# How could $\varnothing$ be unique if it could be vacuously false?

The argument that people use to prove that empty set is unique is that: Let $$A$$ and $$B$$ be two empty sets then $$\forall z : z \in A \implies z \in B$$ since there is no such $$x\in A$$ hence this statement is vacuously true. Also the converse is true therefore $$A=B$$. My objection is we could have equally stated $$\forall z : z \in A \implies z \notin B$$ and thus conclude that $$A\neq B$$.

• There is no way however that the antecedent $z\in A$ is true. A false $\rightarrow$ false conditional is true. – Robert Wolfe Jun 17 at 12:21
• – Mauro ALLEGRANZA Jun 17 at 12:34
• Or, $\forall z : z \in A \implies z \notin B$ doesn't imply that $A\neq B$. – immibis Jun 18 at 0:07

But that's not how you negate $$A = B$$. For two sets to be non-equal, you have to actually find an element which is in one of the two sets, and not the other. The negation of a statement beginning with $$\forall$$ is a statement beginning with $$\exists$$. And if you carefully state "$$A\neq B$$" with the correct quantifiers and implications, and assume $$A$$ and $$B$$ are both empty sets, then you will find that $$A\neq B$$ is actually not true.

• Indeed: It follows from the negation of the definition for equality, via deMorgan's Laws, quantifier duality, and negation of implication. \begin{align}A\neq B \iff&~ \lnot~\big(\forall x~(x\in A\to x\in B)\land \forall x~(x\in B\to x\in A)\big)\\[2ex]\iff&~ \exists x~(x\in A\land x\notin B)~\lor~\exists x~(x\in B\land x\notin A)\end{align} – Graham Kemp Jun 19 at 0:03

Indeed, $$\forall z:z\in A\implies z\notin B$$. And, by the same argument, $$\forall z:z\in B\implies z\notin A$$. But all that you deduce from this is that $$A\cap B=\emptyset$$. There is no contradiction here.

Two sets are different if at least one of them has at least one element that the other one do not have. If there are two different empty sets then at least one of them has at least one element that the other one do not have, but that is not true since none of them has elements, so all empty sets are equal, so there is only one empty set.

The objection raised here is interesting and allows, once answered, to express a paradoxical but true statement about empty sets: two empty sets are necessarily identical, and disjoint.

Your reasoning is as follows :

(1) A and B are empty sets

(2) for any arbitrary object x, the statement "x belongs to A --> x does not belong to B" is true for the antecedent is false and the consequent is true ( and in the False-True case, a conditional statement is true)

(3) therefore, A and B are disjoint ( their intersection is empty)

(4) therefore, A and B are distinct

The mistake lies in the inference from (3) to (4) : the fact that the intersection of A and B is empty does not mean that :

             for all x , ( x belongs to A <--> x belongs to B ) is false,


in other words, it does not mean that :

there is an x such that (x belongs to A but not to B OR x belongs to B but not to A).

Statement (3) about A and B being disjoint simply means :

            for all x , ~ ( x belongs to A & x belongs to B)


Note. The form of the sentences is not the same. Compare :

(1) " ~ for all x, phi(x) "

and

(2) " for all x , ~ phi (x) "

Note. The objection raised in this post relies on the ancient square of oppositions. But set theory relies on modern square of oppositions. ( See : https://plato.stanford.edu/entries/square/)

If ancient square of oppositions were applied to sets, it would give the diagram below, but this diagram is NOT CORRECT in the context of set theory