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I carry on following what is written at the 11th chapter of the book "The axiom of Choice" by Thomas J. Jech.

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

LEMMA 11.6

If $k$ is an infinite cardinal and $\aleph$ is an aleph, and if

(11.8) $\quad\quad\quad\quad\quad\quad\quad k+\aleph=k*\aleph$

then either $k\ge\aleph$ or $k\le\aleph$.

In particular, if

(11.9) $\quad\quad\quad\quad\quad\quad\quad k+\aleph(k)=k*\aleph(k)$

then $k$ is an aleph.

PROOF. Let $k=|K|$ and let $W$ be a well-ordered set such that $\aleph=|W|$. By (11.8), there exist two disjoint sets $K_1$ and $W_1$, such that $K\times W=K_1\cup W_1$ and $|K_1|=k$, $|W_1|=\aleph$. Either there exists $\mathsf k\in K$ such that $(\mathsf k,w)\in K_1$ for every a $w\in W$ and then $k\ge\aleph$ because $K_1\supseteq[(\mathsf k,w):w\in W]$. Or, for every $\mathsf k\in K$ let $w_{\mathsf k}$ be the least $w\in W$ such that $(\mathsf k,w)\in W_1$, and then $k\le\aleph$ because $[(\mathsf k,w_{\mathsf k}):\mathsf k\in K]\subseteq W_1$. In the particular case (11.9), $k\ge\aleph(k)$ is impossible, and $k\le\aleph$ implies that $k$ is an aleph.

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Well I don't understand why the fact that it is $k\le\aleph$ implies that $k$ is an aleph. Could someone explain to me this formally?

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If $k \le \aleph$ then there is an injection $k \to \aleph$, which well-orders $k$. Thus $k$ is an aleph number.

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  • $\begingroup$ Okay, infact somewhere at 2th chapter of the same text I read that "The well- orderable sets are equivalent to ordinal numbers and so we can define cardinals of well-ordered sets, using initial ordinals (i.e. those which are not equivalent to smaller ordinals). As is customary, $\aleph_\alpha=\omega_\alpha$ is the $\alpha^{th}$ infinite cardinal number". However in others books I read that "an aleph is a infinite initial ordinal": using this more formally definitions, how can I see that the fact that it is k≤ℵ implies that k is an aleph? $\endgroup$ Jan 14 '20 at 13:54
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    $\begingroup$ @AntonioMariaDiMauro: There is exactly one cardinal number for each bijection-class of sets, and every well-orderable set is in bijection with a unique initial ordinal. Putting these facts together: $k$ is well-orderable, so $k \cong \omega_{\alpha}$ for some ordinal $\alpha$, and so $k=\omega_{\alpha}~{(=\aleph_{\alpha})}$ since $k$ is a cardinal number. $\endgroup$ Jan 14 '20 at 14:24

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