So, my line of thinking is that a set that contains all sets that do not contain themselves is a paradox. And the opposite of that is a set that contains all sets that contain themselves, and, while it not decidable if it contains itself, neither state leads to a contradiction.

So, I would ignorantly say that the Co-Paradox of the set of all set that do not contain themselves is itself not a Paradox.

Is this close to something in mathematics? Are there Co-Paradoxes that are also Paradoxes?

  • $\begingroup$ nice variation! $\endgroup$ – Ittay Weiss Aug 26 '14 at 8:59
  • $\begingroup$ Of course, whether this “co-paradoxical" set can be assumed in a broader set theory is a matter of details. In ZF it exists and is empty; in positive set theory it exists and contains at least the universe; in NF(U), it engenders paradox because it's complement must exist. $\endgroup$ – Malice Vidrine Aug 26 '14 at 10:12
  • $\begingroup$ The first set u describe leads to a contradiction (u call it a paradox). The second one not necessarily leads to a contradiction (so its not paradox?). Could you specify more what u mean by co-paradox and the opposite? $\endgroup$ – FWE Aug 26 '14 at 10:20
  • $\begingroup$ U might f.i. think of something like this: If I would ask u to give me 1 mio dollar would your answer be the same as ur answer to this question? this is somehow paradox (in any case if u answer the question u ve to give away 1 mio dollar). U can do variations (oppsites?) of this still staying paradox - since paradoxy is not unusual for self.-referential propositions. Paradoxes often base on self-reference or category error (i.e. mixing up things). $\endgroup$ – FWE Aug 26 '14 at 10:31
  • $\begingroup$ Thanks everyone. It seems that my question isn't rigorous enough to have a satisfactory answer. I did find a similar question, but I think for the most part I got paradox confused with undecidability. $\endgroup$ – user2063685 Aug 26 '14 at 21:41

the concept of paradox is not a mathematical concept. The meaning of the word paradox is something that exhibits a strong counter intuitive behaviour. Something that looks ok but upon closer inspection reveals itself to be highly problematic, to the point that it is totally hopeless.

Russell's paradox is a seemingly legit definition of a set, i.e., the set of all sets that do not contain themselves as elements, which upon closer inspection leads to an unrepairable situation; no answer to the question "does that set contain itself as an element?" is consistent, and thus the question is unresolved.

The paradox you describe above is similarly a seemingly legit definition of a set, i.e., the set of all sets that do contain themselves as elements, which upon closer inspection reveals to be problematic; the question "does that set contain itself as an element?" does not seem to be answerable from the definition of the set, and thus the question is unresolved.

Notice however that the paradox you describe is weaker than Russell's. Russell's paradox leads to a contradiction. The situation you describe is not a contradiction. It just shows that no answer to the question immediately leads to a contradiction, and thus none can be immediately discarded. However, it leaves open the possibility that a more clever argument can resolve the question. Of course, you will have to carefully axiomatize your paradox in order to really answer such a question.

  • $\begingroup$ I think it's odd to call the Russell case "unresolved"; it is simply false that anything is a member of the Russell set. There are no models of "the Russell set exists," unlike the case for its complement. $\endgroup$ – Malice Vidrine Aug 26 '14 at 20:12
  • $\begingroup$ @MaliceVidrine I meant that naively (and Russell's paradox is a paradox of naive set theory) the question whether the Russell sets belongs to itself or not is unresolved. I did not mean that the paradox is unresolved as clearly any axiomatization of set theory proves that if a model exists, then the Russell "set" does not exist. $\endgroup$ – Ittay Weiss Aug 26 '14 at 21:53

Proposition (There exists no universe)

Say $A$ is an arbitrary set (or set of discourse or universe).

Then there exists a set $B$ that is not in $A$.

Proof: Say $A$ is a set. Define $$B:=\{x\in A | x\not\in x\}$$

Then $B$ can not be in $A$:

Assume $B\in A$.

Then we have either

$B\in B$ and therefore by definition of $B$ it follows that $B\not\in B$

or we would have

$B\not\in B$ and therefore by definition of $B$ it follows that $B\in B$

so in any case a contradiction. $\diamond$

The point here is that the set $A$ was chosen arbitrarily and what we showed is, that we can always find something not in that set $A$. So this means nothing contains all or u could also say more spectacular there is no universe (universe meaning universe of discourse).

  • $\begingroup$ I find the language in the question too unprecise and this post introduces a way of thinking about the issue more meaningful. (Can you show me such model?) $\endgroup$ – FWE Aug 26 '14 at 9:03
  • $\begingroup$ Goedel-Bernays set theory is not a model of set theory but an axiom system. Can u show me in NBG or ZFC that the set of all sets exists? $\endgroup$ – FWE Aug 26 '14 at 9:07
  • $\begingroup$ But if what u take is a model of set theory axioms (lets say NBG or ZFC) then this means that the axioms interpreted in that model become true. And so the litte statement from above becomes true in that model. That is called correctness theorem. So I doubt there is such a model (but am curious to see it). $\endgroup$ – FWE Aug 26 '14 at 9:15
  • $\begingroup$ I'm sorry, but I think we are having a mis-communication here. I'll just say then that the way we use the term universe is meant to talk about all sets or elements of interest to a particular situation. Not that there is a universe which contains all sets in a particular model of set theory. Moreover, there is a well-developed theory of Grothendieck universes. I don't think though OP was asking about any of that. $\endgroup$ – Ittay Weiss Aug 26 '14 at 9:24

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.