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

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The definition of an ordinal in the lecture notes is as follows: enter image description here

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

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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!


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Ansar
    Commented Feb 11, 2023 at 22:15
  • $\begingroup$ @Ansar Thanks for your comment. In that case, does my proof make sense to you? $\endgroup$ Commented Feb 12, 2023 at 7:30
  • $\begingroup$ it looks correct to me. $\endgroup$
    – Ansar
    Commented Feb 12, 2023 at 10:55


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