Is it true that any set which contains elements with contradictory properties is the empty set? Here's what I've "discovered" but can't verify.
Let's take a set $A$ defined as "the set of all $x$ which don't belong to $A$".
Obviously this definition wouldn't hold in particular axiomatic systems, but it would in others.
Let's analyze this in NST. $X\in A$ iff $X\not\in A$.
This is a logical contradiction, but it still doesn't prove that $A$ doesn't exist. In fact, I would argue that $A$ is the empty set.
If put all of the elements of $A$ into the empty set, we get $(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$.
This is a tautology for whatever $X$ and $A$. Now, if we switch two of the atomic sentences, we get $X\not\in A \text { iff } (X\in A \text { iff } X\in\emptyset$). We know this to also always be true, which means that either $A$ doesn't exist or $A$ is the empty set.
Therefore, the empty set does contain every element belonging to any set it doesn't belong to.
Is there some error in my proof?
 A: The statement $$(*)\quad(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$$ does not mean what you seem to think it means (though I'm not sure exactly what you think it means).  If $A=\{X:X\not\in A\}$, then $(X∈A \text { iff } X∉A)$ must be true for all $X$.  When you assert $(*)$ you are asserting that $(X∈A \text { iff } X∉A)$ is true if $X\in\emptyset$, but also that it is false if $X\not\in\emptyset$.  So if $A$ actually is equal to $\{X:X\not\in A\}$, then this statement (with an implicit universal quantifier on $X$) is false, since $(X∈A \text { iff } X∉A)$ would be true for all $X$, even $X$ such that $X\not\in\emptyset$.
Again, I'm not entirely sure what your train of thought is, but it sounds like you are confusing $$A=\{X:X\not\in A\}$$ with $$A=\{X:X∈A \text { iff } X∉A\}.$$  If $A$ was supposed to satisfy the latter equation, then $A=\emptyset$ would be equivalent to $(*)$ (and indeed, your argument then correctly shows that $A=\emptyset$ satisfies the latter equation).  But the first equation is quite different.
A: Suppose $A$ is a set such that $x\in A$ if and only if $x\notin A$.
Now it's certainly true that from $x\in A$ we can deduce $x\in\varnothing$.
Also, vacuously, from $x\in\varnothing$ we can deduce $x\in A$.
This does not mean $A$ exists and is $\varnothing$. All we have deduced is that if $A$ exists then it must be $\varnothing$. But going back and substituting $A=\varnothing$ doesn't work: taking $x=\varnothing$ too we have $\varnothing\notin\varnothing$, which should imply $\varnothing\in\varnothing$, but the latter is false. The only explanation is that our original assumption - that such an $A$ exists - was wrong.
This is analogous to the following argument.
Suppose there is a real number $x$ such that $x^2+x+1=0$. Multiplying both sides by $x-1$, we get $x^3-1=0$, i.e. $x^3=1$, and the only real number satisfying this is $x=1$.
What we have just done is to deduce that if such a real number exists then it must be $1$ - but it doesn't, so it isn't.
A: NO. If we assume $\exists x: [x\in A \land [P(x) \land \neg P(x)]]$ we can easily obtain a contradiction. Therefore, such a set $A$ cannot exist.
EDIT: In your example, if we assume $\exists x: [x\in A \land [x\in A \iff x \notin A]]$ we can similary obtain a contradiction. Therefore, in this case, too, such a set $A$ cannot exist.
BTW, when you say that a set contains any elements, you rule out that it is empty.
EDIT: Also, if you assume $x\in A \iff x\notin A$, this is already a contradiction. We can only conclude that there cannot exist such a set $A$ and element/object $x$.
We can, however, define an empty set $E$ given set A as follows:
$\forall x: [x \in E \iff x \in A \land x \notin A]$
