# An exercise in Kunen. Getting Axiom of Replacement from set-like transitive closure.

I am studying Kunen's Set Theory (2011 edition) on my own. I got stuck at the excercise I.9.6 which is:

Excercise I.9.6. Derive the axiom of replacement from lemma I.9.5.

And the mentioned lemma is this:

Lemma I.9.5. For a relation R and a class A, if R is set-like on A, then R* is set-like on A.

Here R* is the transitive closure of R. Also, a relation R is set-like on a class A if { x\in A : xRy } is a set for all y\in A.

Help appreciated.

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I have to ask, is there a difference between the 2011 edition and the old edition? – Asaf Karagila May 5 '12 at 22:57
@Asaf: Yes. Quite a few. – Andrés Caicedo May 6 '12 at 2:11
The 2011 version contains a lot more. Also the approach is a bit different. In the new version Kunen does not hesitate to use model theory and topology to get results. This condenses some parts of the old book. Still, the new book is 75 pages longer. This means that there is a lot of new material. Especially, the chapter on infinitary combinatorics contains much more (including sections on small cardinals and elementary submodels). The iterated forcing chapter contains a section on proper forcing. – Ali Kare May 6 '12 at 9:03
Also, in the new version the exercises are not collected at the chapter ends but rather sprinkled throughout the text. I find this to be better. Solving exercises at the spot facilitates learning. Also, facing a long list of exercises at the chapter end can be a bit daunting. (I wish Jech didn't move all exercises to the chapter ends in the third edition. What was he thinking?) – Ali Kare May 6 '12 at 9:04

Suppose that the lemma holds and that we are considering an instance of the replacement axiom, so we have a set $X$ and for some parameter $z$ and for every $x\in X$ there is a unique $y$ such that $\varphi(x,y,z)$. Fix any set $w$ not in $X$, and let $R$ be the class relation such that $R(x,w)$ for each $x\in X$, and such that $R(y,x)$ whenever $\varphi(x,y,z)$. That is, the children of $w$ are exactly the members of $X$, and the child of any $x\in X$ is precisely the corresponding $y$. Thus, the relation $R$ is set-like, since $X$ is a set, and $\{y\}$ is a set for each $y$ that arises. But the transitive closure of $R$ will relate all the $y$'s that arise from any $x\in X$ to $w$. And so if the transitive closure of $R$ is set-like, then the set $\{y\mid \exists x\in X\, \varphi(x,y,z)\}$ will be a set, thereby verifying this instance of the replacement axiom.