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I'm reading Introduction to Modern Set Theory by Judith Roitman, and I'm having some trouble with the section on the axiom of choice. In particular, I'm confused about a step in the proof that Well Ordering implies Zorn's Lemma. I think my problem boils down to not understand how to define functions recursively over the ordinals. Rather than repeat the proof, I'll just start with my confusion.

The setup: We have a partial order $P$ and by Well Ordering we have written $P$ as a sequence of elements indexed by ordinals, $P = \{ p_\beta \vert \beta < \alpha \}$ for some ordinal $\alpha$. We wish to define a function $f(\beta)$. We suppose we know the function for $\gamma < \beta$, which is the same as saying we know the image $f[\beta]$ since $\gamma \in \beta$ when $\gamma < \beta$. We then want to define $f(\beta)$.

The question: Roitman defines $f(\beta) = p_\delta$ where $\delta$ is the least ordinal such that $p_\delta > p$ for all $p \in f[\beta]$. How do we know that such a $\delta$ exists?

My approach: Form the collection $\{\delta \vert p_\delta > p \forall p \in f[\beta] \}$. If this is a set of ordinals, then we can take the minimum. But, how we do know this is a set? I can't use separation because I am not taking the $\delta$ from another set, i.e. I don't have $\{\delta \in \alpha \vert \cdots \}$. Or does the construction go through even if $\{\delta \vert \cdots \}$ is a class and not a set?

Thanks for any help you can provide.

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You can use seperation in exactly the way you suggested! Notice that $p_\delta$ is only defined for $\delta<\alpha$ and hence $$\{\delta\mid \forall p\in f[\beta]\ p_\delta>p\}=\{\delta\in\alpha\mid \forall p\in f[\beta]\ p_\delta>p\}$$ is a set. In any case, it is not necessary for this particular construction that this is a set, as you have a canonical way of choosing an element, namely taking the minimum. To take the minimum however, this set must not be empty. At some point, all possible $\delta$ will be exhausted, so eventually this set is empty for some $\beta$ and you cannot prolong this construction. In that case, you have to invoke the asumption of $P$ being inductive to complete the proof.

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  • $\begingroup$ Ahh. That makes sense now that you tell me. Of course $\delta < \alpha$ and I can use $\{ \delta \in \alpha \vert \cdots \}$. Roitman starts the proof by defining $f(\beta)=p_0$. Does this get around the non-empty problem? Also, I'm not sure what you mean by "invoke the assumption that $P$ is inductive." $\endgroup$ – Robert Singleton Mar 1 at 15:55
  • $\begingroup$ The set being nonempty is actually not a problem, you want this to happen! What you are trying to do is finding a maximal element of $P$. If at some point $\beta$, $f(\beta)$ is a maximal element, it is impossible to find $f(\beta+1)$! So if this set is empty, the hope is that you have come across a maximal element of $P$. If you have not, then the range of $f$ would be an increasing chain in $P$ without an upper bound, which contradicts the assumption of $P$ being inductive. $\endgroup$ – Andreas Lietz Mar 1 at 16:01
  • $\begingroup$ Ok. That makes sense. BTW, I should have written $f(0)=p_0$ above. Thanks again. In fact, now that I think about it, I've been confused about this very point. I just didn't know how to articulate the question. $\endgroup$ – Robert Singleton Mar 1 at 16:05
  • $\begingroup$ @AdreasLietz Does this mean that showing $f(\beta + 1)$ is empty is part of the proof/definition? Or does it follow trivially from choosing $\delta$ to be the least ordinal such that some condition is met (in this case $p_\delta > p$)? $\endgroup$ – Robert Singleton Mar 5 at 16:20

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