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I carry on following what is written at the 9th chapter of "Introduction to Set Theory" by Karel Hrbacek and Thomas Jech

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1.3 Theorem

Let $\lambda$ an infinite cardinal, let $k_{\alpha<\lambda}$ be nonzero cardinal numbers, and let $k=sup[k_{\alpha<\lambda}]$. Then

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \sum_{\alpha<\lambda}k_\alpha=k*\lambda$.

Proof. On the one hand, $k_\alpha\le k$ for each $\alpha<\lambda$, and so $\sum_{\alpha<\lambda}k_\alpha\le\sum_{\alpha<\lambda}k=k*\lambda$. On the other hand, we notice that $\lambda=\sum_{\alpha<\lambda}1\le\sum_{\alpha<\lambda}k_\alpha.$ We also have $k\le\sum_{\alpha<\lambda}k_\alpha$: the sum $\sum_{\alpha<\lambda}k_\alpha$ is an upper bound of the $k_\alpha$'s and $k$ is the least upper bound. Now since both $k$ and $\lambda$ are $\le\sum_{\alpha<\lambda}k_\alpha$, it follows that $k*\lambda$, which is the greater of the two, is also $\le\sum_{\alpha<\lambda}k_\alpha$. The conclusion of Theorem 1.3 is now a consequence of the Cantor-Bernstein Theorem.

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Well I don't understand why since both $k$ and $\lambda$ are $\le\sum_{\alpha<\lambda}k_\alpha$, it follows that $k*\lambda$, which is the greater of the two, is also $\le\sum_{\alpha<\lambda}k_\alpha$. Could someone explain to me this formally?

Anyway it seems to me that it could be possible to prove it in this way: since $\lambda$ is an infinite cardinal, $k$ must have this property and then it's $k*\lambda=k+\lambda \le\sum_{\alpha<\lambda}k_\alpha+\sum_{\alpha<\lambda}k_\alpha=\sum_{\alpha<\lambda}k_\alpha$; is it correct?

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Exercise.
Show if a and b are infinite cardinals, then a×b = max{a,b}.
That will answer your question.
This his comment "which is the greater of the two."

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  • $\begingroup$ Of course; I already knew this: if $\alpha$ and $\beta$ are infinite cardinals then $\alpha*\beta$=$max${$\alpha,\beta$}=$\alpha+\beta$ so the results is immediate considering that $\lambda$, $k$ and $\sum_{\alpha<\lambda}k_\alpha$ are infinite cardinals: infact it's result $k*\lambda=k+\lambda\le\sum_{\alpha<\lambda}k_\alpha+\sum_{\alpha\lambda}k_\alpha=\sum_{\alpha<\lambda}k_\alpha$; or as you said $k*\lambda=$$max${$k,\lambda$}$\le\sum_{\alpha<\lambda}k_\alpha$. Maybe I was wrong? $\endgroup$ Jan 14 '20 at 20:47
  • $\begingroup$ You are right, a+b = the larger. You way requires extra, unneeded steps. $\endgroup$ Jan 14 '20 at 21:47

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