# If $k$ is a infinite cardinal and if $\aleph(k)$ is a Hartog's number of $k$ and if $k+\aleph(k)=k*\aleph(k)$, is $k$ an aleph?

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?

If $$k \le \aleph$$ then there is an injection $$k \to \aleph$$, which well-orders $$k$$. Thus $$k$$ is an aleph number.
• 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? Jan 14 '20 at 13:54
• @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. Jan 14 '20 at 14:24