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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 E. 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
up vote 12 down vote accepted

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.

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Nice spolier. May I suggest in future less explicit answers for text exercises so the poster has more to do? e.g. "See if you can find a set-like class relation R so that R* will yield a set containing the desired set in the replacement axiom; for starters, you could have R(x,y) for every x in X such that there is a unique y so that phi(x,y,z), where phi is the formula used in the replacement axiom with some parameter z; what else would keep R set-like and have R* help you to show (y | there is x in X such that phi(x,y,z) ) is a set?" Gerhard "Ask Me About System Design" Paseman, 2012.05.05 – Gerhard Paseman May 6 '12 at 0:05
I don't agree with that perspective, Gerhard. (See for a further explanation of my views.) If someone asks a good math question, as here, then we should simply try to give the best possible answer. Math.SE in particular has numerous instances of totally unsatisifying half-answers to questions, in the style you suggest, but I find them simply to be poor answers. Let a student try hard on his or her own, but when they come for assistance, let us all try to give the best answer we possibly can, mustering whatever elegance and clarity we are able. – Joel David Hamkins May 6 '12 at 0:17
I will visit the meta thread and suggest an alternate approach there. Thank you for the response Joel. Gerhard "Not Yet Agreeing To Disagree" Paseman, 2012.05.05 – Gerhard Paseman May 6 '12 at 1:04
@ Professor Hamkins: Thank you for taking your time to answer my question. You have been most helpful. – Ali Kare May 6 '12 at 8:58
@ Prof. Paseman: Sometimes it is a good student who already knows the topic but nevertheless gets stuck at an easy problem. It happens. I've already worked on it long enough and unfortunately I do not have any more time to spare. So please do not prolong my misery by subjecting me to the socratic method. It is already embaressing for me to ask such a simple question. What I need is to be shown the answer so that I can see what I missed and move on. I guarentee you that I will learn from the answer given, especially because I am upset that I couldn't think of it myself. – Ali Kare May 6 '12 at 9:00

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