0
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

The statement "every infinite cardinal is an aleph" means that every infinite set can be put in bijection with an aleph (which is a particular ordinal). Thus, every set can be well-ordered. The well-ordering principle is well-known to be equivalent to the axiom of choice.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.