# Is there conflict between the law of excluded middle and “no set is its own member”?

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.

• 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. – littleO Sep 16 '18 at 9:22
• You have an implicit assumption that there is a universal set. Assuming ZF(C), there is none. – Asaf Karagila Sep 16 '18 at 9:38
• @AsafKaragila Does it mean that there are no absolute complements in ZFC? – user161005 Sep 16 '18 at 9:54
• 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. – Mauro ALLEGRANZA Sep 16 '18 at 10:10
• @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. – M. Winter Sep 16 '18 at 10:25

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 $P$.