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This is a fairly soft, general question, but whilst attending a lecture on basic set theory, inevitably Russell's Paradox was brought up. The speaker used the analogy of two books, one which contains all books which contain themselves, and one which contains all the books which do not, the secondly obviously acting as the analogy for the paradox.

In hindsight I began to think about the other book talked about, which in formal set theory (I believe) would be defined:

$$A=\{B \ | \ B \in B\}$$

The statement: $A \in A$ clearly also has no solution, since there is no contradiction either way (if $A \in A$ then it is true because it should be; if $A \notin A$ then clearly it is also ok because it shouldn't be).

Obviously no huge mathematical issues are raised here (hence why the other example has much more press), however I was wondering if there is any significance to the fact that this statement has two (tautological) solutions, either of which can be taken as valid (however not both).

Apologies if this question is too broad, vague, or simply meaningless, and all input is much appreciated!

[NB: I'm not sure if I'm quite using non-decidable in its correct formal usage; in this question what I mean by it is a statement for which the truth value can't be ascertained.]

aidangallagher4
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  • I guess I was more thinking $\subset$ was the one to use oh well. –  Sep 15 '17 at 21:08
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    @RoddyMacPhee I believe it is defined using $\in$ – aidangallagher4 Sep 15 '17 at 21:10
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    @RoddyMacPhee It's definitely $\in$. Think about it: the set of all sets which aren't subsets of themselves is just $\emptyset$, since every set is a subset of itself; and since $\emptyset$ is a subset of itself, there's no demand to have $\emptyset\in\emptyset$, so there's no contradiction at all. – Noah Schweber Sep 15 '17 at 22:11

1 Answers1

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It is easy to confuse oneself into thinking this is deeper than it really is.

Let's start by briefly recalling Russell's paradox: $$ R = \{ B \mid B \notin B \} $$ The particularly striking thing about this is that it can be identified as nonsense knowing nothing else than that $\in$ is some relation that can either hold between two "things" or not, and that the set builder notation means $$ \forall B(\; B\in R \iff \neg(B\in B) \;)$$ This is all -- in particular we don't need to know anything about what sets are, how they behave, or how the $\in$ relation behaves in order to see that there cannot be any $R$ with this property. It would be the same with any other symbol instead of $\in$: $$ \forall B(\; B\succeq R \iff \neg(B\succeq B) \;)$$ is also a contradiction, as a matter of pure logic.

However, unfolding the set builder in your definition gives $$ \forall B(\; B\in A \iff B\in B \;) $$ Here we are not in the Russell-like situation that pure logic alone tells us whether there is an $A$ that satisfies the condition -- and if so, which properties it has. This is a much more common situation. In order to investigate that, we need to use some actual knowledge (or assumptions, according to your temperament) about how sets behave: the axioms of set theory.

The axioms of set theory that are most often used are known as ZFC, but unfortunately doesn't quite settle our question, because there are a few slightly different systems that all may be described as ZFC. But the most common variant includes an axiom called the Axiom of Regularity, which implies that (1) your $A$ exists, and (2) it is in fact the empty set (and therefore $A\notin A$).

(Variants of ZFC without the Axiom of Regularity can't even tell you whether your $A$ exists, nor whether it will contain itself if it does).

There are other set theories with different axioms and different consequences. Some of them (e.g. NF or AFA) prove that there cannot be an $A$ that satisfies your condition.

I'm not immediately aware of any natural proposed axioms that prove that $A$ exists and must necessarily contain itself, but there might well be.

hmakholm left over Monica
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    +1. Just a quick addition (not worthy of a separate answer): (a) Positive set theory shows that the set of all sets containing themselves exists; I don't know whether it decides whether that set is an element of itself, though. (b) A parallel question could be asked about the Godel sentence - what is the status of the sentence "This statement is provable"? It turns out that this statement is outright provable - this surprising result is due to Lob. (cont'd) – Noah Schweber Sep 15 '17 at 22:22
  • Lob's theorem does demand that the way we express "is provable" be reasonable, and if we allow "pathological" expressions of "is provable" his result breaks down; see this MSE question for a bit more about this. In this way the situation with the Godel sentence is very different than with Russell's paradox or the Liar - whether the set of all sets which do contain themselves contains itself, or whether "This statement is true" is true, don't seem to be decided by any nice argument. I think this is pretty neat! – Noah Schweber Sep 15 '17 at 22:24
  • @NoahSchweber Do you know why $\implies$ is not considered a component of positive formulas in positive set theory? – DanielV Sep 16 '17 at 04:04
  • @DanielV Because $A\implies B$ is equivalent to $\neg A\vee B$ - it has a "hidden" negation. – Noah Schweber Sep 16 '17 at 04:08
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    @DanielV: Specifically, if you allowed that, you could prove $\forall x,y(x\in y)$. Namely if $x\notin y$, then ${z\mid z\in z \Rightarrow x\in y}$ would be a Russell set, which is impossible. – hmakholm left over Monica Sep 16 '17 at 10:21
  • I see, thanks. Interesting how positive formulas in that context is different than positive partial fragment of FOL. – DanielV Sep 16 '17 at 10:31
  • I'm still a bit confused - why do you say that ZFC is poorly defined? This isn't the case at all, and regularity/foundation is an axiom of ZFC, not just the "most common variant." (More precisely: there are variations in how ZFC is presented, e.g. collection vs. replacement, but each presentation proves all the axioms in each other presentation so they're all equivalent.) – Noah Schweber Sep 16 '17 at 19:25
  • @NoahSchweber: From Mendelson, Introduction to Mathematical Logic 4th ed (1997), p. 288: "In recent years ZF is often assumed to contain a set-theoretic regularity axiom (Reg): $x\ne \emptyset\Rightarrow (\exists y)(y\in x\land y\cap x=\emptyset)$. The reader should always check to see whether (Reg) is included within ZF." – hmakholm left over Monica Sep 16 '17 at 19:47
  • @HenningMakholm Wow! That's incredibly surprising to me. I've just asked a question about the history of regularity. Certainly it's true that all remotely modern work in set theory adopts regularity into ZF; I'm surprised, though, that there seems to have been ambiguity on this point even into the 60s. – Noah Schweber Sep 16 '17 at 20:54