof of the fact about addition.