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:
- How can I show that $\gamma = \bigcup X$ is the least ordinal with this property?
- 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.