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Following what is written at the 11th chapter of "The Axiom of Choice" by Thomas Jech.

For every infinite cardinal numer $\kappa$, 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 an aleph, and if

$$\kappa+\aleph=\kappa*\aleph,\tag{11.8}$$

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

$$\kappa+\aleph(\kappa)=\kappa*\aleph(\kappa)\tag{11.9}$$

then $\kappa$ is an aleph.

Theorem 11.7

If $\kappa^2=\kappa$ for every infinite cardinal number $\kappa$, then the Axiom of Choice holds.

Proof. We will show that under the assumption of the theorem, every infinite cardinal is an aleph. To do so, it suffices to show that

$$\kappa+\aleph(\kappa)=\kappa*\aleph(\kappa).$$

Since $\kappa+\aleph(\kappa)\le\kappa*\aleph(\kappa)$, we have only to show that $\kappa+\aleph(\kappa)\ge\kappa*\aleph(\kappa)$.

This is proved as follows:

$$\kappa+\aleph(\kappa)=(\kappa+\aleph(\kappa))^2=\kappa^2+2\kappa*\aleph(\kappa) +(\aleph(\kappa))^2\ge\kappa*\aleph(\kappa).$$

So I don't understand the proof of theorem 11.7: could someone explain to me why what is here written prove the theorem?

I have understood that if every infinite cardinal number is an aleph also every infinite set can put in bijection with an aleph, which is a well ordered set so this implies the well-ordering theorem and this is equivalent to the Choice Axiom; but how to prove the opposite implication?

Then I don't understand why $\kappa+\aleph(\kappa)\le\kappa*\aleph(\kappa)$.

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    $\begingroup$ @John Bentin: PLEASE don’t use in-line displays with manual spaces to fake displayed equations. Displayed equations center automatically according to each user’s interface, in-line displays with quads do not. $\endgroup$ – Arturo Magidin Jan 17 at 22:32
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    $\begingroup$ Seeing your recent questions, one has to wonder if you're not trying to bite much more than you can chew... $\endgroup$ – Asaf Karagila Jan 17 at 22:35
  • $\begingroup$ @ArturoMagidin : I take your point; but I didn't touch the OP's spacing. My editing was confined to the English (spelling, punctuation, and some wording). $\endgroup$ – John Bentin Jan 17 at 22:38
  • $\begingroup$ @JohnBentin: You seem to have rolled back my edit to the OPs version and then edited that. I do not know why that happened, but you undid my formatting. From what you say, it wasn’t on purpose but perhaps some quirk of the interface. My apologies for attributing intent where none existed. $\endgroup$ – Arturo Magidin Jan 17 at 22:40
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    $\begingroup$ @AntonioMariaDiMauro A proof in ZF that $\kappa^2=\kappa$ for alephs is given in Hrbacek and Jech chapter 7 theorem 2.1 or Kunen Chapter 1 Theorem 10.2. $\endgroup$ – spaceisdarkgreen Jan 18 at 2:06
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The strategy for proving 11.7 is to show that if $\kappa^2=\kappa$ for every infinite cardinal $\kappa$, then any infinite cardinal is an aleph, since this implies that every set can be well ordered, which is equivalent to $\mathsf{AC}$.

In particular, by 11.6, to show that every infinite cardinal $\kappa$ is an aleph it suffices to show that $\kappa+\aleph(\kappa)=\kappa\ast\aleph(\kappa)$ for every $\kappa$.

The inequality $\kappa+\aleph(\kappa)\leq\kappa\ast\aleph(\kappa)$ is clear from the definitions: $\kappa+\aleph(\kappa)=|(\kappa\times\{0\})\cup(\aleph(\kappa)\times\{1\})|$, while $\kappa\ast\aleph(\kappa)=|\kappa\times\aleph(\kappa)$| and so there is a clear injection from the former to the latter, given by $(\gamma,0)\mapsto(\gamma,a)$ and $(\eta,1)\mapsto(\eta,b)$, where $a,b$ are arbitrary distinct elements of $\kappa$ and $\aleph(\kappa)$ respectively.

The chain of inequalies on the last line of the quoted text shows that $\kappa+\aleph(\kappa)\geq\kappa\ast\aleph(\kappa)$, which together with the previous inequality establishes $\kappa+\aleph(\kappa)=\kappa\ast\aleph(\kappa)$, and this is enough to obtain $\mathsf{AC}$ by the proof strategy I outlined at the beginning of this answer.

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  • $\begingroup$ Okay, and why the Choice Axiom implies that for every infinite cardinal $\kappa$ it results that $\kappa^2=\kappa$? $\endgroup$ – Antonio Maria Di Mauro Jan 17 at 22:54
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    $\begingroup$ $\mathsf{AC}$ implies that every cardinal is an aleph, and $\mathsf{ZF}$ already shows that $\aleph_\alpha=(\aleph_\alpha)^2$ for every $\alpha$. There's different ways to show this, for example one can show that Gödel's pairing function well orders $\kappa^2$ in order type $\kappa$. Anyway no offense meant but if your not familiar with cardinal arithmetic with choice it probably would be good to learn about that before working in $\mathsf{ZF}$, especially considering the book you're using. $\endgroup$ – Alessandro Codenotti Jan 17 at 22:59
  • $\begingroup$ I've studied that every cardinals are initial ordinals and fortunately the ordinals are well ordered sets and so it isn't necessary to invoke the Axiom of Choice to say that the ordinals can be well ordered, because we exactly know the well order for the ordinals; then I've studied that for definition an aleph is a cardinal ($\aleph_\alpha:=\omega_\alpha$). So why does AC implies that every cardinal is an aleph? Anyhow I've studied the cardinal arithmetic: I'm studing from "Teoria de Cojuntos, una introduccion" by Fernando Hernandez. $\endgroup$ – Antonio Maria Di Mauro Jan 18 at 0:13
  • $\begingroup$ Anyway I colud use this tautology: the AC implies that every cardinal (initial ordinal) are equipotent to an unique initial ordinal that for definition is an aleph and so using ZF it's possible to demonstrate that $\aleph_\alpha*\aleph_\alpha=\aleph_\alpha$; is this the correct reasoning? $\endgroup$ – Antonio Maria Di Mauro Jan 18 at 0:21
  • $\begingroup$ Cardinals are initial ordinals when assuming AC, in ZF that's not necessarily the case and that's where all the issues come from, you can't assume cardinals to be well ordered if you don't have choice. $\endgroup$ – Alessandro Codenotti Jan 18 at 0:24

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