I was playing around with a theorem in combinatorics about finite sets and was trying to find a meaningful generalization of it to non-finite sets. At one point without thinking I wrote that:

$$\text{ The power-set of }X=\mathcal{P}(X)=\bigcup_{n=0}^{\infty}\{S\subseteq X:|S|=n\}$$

Then about five seconds later I realized it was false, because my set $X$ might contain a subset whose cardinality is not a natural number. So I tried to salvage this partially, using my essentially zero knowledge on ordinal numbers and more advanced set theory, basically I wrote something akin to: $$\mathcal{P}(X)=\bigcup_{\substack{\gamma\leq |X|\\\gamma\text{ a cardinal number}}}\{S\subseteq X:|S|=\gamma\}$$

However I'm not even sure if this makes sense. To make it formal I would have to reform that lower index to something like "the set of cardinals", but I don't even know if that's a set?? I'm starting to feel really stupid, I mean throughout most of my time studying I've never really made use of any sets with cardinality say greater then $|\mathbb{R}|$ maybe $|\mathcal{P}(\mathbb{R})|$ when working with function spaces, though even then I never really made use of it.

Also I've tried to play around with stuff like this before, where I see something cool and want to make a generalization for non-finite sets and sometimes I can manage by manipulating things weirdly and finding bijections to establish equinumerosity. But I imagine if I had some more tools up my sleeve like a familiarity with cardinal arithmetic etc. I could probably do this much faster and likely find results I couldn't have got before because I lacked the ability to manipulate sets with arbitrary cardinality in a nice way.

So in short can someone recommend me some reading on "advanced set theory" (no idea what to call it, just want to make sure its not a book on naive set theory etc. which I'm fine with).

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    $\begingroup$ Why can’t you just say that the power set is the set of all subsets and be done with it? $\endgroup$ – Randall Jan 7 '18 at 3:50
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    $\begingroup$ Say I have a set $X$ equipped with an equivalence relation $\sim$ and I want to find all the permutations $\sigma\in \text{Sym}(X)$ such that $x\sim y\iff \sigma(x)\sim \sigma(y)$. Now these permutations form a group under composition which can be explicitly written out as a wreath product BUT the expression involves pairing up all the sets $S\in X/\sim$ by their cardinality. $\endgroup$ – user3865391 Jan 7 '18 at 3:53
  • $\begingroup$ @user3865391 en.wikipedia.org/wiki/Axiom_of_power_set $\endgroup$ – David Reed Jan 7 '18 at 4:02
  • $\begingroup$ @DavidReed The powerset was just an example. On occasion I find it useful to partition out sets by their cardinality, like in my previous example involving $\sim$ preserving automorphisms. More generally when working with say infinite permutation groups if one finds a block system en.wikipedia.org/wiki/Block_(permutation_group_theory) and wants to separate out the blocks. Breaking up all the blocks into subsets by cardinality seems like a good first step to prune out what is being interchanged/swapped by the group action. Or at least thats what I might do, I don't know I'm no expert. $\endgroup$ – user3865391 Jan 7 '18 at 4:06
  • $\begingroup$ @user3865391 Ok. I was also thinking you could hit some of the links from that page to ZFC and axiomatic set theory to get a taste of it. Really the entire subject requires tiresome development which is why I limited my response to references. To answer your other question, the class of all cardinal numbers is not a set. (i.e. the set doesn't exist $\endgroup$ – David Reed Jan 7 '18 at 4:10

Axiomatic Set Theory is the term you are looking for. Technically speaking you should really make sure you have a strong background in first-order logic first, as ZFC(Zermelo-Frankel Set Theory with Choice-the "standard" set theory construction) is formulated in FOL. However, you could probably get away without it if you are familiar with the basics of quantifiers and logical symbols and are just looking to take the conceptual approach. If that's the case a good one would be Axiomatic Set Theory by Suppes. A good intro to FOL book is "Computability and Logic" by Boolos. Alternatively, if you search "Axiomatic Set Theory" on amazon a bunch of books will come up that you can read user reviews of.


