0
$\begingroup$

About a year and a half ago, while I was looking on the Web for papers regarding the Russell paradox, I chanced to find an interesting concept. This concept was contained in what (for want of a better term) was a two-page 'advertisement' for Math 582, "Introduction to Set Theory", taught at the University of Michigan (the 'advertisement' had the title "The Joy of Sets" and can still be found on the Web under the title "The Joy of Sets--University of Michigan"). The author of the 'advertisement' was Karl Liechty, who is an assistant professor there. The concept (which Prof. Liechty called "the MOST IMPORTANT PROPERTY a set S has...") is the following (I quote the sentence it is contained in verbatim):

"For us, the MOST IMPORTANT PROPERTY a set S has is this: if x is an object, then either x $\in$ S or x $\notin$ S, but not both."

This concept can be made purely symbolic as follows:

(x)((x $\in$ S $\lor$ x $\notin$ S) $\land$ $\lnot$(x $\in$ S $\land$ x $\notin$ S)) ( it is interesting to note that the concept, formulated as such, seemingly allows the universal quantifier to be unrestricted--that is, to mean 'for all possible objects'....).

Prof. Liechty uses this concept as follows:

"Russell's paradox provides a non-example of a set. Consider

{S is a set| S $\notin$ S}.

Call this candidate for set-hood T. As you should verify, we have both T $\in$ T and T $\notin$ T. Thus, T does not have the MOST IMPORTANT PROPERTY, and so is not a set."

Can the other "candidates for set-hood" that generate the Curry paradox, The Burali-Forti paradox, Cantor's paradox, etc. be shown to be nonexamples of sets because they, too, do not have the "most important property" (it should be noted that since the candidates for set-hood that generate the paradoxes can be formulated using set-builder notation (i.e. {x|$\phi$(x)}) this notation can be reserved for mere collections (or classes, if you will))? Also if the paradoxical candidates for set-hood can be shown to be non-examples of sets by 'the most important property', can Naive set theory, when restricted to collections deemed sets by 'the most important property', be rendered free of paradox?

$\endgroup$
  • $\begingroup$ The part that says that both $T\in T$ and $T\notin T$ seems dodgy: it assumes that $T$ is a set. Rather, if we assume by contradiction that $T$ is a set, then it would follow (in classical logic, anyhow) that $T\in T$ and $T\notin T$, which is a contradiction, so $T$ is not a set. In particular, $T\notin T$ and not $T\in T$. $\endgroup$ – tomasz Jul 6 '14 at 14:54
  • $\begingroup$ @tomasz: Could it be that Prof. Liechty, in saying that T(={S is a set|S $\notin$ S}) is a "candidate for set-hood", is assuming by contradiction that T is a set? $\endgroup$ – Thomas Benjamin Jul 7 '14 at 7:58
  • $\begingroup$ In that case, he should have written that "we would have...". Anyway, this "most important property" is nothing but (a special case of) law of excluded middle. $\endgroup$ – tomasz Jul 7 '14 at 12:00
1
$\begingroup$

No, that doesn't tell us anything relevant about sets.

Since $x\notin S$ is just an abbreviation for $\neg(x\in S)$, we can abbreviate your property $$ \forall x. \; (x\in S \lor \neg(x\in S))\land \neg(x\in S\land \neg(x\in S)) $$ as $$ \forall x. \; (A\lor \neg A)\land \neg(A\land \neg A) $$ where $A$ is the proposition $x\in S$.

And this is a logical tautology -- it is true as a matter of pure logic no matter what the proposition $A$ it, and therefore also independently of what $x\in S$ means or how sets behave in general.

It's just a simple consequence of the fact that we want to reason about sets using ordinary classical logic.


At best, the statement could be taken as a roundabout way to say that $x\in S$ is a logical proposition that obeys the rules that logical propositions do -- in particular whether $x\in S$ is true does not depend on anything else than what $x$ and $S$ are. For example, the truth of $x\in S$ is not allowed to depend on which context we ask the question in, or be different if we ask the same question at different times -- unless the different context assigns different meaning to one or both of the letters $x$ and $S$.

