I'm reading Jech's Set Theory and after his introduction of the canonical well-ordering of $\alpha\times\alpha$ (to prove $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$) I found the following statement (p. 31):
As a corollary we have $$ \aleph_\alpha+\aleph_\beta=\aleph_\alpha\cdot\aleph_\beta=\max\{\aleph_\alpha,\aleph_\beta\} $$
I understand $\aleph_\alpha\cdot\aleph_\beta=\max\{\aleph_\alpha,\aleph_\beta\}$ well, and have come up with a proof for the addition part, about which I'm still not sure:
Proof (without AC). Suppose $\aleph_\alpha\leq\aleph_\beta$. Write $A$ for $\{0\}\times\omega_\alpha$ and $B$ for $\{1\}\times\omega_\beta$. Then there is an injection $f:A\to B$. Define the extension of $f$
$$g(x) = \begin{cases}
f(x) & \text{if $x\in A$} \\
x & \text{if $x\in B$}
\end{cases}
$$
Then $g$ is a bijection from $A\cup B$ onto $B$, and so $\aleph_\alpha+\aleph_\beta=\aleph_\beta$.
Is this correct?
Best Answer
The function you defined is clearly not a bijection. Take any $x\in A$ and let $f(x)=y$. Then by definition $g(x)=g(y)$, while clearly $x\ne y$. So $g$ is not injective.
Anyway, if you know that $\aleph_{\alpha}\cdot\aleph_{\alpha}=\aleph_{\alpha}$ then you can easily prove your result. If $\aleph_{\alpha}\leq\aleph_{\beta}$ then:
$\aleph_{\beta}\leq\aleph_{\alpha}+\aleph_{\beta}\leq\aleph_{\beta}+\aleph_{\beta}=2\aleph_{\beta}\leq\aleph_{\beta}\cdot\aleph_{\beta}=\aleph_{\beta}$
And so by Cantor-Schroder-Benstein it follows that $\aleph_{\alpha}+\aleph_{\beta}=\aleph_{\beta}$. And the proof of $\aleph_{\alpha}\cdot\aleph_{\beta}=\aleph_{\beta}$ is pretty much the same.