# If $\lambda=\omega_\overline\alpha$, $k_{\alpha<\lambda}\neq0$ and $k=\sup k_{\alpha<\lambda}$ then $\sum_{\alpha<\lambda}k_\alpha=k*\lambda$?

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?

• Sorry, could you explain better? Jan 14 '20 at 20:31
• Why? Sorry but I don't understand. Jan 14 '20 at 21:44

• 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? Jan 14 '20 at 20:47