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In axiomatic set theory, we have Axiom of empty set: $\exists \varnothing \forall x ( x\notin \varnothing)$. Is there any equivalent statement without the use of quantifiers? For example, $ \exists x\in y (P (x)) $ is equivalent to $ x\in y\: \And\: P (x) $ and $\forall x\in y (P (x) $ is equivalent to $ x\in y\Rightarrow P (x) $ (or are they?). Any try to clear things up is appreciated.

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$\exists x\in y\,P(x)$ is not equivalent to $x\in y\,\And\, P(x)$. In the former, $x$ is a bound variable, the statement is the same as $\exists w\in y\,P(w)$. In the latter, $x$ is a free variable, and cannot be replaced with $w$. Same issue occurs with the $\forall$ example. –  Andres Caicedo Aug 1 '13 at 14:34
    
If you allow arbitrary subsets of any set, then you don't need an empty-set axiom, because every set would have an empty subset. And all empty sets are identical. –  Dan Christensen Aug 1 '13 at 17:43
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4 Answers

up vote 5 down vote accepted

1) '$\exists \varnothing \forall x ( x\notin \varnothing)$' is ill-formed. '$\varnothing$' is a constant, not a variable.

2) Extensionality tells you that any sets $a$ and $b$ which lack members, if such sets exist, are the same set. So adopt the axiom that there is a set which lacks members: $\exists y \forall x ( x\notin y)$. It will now follow, given that consequence of extensionality, that $\exists! y \forall x ( x\notin y)$ -- there is a unique empty set. So that justifies introducing a constant `$\varnothing$' to denote the unique empty set.

3) Note, in introducing the empty set that way, you do need an existential axiom.

4) You could, I suppose, instead build set theory using a classical first-order language with the signature $\{\in, \varnothing\}$ from the start, and then just make do with the universal axiom $\forall x\, x \notin \varnothing$: but that would be a bit cheaty -- as you'd be presupposing the useful but not compulsory convention that all constants denote.

5) Re: 'A statement must include some form of (implicit or explicit) quantifier.' Really? '$\neg (\varnothing \in \varnothing)$' is a perfectly good quantifier free statement if $\varnothing$ is introduced -- as is often the case -- as a constant. [The justification for the introduction is something quantificational, but what is introduced is a constant.]

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(If we do not use constants in the language, then even $\lnot(\emptyset\in\emptyset)$ has quantifiers.) –  Andres Caicedo Aug 1 '13 at 14:32
    
(Distinguo: We could introduce $\varnothing$ as a new constant [not in the original signature], or as a descriptive term to be eliminated by Russell's Theory of Descriptions. In the latter case, it indeed is a disguised quantifier!) –  Peter Smith Aug 1 '13 at 15:13
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What you wrote about bounded quantifiers is very wrong.

$\forall x\in y\varphi(x)$ is not $x\in y\rightarrow\varphi(x)$. It's an abbreviation for $\forall x(x\in y\rightarrow\varphi(x))$. Similarly $\exists x\in y\varphi(x)$ is abbreviation for $\exists x(x\in y\land\varphi(x))$.

The axiom asserts existence, so you cannot avoid $\exists x(\ldots)$ in its form (unless you prefer $\lnot\forall x(\ldots)$ instead).

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But how is "existence" defined? –  Constantine Aug 1 '13 at 14:36
    
Constantine, I'm not sure I understand your question. The statement $\exists x\varphi(x)$ is true in a model $M$ if and only if there exists some $m\in M$ such that $M\models\varphi(m)$. –  Asaf Karagila Aug 1 '13 at 15:09
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Any statement without quantifiers, e.g. $x\cdot x=x^2$ is at most equivalent to a corresponding statement with allquantors (e.g. $\forall x\colon x\cdot x=x^2$) because the usual rules of inference include both $\phi(x)\vdash \forall x\colon \phi(x)$ and $\forall x\colon \phi(x)\vdash \phi(x)$, but cannot guarantee the existence of an object. The main purpose of an Axiom of Empty set is the existence of a set at all, for once you have any set $a$ (which may also follw from other axioms such as the Axiom of Infinity, if included) you can define $\emptyset := \{\,x\in a\mid \neg (x=x)\,\}$ using the other axioms. But you can never get existence from a formula introduced with only allquantors (or a formula without quantors, having implicit allqantors)

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Deduced from your given example, one could form $$ x \text{ is an Element} \Rightarrow x\notin\emptyset$$ but that actually just omits $\forall \text{Elements } x\ : x\notin\emptyset$. As said in the Comments, a statement must include some form of (implicit or explicit) quantifier.

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You are just muddying the waters here! Neither of your formulas is well-formed. And if you meant the formulas $x \notin \varnothing$ and $\forall x\ (x\notin\varnothing)$, then these are not the same. –  TonyK Aug 1 '13 at 14:05
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