0

Let $R$ be a relation on a class $X$ then a path in $(X,R)$ is a function $p:n+1\to X,n\geq 1$ such that $p(i)Rp(i+1)$. Given a subset $Y\subseteq X$, is the class $Z$ of all $x\in X$ such that there is a path $p$ in $(X,R)$ connecting some elements in $Y$ and $x\in\text{ran}(p)$ a set?

My question is related to this post about the well-foundedness of the transitive closure of a well-founded relation. The OP does not specify if $R$ is a relation on a set or at least set-like. Then I do not see how the $B$ in the proof by hmakholm has an $R$-minimal element, given a relation is well-founded iff every subset has a minimal Element (check Endertons definition, which OP is refering to). If $R$ is a relation on a set, then sure; if $R$ is at least set-like, then every subclass has a least element, and so it does not really matter if $B$ is a set.

For example: if $X=Ord$ then for every subset $Y\subseteq X$ the images of such paths lie in $\bigcup Y+1=U$ and so by replacement, powerset and union $O=\bigcup\{\mathcal{P}(n\times U):n\in\omega\}$ is a set and so $T=\{p\in O:p\text{ is a path in }(X,R)\land p(0),p(n)\in Y\}$ is a set by comprehension and finaly $Z=\bigcup\{\text{ran}(p):p\in T\}$ is a set by replacement and union. Hence for $Z$ in the original question to be a set, it would be enough, if the images of the paths lie in some subset.

Now this post is claiming that if $R$ is well-founded on $X$ then every subclass has a least element, but B. M. Scott is using $\text{rank}$ defined by the Von-Neumann-Hierarchy in his answer and so implicitly assuming foundation. In this post E. Wofsey remarks that using choice, one can construct a sequence, where well-foundedness fails for a set. So I am assuming for the well-founded property on arbitrary subclasses, you need one of the following conditions: 1) $R$ is a relation on a set 2) $R$ is set-like 3) foundation 4) choice.

This post became longer than I anticipated, so feel free to adress any of the mentioned stuff. To summarize my questions:

  1. Is the class $Z$ in the first question a set?
  2. Are there other sufficient condicitons for the well-foundedness on arbitrary subclasses?
Lukas Chlebovec
  • 59
  • Let $X=Ord\cup{x}$ for some set $x\not\in Ord$ and $(\alpha,\beta)\in R$ iff $(\alpha\in Ord\land \beta=x)\lor(\alpha=x\land\beta\in Ord)$, then the class $Z$ of Elements of paths conencting elements in ${x}$ is $Z=X$, since $xR\alpha Rx$ is a path for every $\alpha\in Ord$. But this relation is not well-founded, nor set-like, since ${y\in X:yRx}=Ord$ and ${x,\alpha}$ does not have a least element, so generally no, but it still does not answer the question. Although given Enderton states transfinite recursion with sets, I assume the statement in the post assumes $R$ is a set-relation. – Lukas Chlebovec Jun 04 '25 at 06:55

1 Answers1

0

Now that I gave it a bit thought, I think I already answered my questions in the post. In general, $Z$ is not a set. I tried to get that if for each subset $Y$ the set $Z$ is a set, then $R$ is set-like, but I do not think that works, because it only says something about elements in between and not underneath.

Lukas Chlebovec
  • 59