1
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ $\exists x\in y\,P(x)$ 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. $\endgroup$ Aug 1 '13 at 14:34
  • $\begingroup$ 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. $\endgroup$ Aug 1 '13 at 17:43
5
$\begingroup$

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.]

$\endgroup$
2
  • 1
    $\begingroup$ (If we do not use constants in the language, then even $\lnot(\emptyset\in\emptyset)$ has quantifiers.) $\endgroup$ Aug 1 '13 at 14:32
  • $\begingroup$ (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!) $\endgroup$ Aug 1 '13 at 15:13
3
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ But how is "existence" defined? $\endgroup$ Aug 1 '13 at 14:36
  • $\begingroup$ 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 $M\models\varphi(m)$. $\endgroup$
    – Asaf Karagila
    Aug 1 '13 at 15:09
0
$\begingroup$

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)

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$
    – TonyK
    Aug 1 '13 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.