Existence of supremum of a set of ordinals

I'm trying to understand why the statement in yellow below is true.

The definition of an ordinal in the lecture notes is as follows:

I am also adding Theorem 29, Lemma 33 and Lemma 34 for reference:

This is what I got so far. If $$X$$ is a set of ordinals, then there is an ordinal $$\gamma = \bigcup X$$ such that $$\beta \leq \gamma$$ for all $$\beta \in X$$. To see this let $$\beta \in X$$ and assume that $$\gamma \in \beta$$. Then $$\gamma \in \bigcup X = \gamma$$ by the definition of the union of a set. This is a contradiction, and so by Theorem 29, $$\beta \leq \gamma$$ for all $$\beta \in X$$.

Now my questions are the following:

1. How can I show that $$\gamma = \bigcup X$$ is the least ordinal with this property?
2. Why do the lecture notes claim that the statement follows from Lemma 35 (the only part I've used is that $$\gamma$$ is an ordinal).

Thank you very much!

Edit:

I think I've found the solution for 1). As shown above $$\gamma$$ is an "upper bound" for $$X$$, i.e. $$\beta \leq \gamma$$ for all $$\beta \in X$$. Now let $$\alpha < \gamma$$, then $$\alpha \in \gamma = \bigcup X$$ by definition of $$<$$. So by the definition of the union of a set, $$\alpha \in y$$ for some $$y \in X$$. This means that $$y \not \leq \alpha$$, and hence $$\alpha$$ cannot be the supremum of $$X$$. It follows that $$\gamma$$ is the supremum.

I would appreciate if someone could confirm this is correct.

• Yeah, saying that the existence of the supremum follows from the lemma seems odd. Maybe they meant to say that it follows from the proof of the lemma - this would make more sense. Commented Feb 11, 2023 at 22:15
• @Ansar Thanks for your comment. In that case, does my proof make sense to you? Commented Feb 12, 2023 at 7:30
• it looks correct to me. Commented Feb 12, 2023 at 10:55