Thank you so much for your answer! It seems to me that you meant $\alpha + \sigma = \alpha + \underset{\eta < \sigma}{\lim}\eta$ rather than $\underset{\eta < \sigma}{\lim}\alpha+\eta$, and $ \alpha + \underset{\eta \in A}{\lim}\eta$ rather than $\underset{\eta \in A}{\lim} \alpha + \eta$. On the basis of your answer, I have presented a detailed proof and posted it as an answer below. If you don't mind, please have a look at it. Thank you so much!

Thank to your elaboration, I got what that sentence's meaning. But I think the author's proof is not sufficient and not complete. Although he has proved that all ordinals less than $\sup\limits_{β∈A}(α+β)$ is also less than $α+\sup\limits_{β∈A}(β)$, he has done NOTHING in proving the converse direction that all ordinals less than $α+\sup\limits_{β∈A}(β)$ is also less than $\sup\limits_{β∈A}(α+β)$. As a result, he can not claim that Therefore the ordinals less than $\sup\limits_{β∈A}(α+β)$ and the ordinals less than $α+\sup\limits_{β∈A}(β)$ are the same. Thus he has not completed the proof.

@LeAnhDung, about the second comment, you are right! But the converse direction is easier, as you yourself noted. Maybe the author should have explicitely said that, anyway.

About the first comment, $\alpha + \sigma =\underset{\eta < \sigma}{\lim} \alpha + \eta$ by definition of ordinal addition for limit ordinals. (Of course, it is also equal to $\alpha + \underset{\eta < \sigma}{\lim} \eta$)

Honestly, I have not been able to prove the converse direction. I quite don't understand what you meant by But the converse direction is easier, as you yourself noted. Could you explain how to prove that all ordinals less than $α+\sup\limits_{β∈A}(β)$ is also less than $\sup\limits_{β∈A}(α+β)$?

I'll edit my answer.

Hello! Feel free to talk if you want to discuss the exercise

Thank you so much!

Although I have successfully proved the theorem, I do it by another approach, which is different than that of the author.

so, you are right, the converse direction is not the trivial one

right, I got that

But I'm still interested in the author's proof

you know, if you prove the converse inequality, i.e., RHS $\leq$ LHS, then automatically you have $RHS \subseteq LHS$

(RHS is Right Hand Side, and LHS is Left Hand Side)

Yeap, I got that. I'm trying to prove that

:)

oh, you'd like to take $\beta < \alpha + \sigma$

and work from there?

OMG I'm able to prove it

sorry, $\gamma < \alpha + \sigma$

cool!

In the same direction you suggested

great

Hi

I have finished typing my proof at math.stackexchange.com/a/2987245/368425. If you don't mind, please have a check on it. Thank you so much!

ok! in an hour I'll take a look at it

Best wish for you!