It would be somewhat more meaningful to say that the ONLY property a set $S$ has is the totality of answers to the question "does $x\in S$" for all possible $x$s. That's formalized as the axiom of extensionality, saying that if $\forall x.x\in S\leftrightarrow x\in T$, then $S=T$.

$\endgroup$
  • $\begingroup$ Was Prof. Liechty wrong then in saying that since "{S is a set| S $\notin$ S}" ( a direct quote from his 'advertisement') does not satisfy that tautology (unless I formulated his statement incorrectly), then T is not a set? $\endgroup$ – Thomas Benjamin Jul 6 '14 at 13:18
  • $\begingroup$ @ThomasBenjamin: I hold that he's at least being needlessly confusing. The usual argument is the contradiction shows that there is no thing whose members are exactly everything that is not a member of itself. It's not that such a thing exists and doesn't satisfy a property; it is not there in the first place. $\endgroup$ – Henning Makholm Jul 6 '14 at 14:36
  • $\begingroup$ In some set theories we can speak of a "thing" (a proper class) that has as members every set that doesn't contain itself, but it is false and confused to say that this thing is a non-set because it is both a member of itself and not a member of itself. It is clearly and unambiguously not a member of itself; it is false to say that it is a member of itself. The reason why it isn't a set is that it just isn't. We know it is not a set because it would lead to a contradiction if it were. $\endgroup$ – Henning Makholm Jul 6 '14 at 14:37
  • $\begingroup$ ...and that contradiction is not allowed by the tautology the "most important property a set S has" in fact is (as you have correctly shown)? As regards the Curry, Burali-Forti, and Cantor paradoxes, would the tautology "the most important property a set S has," is (I realize that "the most important property a set S has" is, in fact, a misnomer because it is a tautology), show that the 'candidates for set-hood' that cause these paradoxes to exist are, in fact, not sets but mere collections? $\endgroup$ – Thomas Benjamin Jul 7 '14 at 8:12
  • 1
    $\begingroup$ @ThomasBenjamin: Tautologies show nothing, no matter what you call them. If you derive a contradiction, you have a contradiction. You don't need to appeal to a specific tautology to do that; it will be built into the rules of whatever system of logic you do your reasoning in. (Except, in Hilbert-like system, the particular tautologies that have been selected as logical axioms, but the one you present here is unlikely to be one). $\endgroup$ – Henning Makholm Jul 7 '14 at 10:32
0
$\begingroup$

In tricky situations such as this, set-builder notation is probably not the best choice. Instead of {S is a set| S $\notin$ S}, with it's inherent ambiguity (being neither true nor false), you should write something like, $\forall s:[s\in R\iff s\notin s]$. So the "resolution" of Russell's Paradox might (depending on your rules of inference and set theory) go something like the following proof by contradiction:

  1. $\forall s:[s\in R\iff s\notin s]$ (Premise)
  2. $R\in R \iff R\notin R$ (Universal Specification from 1)
  3. $\neg \forall s:[s\in R\iff s\notin s]$ (Conclusion from 1 and 2)
$\endgroup$
  • $\begingroup$ Christiansen: Interesting. I chose to use set-builder notation for 'collection/class' because of the intuition 'you can enclose any list of objects you want between a left bracket and a right bracket and form a 'collection''. $\endgroup$ – Thomas Benjamin Jul 12 '14 at 15:20
  • $\begingroup$ Actually, the above should read 'you can enclose any objects (actually, symbols representing objects) you want between a left bracket and a right bracket and form a 'collection'.' The act of listing is (possibly) the act of forming a well-ordering and that is not completely general. $\endgroup$ – Thomas Benjamin Jul 13 '14 at 4:31
  • $\begingroup$ Also, one probably would want to say that the Axiom of unrestricted comprehension defines the notion of 'collection' rather than 'set'. $\endgroup$ – Thomas Benjamin Jul 13 '14 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.