## Exercises in Set Theory and Teen Reminiscences

By fredw on Monday, November 19 2012, 19:55 - Permalink

I recently spent some time to solve some exercises from Thomas Jech's book "Set Theory" and to typeset my solutions. I'm now done with the six first chapters, that is half of the first part of the book... entitled "Basic Set Theory". Well, I'll probably have to finish the second half to be able to claim that it is "basic" and then I'll still have a lot to study in the two other more advanced parts ;-) New exercises are essentially for chapter 4, as I was already almost done with the other chapters before. That was good to do some topology again! (without using the axiom of choice!)

It was also funny to solve exercise 5.4: if a set $A$ can be well-ordered then $P(A)$ can be totally ordered. It recalls me when I was 13-14 and was trying to well-order sets constructed from other well-ordered sets. Of course (*), I was stuck on how to build a well-ordering on the powerset of a well-ordered set but I had however been able to construct the linear order described in exercise 5.4. Nowadays, my old proof looks overcomplicated although it could probably be simplified if one uses the symmetric difference and relies on well-known set properties. My most recent proof is much simpler but still not really straightforward as I do not see other ways than testing each case. If someone has a better solution, please post a suggestion...

Here are the links to each chapter:

- Chapter 1: Axioms of Set Theory
- Chapter 2: Ordinal Numbers
- Chapter 3: Cardinal Numbers
- Chapter 4: Real Numbers
- Chapter 5: The Axiom of Choices and Cardinal Arithmetic
- Chapter 6: The Axiom of Regularity

(*) Some years later, someone congratulated me by email for my wonderful proof of how to well-order $P(A)$ from a well-ordering on $A$. However, my proof had never been finished and will remain uncomplete. If we can prove in ZF that for any well-orderable set $A$, its powerset $P(A)$ is well-orderable then it is easy to prove by ordinal induction that all ${V}_{\alpha}$'s are well-orderable and thus the theorem of Zermelo, which is equivalent to the axiom of choice. But this one is well known to be independent from the other axioms of ZF...

## Comments

Alternative solution to exercise 5.4, using the same notations. We have $X\u25b5Y=(X\u25b5Z)\u25b5(Z\u25b5Y)$, so ${m}_{1}\ge {m}_{3}\vee {m}_{1}\ge {m}_{2}$. Similarly, $Y\u25b5Z=(Y\u25b5X)\u25b5(X\u25b5Z)$, so ${m}_{2}\ge {m}_{1}\vee {m}_{2}\ge {m}_{3}$. Hence we have $({m}_{1}\ge {m}_{3}\vee {m}_{1}\ge {m}_{2})\wedge ({m}_{2}\ge {m}_{1}\vee {m}_{2}\ge {m}_{3})$ which implies ${m}_{1}={m}_{2}$ or ${m}_{1},{m}_{2}\ge {m}_{3}$. The first case is not possible because ${m}_{1}\notin Y$ and ${m}_{2}\in Y$. If we assume that ${m}_{3}\in Z$ then because ${m}_{3}\le {m}_{2}$ we get ${m}_{3}\in Y$ and because ${m}_{3}\le {m}_{1}$ we get ${m}_{3}\in X$. A contradiction.