Can someone recommend me some reading on "advanced set theory"

There are many detailed recommendations of books on (non naive) set theory -- entry level books in §4.3, and the whole of §7 on rather more advanced books -- in the Teach Yourself Logic Study Guide. There should be enough description of the level/coverage of the various books for you to find what you need.

  • $\begingroup$ That's a nice (though somewhat intimidating) guide you've written there! $\endgroup$ – Stefan Mesken Jan 8 '18 at 16:28

Your second idea is workable, and we do have to avoid the problem you noticed, namely that there is no set of all cardinals. Note that there is a $1$-parameter sentence $C$ over ZFC such that $C(x)$ says "$x$ is a cardinal", which I think you know. Now we have the following:

Let $F = \{ T : T \in P(P(X)) \land T=\{ S : S \in P(X) \land |S|=k \} \land C(k) \land k \le |X| \}$.

Then $F = \{ \{ S : S \in P(X) \land |S|=k \} : C(k) \land k \le |X| \}$.

Thus $\bigcup F = P(X)$.

Some people recommend Kunen's "Set Theory", which would definitely be a rigorous textbook on ZFC set theory, though I have not read it myself. If you are interested in a concise motivation and explanation of ordinals and cardinals in ZFC set theory, you can also read this post.

  • $\begingroup$ How does this answer the question? $\endgroup$ – user170039 Jan 7 '18 at 16:30
  • $\begingroup$ I am answering the mathematical part of the question, which the asker actually asked me directly. Note that the title and tags were changed by someone other than the asker. Furthermore, the asker was satisfied with my answer so that's what matters. $\endgroup$ – user21820 Jan 7 '18 at 17:10
  • $\begingroup$ I think that in that case it will be better to include that information in this post that you are only answering the mathematical part of the question asked to you directly in a comment (and most probably in the Logic room) and not responding to the last line of the post. Otherwise, it still doesn't answer the question stated in the post. $\endgroup$ – user170039 Jan 8 '18 at 13:40
  • $\begingroup$ @user170039: You see fit to criticize my post for not answering the last line of the question. I agree that I did not do so. But so did the top-voted answer as of now. Nor did Carsten S's answer. You did not criticize those posts. Will you apply your standards consistently? $\endgroup$ – user21820 Jan 8 '18 at 14:57
  • $\begingroup$ First of all, I did not criticize your post, only pointed out that your post didn't answer the question as stated in the post. Second, you are wrong, that the top voted answer did mention Suppes's book (read David Read's answer again) and as for Carsten S's answer he did explicitly say at the beginning of the question, "[e]ven though I do not know of a book to recommend, I feel compelled to answer,.." So it is evident (at least for me) that he/she wasn't aiming at answering the question. $\endgroup$ – user170039 Jan 9 '18 at 3:26

Even though I do not know of a book to recommend, I feel compelled to answer, because I think that despite of the current title of the question you do not want an introduction to axiomatic set theory, but a primer in sets and cardinalities for the working mathematician.

You are right that the class of all cardinal numbers is not a set, however all cardinals below a given one certainly form a set, so there is no problem here.

Just try to read a bit about cardinal arithmetic somewhere, and go from there.

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    $\begingroup$ Note that the title was not given to this question by the OP. $\endgroup$ – Asaf Karagila Jan 7 '18 at 10:15

I recommend Set Theory by Kenneth Kunen:


It is not the most easiest book, but it's well written and cheap.

  • $\begingroup$ "It is not the most easiest book" I'd say, among books dealing with advanced set theory, this is one of the easiest to read. But that could very well be due to personal preference -- it certainly is detailed and my understanding relies on getting the details right. $\endgroup$ – Stefan Mesken Jan 8 '18 at 16:57

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