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In his book Notes on Set Theory, Moschovakis begins to list the axioms of $\text{ZFC}$ on p. 24. The first is the Axiom of Extensionality, which he expresses like this:

$$ A = B \Longleftrightarrow (\forall x)[x \in A \Longleftrightarrow x \in B]. $$

He then goes on to list Emptyset Axiom:

There is a special object $\emptyset$, which we will call a set, but which has no members.

He then goes on to note:

The Axiom of Extensionality implies that only one empty set exists, ...

Now, here's my problem. In its formulation, the Axiom of Extensionality uses the membership of some existing "thing" $x$ to establish "sameness" of two sets. But since $x$ is an existing "thing", how is Extensionality even relevant to $\emptyset$, which, by definition, contains no existing "things"? It seems to me that we cannot establish that there exists exactly one $\emptyset$, since the very notion of "sameness" is built upon existing $x$'s being members of sets, and $\emptyset$ contains no existent $x$'s.

Addendum

So if I understand correctly, my misinterpretation of Extensionality lies in this: I interpreted $x \in A$ as a demand for existing $x$'s to be a part of sets in order for them to be the same set. But what Extensionality really says is that:

$A = B \Longleftrightarrow (\forall x)[x \in A \Longleftrightarrow x \in B]$ holds for all $x$'s.

Then indeed, it follows that:

$$ A = \emptyset, \; B = \emptyset' \\ A = B \Longleftrightarrow (\forall x)[x \in A \Longleftrightarrow x \in B], $$

since it holds for both $A$ and $B$ that they have no members, hence every member that is a member of $A$ will be a member of $B$ — that is to say, no member.

While thinking about it, I came up with an interesting "modification" of $\text{ZFC}$. Instead of conceiving of $\emptyset$ as a set without members, let us denote nothing as $\mathfrak{n}$. $\mathfrak{n}$ is an element of every set. More specifically, we define $\emptyset$ as $\emptyset := \{ \mathfrak{n} \}$. With this setup, my interpretation of Extensionality holds for all sets. The only drawback is that such a setup contains atoms (one atom, $\mathfrak{n}$, to be precise), which violates the Principle of Purity. Well, just a thought I wanted to share. ;)

Please correct me if I got something wrong.

Daniel Fischer
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God bless
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    Assume for contradiction that there are two different empty sets: $\emptyset$ and $\emptyset'$. What does it mean that they are different ? By Extensionality that there is some element of the first that is not an element of the second. – Mauro ALLEGRANZA Mar 16 '20 at 11:00
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    Does this answer your question? Proving that the empty set is unique – Mauro ALLEGRANZA Mar 16 '20 at 11:01
  • @MauroALLEGRANZA Granted, I see how the empty set is unique intuitively. What bothers me is that it seems that Extensionality in and of itself is not formulated sufficiently to guarantee the uniqueness of the empty set, since Extensionality presupposes membership of existing "things" in sets. But the empty set contains no existing "things" and thus it seems that the uniqueness of the empty set in the context of Extensionality alone is ambiguous at best. – God bless Mar 16 '20 at 11:12
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    The implication $p\implies q$ is true whenever $p$ is false. Applying this we find that implication $x\in A \implies x\in B$ is true if $A$ is an empty set. If also $B$ is also an empty set then we have $x\in A \iff x\in B$ and in accordance with extensionality $A=B$. – drhab Mar 16 '20 at 11:17
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    "...Extensionality presupposes membership of existing "things" in sets..." No, it does not. – drhab Mar 16 '20 at 11:21
  • It is not only unique "intuitively": it is unique because we prove it so, and the proof needs Extensionality. – Mauro ALLEGRANZA Mar 16 '20 at 13:07
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    Your concern is not really about set theory but instead about logic, in particular the meaning of the universal quantifier, and the meaning of the bi-implication $\iff$. Given two empty sets $A$ and $B$, if you don't understand why the bi-implication "$x \in A \iff x \in B$" is true regardless no matter what you substitute for $x$, then you need to ponder the meaning of $\iff$. – Lee Mosher Mar 16 '20 at 17:28
  • This " modification" envisionned might introduce a contradiction in the system. First in order " nothing" to belong to any set, nothing has to exist, for membership relation ( as any relation) only holds between existing relata. Second, what should we answer to the question : what has the set { n } in it? Ansxer : it has nothing in it which makes it empty; but also, it has something in it, since it has " nothing " as element, which prevents it from being empty. –  Mar 17 '20 at 09:31
  • You say " will be a member of B - hence no member" which suggests that the emptyness of B plays a role here. Actually, regardless of the fact B is empty or not, the fact that $A$ is empty is a sufficient condition for the proposition " if $x$ belongs to $A$ then $x$ belongs to $B$" to be true , for all $x$ ( which means that $A$ is included in $B$). The fact that $B$ is empty plays only a role as to the reverse conditional. –  Mar 17 '20 at 09:47
  • @RayLittleRock It seems that the issue you expose is purely nominal. Of course $\mathfrak{n}$ is not nothing, it is an atom which we've named "nothing" so we can conveniently say that $\emptyset := { \mathfrak{n} }$ contains "nothing". So the connection between $\mathfrak{n}$ and literally nothing is merely nominal and there is no contradiction. – God bless Mar 18 '20 at 09:46
  • I think you are on the right track when you rephrase the issue as a language problem. The surface grammar of the sentence " I have nothing" is " Subject + V+ Object". So language inclines us to think that " nothing " is , in spite of all, an entity,, though of a special kind, namely, the " null entity" or the " null-object". What logic teaches us is that the real grammar of the sentence is " there is no $x$ such that I have $x$", in other words, the open sentence " I have $x$" is true for no value of $x$. –  Mar 18 '20 at 09:56
  • In the same way, the surface grammar of " the set that has nothing it it" inclines us to think that the empty set has in it some " null object" , the " nothing". But logic teaches us to analyze the sentence: " the empty set is the set that has nothing in it" as " S is empty iff the open sentence " $ x \in S$" is false for all possible values of $x$ " –  Mar 18 '20 at 09:58
  • This is a philosophical issue. Hegel is well-known for having done this inference : " Being ("Esse" in latin ) is no being ( " ens"), but the source of all particular beings ( " entia"). Therefore, being in non-being , being is " nihil" ( nothingness). Modern logic does not allow this inference. ( Reference : Hegel's Logic, very beginning). –  Mar 18 '20 at 10:07
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    Other instance of the same illegal inference : " Nothing is without a sufficient reason ( "Nihil est sine ratione", this is Leibniz's principle of sufficient reason ). Therefore, nothingness is that which has no reason." ( Baumgarten, Metaphysica, very beginning). –  Mar 18 '20 at 10:26

