# Is the Axiom of Choice equivalent to say that for two (infinite) cardinal $\kappa$ and $\lambda$ it's result $\kappa+\lambda=\kappa*\lambda$

Following what is written at the 10th chapter of the "Teoria de Cojuntos, una introduccion" by Fernando Herandez (For the sake of completeness I should say that this is also present at the 11th chapter of "The Axiom of Choice" by Thomas Jech).

For every infinite cardinal number $$\kappa$$, let $$\aleph(\kappa)$$ be the Hartogs number of $$\kappa$$, i.e., the least ordinal which cannot be embedded by a one-to-one mapping in a set of cardinality $$\kappa$$. For every $$\kappa$$, $$\aleph(\kappa)$$ is an aleph, viz. the least aleph $$\aleph$$ such that $$\aleph\not\le\kappa$$.

Lemma 10.5

If $$\kappa$$ is an infinite cardinal and $$\aleph$$ is ana aleph, and if

10.1.1 $$\quad\quad\quad\quad\quad\quad\kappa+\aleph=\kappa*\aleph$$,

then either $$\kappa\ge\aleph$$ or $$\kappa\le\aleph$$. In particular, if

10.1.2$$\quad\quad\quad\quad\quad\kappa+\aleph(\kappa)=\kappa*\aleph(\kappa)$$

then $$\kappa$$ is an aleph.

Theorem 10.6

The Axiom of Choice is equivalet to say that for two cardinal $$\kappa$$ and $$\lambda$$ it's result $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\kappa+\lambda=\kappa*\lambda$$.

Proof. We will show that under the hypothesis of the theorem any infinite cardinal is an aleph. Let $$\kappa$$ be an infinite cardinal, by Lemma 10.5 it follows that $$\kappa\le\aleph(\kappa)$$, which says that $$\kappa$$ is an aleph.

So I don't understand the proof of theorem 10.6: could someone explain to me why what is here written prove the theorem?