6
$\begingroup$

And if yes, then how can it be resolved?

As far as I know in standard set theory it's true that "no set is its own member". Also in standard logic the law of excluded middle is true, either $A$ or $\lnot A$.

Now let's consider simple sentence "Entity $X$ is either a real number or not a real number". This can be restated in terms of set theory as "Element $x$ either belongs to set $\mathbb{R}$ or to set non-$\mathbb{R}$".

Looks like a tautology, doesn't it? But I will show you that it's not because there is one thing that isn't a member of either set. Namely, it's set "non-$\mathbb{R}$". This can't belong to set $\mathbb{R}$ because it isn't a real number. This also can't belong to set non-$\mathbb{R}$ (i.e. it can't belong to itself) because of "no set is its own member". Thus "Element $x$ either belongs to set $\mathbb{R}$ or to set non-$\mathbb{R}$" isn't a tautology.

$\endgroup$
  • 10
    $\begingroup$ I suspect the answer is something like "there is no such thing as the set non-$\mathbb R$". A similar idea is that there is no "set of all things", because it would have to contain itself. Trying to introduce a set of all things that are not real numbers is similar to trying to introduce a set of all things. $\endgroup$ – littleO Sep 16 '18 at 9:22
  • 7
    $\begingroup$ You have an implicit assumption that there is a universal set. Assuming ZF(C), there is none. $\endgroup$ – Asaf Karagila Sep 16 '18 at 9:38
  • 3
    $\begingroup$ @AsafKaragila Does it mean that there are no absolute complements in ZFC? $\endgroup$ – user161005 Sep 16 '18 at 9:54
  • 4
    $\begingroup$ The move from $\lnot (x \in \mathbb R)$ to $(x \notin \mathbb R)$ is safe: it is only "logic". But the next move, to $(x \in \mathbb R^C)$ assume that the set $\mathbb R^C$ exists. It needs mathematics, i.e. it needs the proof in the relevant part of math (i.e. set theory) that the existence of the purported set is provable from set axioms. $\endgroup$ – Mauro ALLEGRANZA Sep 16 '18 at 10:10
  • 3
    $\begingroup$ @user161005 You cannot specify such a set because (as you proved in your post) it does not exist. However you can specify everything that is not a real number by the predicate $x\not\in\Bbb R$. Just not every predicate gives you a set. $\endgroup$ – M. Winter Sep 16 '18 at 10:25
12
$\begingroup$

This is similar to Russell's paradox.

The usual approach is to cast blame on the principle of unrestricted comprehension — the hypothesis that for any property $P$ of sets, there is a set of everything satisfying $P$.

ZFC replaces this with the more modest hypothesis that if you're additionally given a set $S$, then you can form the set of everything in $S$ satisfying $P$.

In particular, in the usual formulation of modern set theory, the thing you call "non-R" is not a set. (c.f. "proper class")

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Sep 17 '18 at 15:03

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.