Let $\beta=\sup\limits_{\nu<\gamma}\beta_\nu$. How do the authors prove $\alpha+\beta=\sup\limits_{\nu<\gamma}(\alpha+\beta_\nu)$?

48 Views Asked by At

Here is an excerpt from my textbook Introduction to Set Theory by Hrbacek and Jech. Although I'm able to prove these three equalities, my proof is up to a page long containing even a lemma (If you are interested, you can find it here).

enter image description here

I retype here.

First we observe that the ordinal functions $\alpha+\beta,\alpha\cdot\beta$, and $\alpha^\beta$ are continuous in the second variable: If $\gamma$ is a limit ordinal and $\beta=\sup\limits_{\nu<\gamma}\beta_\nu$, then

$$\alpha+\beta=\sup\limits_{\nu<\gamma}(\alpha+\beta_\nu), \alpha\cdot\beta=\sup\limits_{\nu<\gamma}(\alpha\cdot\beta_\nu), \alpha^\beta=\sup\limits_{\nu<\gamma}(\alpha^{\beta_\nu})$$

This follows directly from Definitions 5.1, 5.6, and 5.9.

I add Definitions 5.1, 5.6, and 5.9 here for reference.

enter image description here

enter image description here

enter image description here

I can not understand why the authors say This follows directly.... From my proof, I don't see it directly at all.

Could you please elaborate/explain what authors want to communicate by the sentence This follows directly from Definitions 5.1, 5.6, and 5.9?

Thank you so much for your help!