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Just / Weese contains the following theorem (p 126):

Theorem 5: Let $\mathbf Z \neq \varnothing$ and let $\mathbf W \subseteq \mathbf Z \times \mathbf Z$ be a wellfounded set-like class. Let $\mathbf G$ be a functional class such that $dom(\mathbf G)$ consists of all pairs of the form $\langle f,z \rangle$, where $z \in \mathbf Z$ and $f$ is a function with domain $I_{\mathbf W}(z)$. Then there exists exactly one functional class $\mathbf F : ]\mathbf Z \to \mathbf V$ such that $$ \mathbf F (z) = \mathbf G ( \mathbf F \mid I_{\mathbf W} (z), z) \hspace{0.5 cm} \forall z \in \mathbf Z$$

On the same page before the theorem they write "...In Chapter 12 we shall see that the existence of a definable, set-like wellorder of the universe is relatively consistent with $ZFC$. ..."

I am not aware of any other places in the book where the word "set-like" is used, in particular not a definition of what it means. Could someone give me the definition of "set-like" class and explain to me what it means? Many thanks for your help.

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It means every initial segment is a set. – Asaf Karagila Dec 22 '12 at 14:36
@Amzoti: How does a monograph about mathematics without infinity helpful to this question? – Asaf Karagila Dec 22 '12 at 15:30
up vote 7 down vote accepted

I believe in this instance it should be "set-like well-founded relational class" or something like that. The point is that for each $x \in \mathbf{Z}$ the class $\{ z \in \mathbf{Z} : z \mathrel{\mathbf{W}} x \}$ is actually a set.

Indeed, on p.125 Just-Weese give the following defintion:

A relational class $\mathbf{W}$ is said to be set-like if $$ (\forall x) (\exists y) ( y = \{ z : \langle z,x \rangle \in \mathbf{W} \} ).$$

Notice that this is precisely what you need in order to assert the existence of such a function(al class) $\mathbf{G}$ in the hypothesis of the theorem. If $I_{\mathbf{W}} ( z )$ were a proper class for some $z \in \mathbf{Z}$, then there could be no pairs of the form $\langle f , z \rangle$ where $\mathrm{dom} ( f ) = I_{\mathbf{W}} (z)$, since $f$ must be a set, and so its domain must be a set, which it cannot be. (Recall that only sets have real existence in ZF(C)). Also, the function(al class) $\mathbf{F}$ that the theorem asserts to exist similarly could not exist because it could not be defined at such a $z$.

For an example of a well-founded relational class which fails to be set-like, consider $\mathbf{W} \subseteq \mathbf{On} \times \mathbf{On}$ defined by $$\langle \alpha , \beta \rangle \in \mathbf{W} \quad \Longleftrightarrow \quad \begin{cases} 0 < \alpha < \beta, &\text{or}\\ \beta = 0 \wedge \alpha \neq 0. \end{cases}$$ This is a "well-ordering" of $\mathbf{On}$ which makes $0$ the unique maximum element (but otherwise retains the usual ordering of $\mathbf{On}$). Clearly $I_{\mathbf{W}} ( 0 ) = \mathbf{On} \setminus \{ 0 \}$ is not a set.

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Yay! I thought you might be on vacation because I didn't get any answers from you for a while. Thank you for this nice answer. : ) – Rudy the Reindeer Dec 22 '12 at 19:42

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