7
$\begingroup$

My question is whether Halmos' book "Naive Set Theory" is still up-to-date concerning contemporary mathematics; that is, is it outdated or not?

I really love the book so far, and while it's clear the notation is outdated, so far nothing contradicts what we had in the lectures. My assistant in calculus warned me about the book not being the newest though, so I'm a little confused right now.

Can I safely continue reading it without learning anything wrong or should I look for another book? If the latter is the case, can anyone of you give me a suggestion which book would be better suited?

Thanks in advance!

Sincerely, SDV

$\endgroup$
2
  • 1
    $\begingroup$ I haven't read it, or thoroughly reviewed it (with the exception of several questions about the Zorn-choice equivalence, as I remarked below Ittay's answer); but I can say that I looked over Enderton's book and it looks really nice. If you want to learn to read Hebrew, Azriel Levy has excellent notes in Hebrew about naive set theory (unrelated to his book which is a bit more advanced). In any case, the fundamentals of set theory haven't change since Halmos wrote the book, so you don't need to worry about learning wrong things. $\endgroup$
    – Asaf Karagila
    Oct 19, 2014 at 15:23
  • $\begingroup$ The book becomes aa bit difficult for self study at a point ( because explanations are too concise). . I've found Enderton 's book more reader friendly. ( Also, for a complete beginning, you may have a look at Seymour Lipschutz' Theory And Problems Of Set Theory" , Schaum's series, available at archive.org. The author makes concepts very clear.). $\endgroup$
    – user655689
    Mar 25, 2020 at 18:04

1 Answer 1

7
$\begingroup$

It's still an excellent book. You will certainly not learn anything wrong by reading it. The notation is quite the standard one and the approach is crystal clear and highly relevant. The only negative thing I have to say about the book is about its proof of Zorn's Lemma from the Axiom of Choice. It's technically quite involved to my taste, especially considering that there are other very elegant proofs. For instance, the proof given in "Sets, Models and Proofs" by Moerdijk and van Oosten, which is generally a good supplement to Halmos' Naive Set Theory.

$\endgroup$
8
  • 1
    $\begingroup$ I haven't read Halmos' book. I do know the Zorn-choice equivalence from the myriad of questions on this site regarding the proof. It's quite horrible indeed. It's like insisting to use only nails to build a house, when there are ready piles of lumber, cinder blocks and cement. $\endgroup$
    – Asaf Karagila
    Oct 19, 2014 at 15:21
  • $\begingroup$ It does use one notation that I consider old-fashioned: $A-B$ for $A\setminus B$. $\endgroup$ Oct 19, 2014 at 19:29
  • $\begingroup$ What's wrong with $A-B$? $\endgroup$
    – Nishant
    Oct 19, 2014 at 20:32
  • $\begingroup$ When $A,B$ are sets of, e.g., numbers, then it is natural to write $A+B=\{a+b\mid a\in A,b\in B\}$. So, @Nishant, what would $A-B$ mean? $\endgroup$ Oct 19, 2014 at 20:36
  • 1
    $\begingroup$ Hmm, that's a good point. The reason I don't like $A\setminus B$ is because I feel like I may write it wrong and/or confuse it with a quotient of $A$ by $B$... $\endgroup$
    – Nishant
    Oct 19, 2014 at 20:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .