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