2 Answers2

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The notion of "sameness" is not built upon existing $x$'s being members of sets. All we care about are the implications if $x\in A$, then $x\in B$ and if $x\in B$, then $x\in A$. If $A$ and $B$ are empty sets $x\notin A$ and $x\notin B$, hence these implications are (vacuously) true. I guess the concept of vacuous truth is what you are looking for.

There is a problem with this modification: we still can construct the empty set (in the old sense) due to the axiom of specification. For instance, consider $\{x\ |\ x\in\{\mathfrak{n}\}\text{ and }x\neq x \}$. So the requirement that every set contains $\mathfrak{n}$ doesn't hold.

Ansar
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Short answer : saying that "set $A$ = set $B$" is not a positive assertion as to the existing members of $A$ and of $B$; it is a negative statement as to the non-existence of any $x$ belonging to one set but not to the other.

  • The temptation to over-interpret the conditional ( and consequently the biconditionnal) operator featuring in the definition of set identity must be resisted.

  • In fact, the proposition " for all $x$ , $x$ is in $A$ $\rightarrow$ $x$ is in $B$ " has no existential import. Actually ( and that can be seen by inspecting the truth table of the "if...then..." operator) a conditional says very little, and what it says is purely negative: asserting "$A\rightarrow$ $B$" amounts to saying " we are not on line $2$ of the truth table" , that is, " it is not the case that $A$ is true while $B$ is false".

  • So " for all x, $x$ is in $A$ $\rightarrow$ $x$ is in $B$ " simply means :

    "it is not the case that there exists an $x$ such that $x$ is in $A$ while $x$ is not in $B$"***

  • Therefore, the test for sameness amounts to asking 2 questions :

(1) does any existing $x$ belong to $A$ but not to $B$?

(2) does any existing $x$ belong to $B$, but not to $A$?

and a positive answer to the whole test requires 2 negative answers to these questions. ( that is, 2 assertions of non-existence).

  • So let's say that Empty$_1$ and Empty$_2$ are two ( alledgedly distinct) empty sets.

(1) can you see an existing $x$ that belongs to Empty$_1$ but not to Empty$_2$ ?

and

(2) can you see an existing $x$ that belongs to Empty$_2$ but not to Empty$_1$?

  • In case you have answered "no" to these 2 questions, you know that these two empty sets are, in fact, one and the same set.
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