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I am studying Elements of Set Theory by Enderton and don't understand how the existential quantifier in the Empty Set Axiom guarantees its existence.

The axiom reads

$$\exists B(\forall x \neg(x\in B)).$$

From my logic book, I understand this quantifier to represent the disjunction of the wff (here, $\forall x \neg(x\in B))$) over the domain of quantification, i.e.

$$\bigvee B(\forall x\neg(x\in B)).$$

If the domain is void, then there are no disjuncts for the above axiom, and I would assume it's truth value is undefined, or that it is not even a sentence in the formal language.

To restate more colloquially, how are we bootstrapping a void domain with an object that by definition comes into existence by the use of a quantifier that must range over this void domain?

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    $\begingroup$ The value of the empty disjunction is false. As a side note, the Empty Set Axiom follows from other axioms. Specifically, one can first prove that there is a set using logical axioms and then prove that the empty set exists using the axiom of specification. $\endgroup$ – Yury Dec 26 '13 at 18:52
  • $\begingroup$ Yury, can you provide me a reference on how to prove set existence using logical axioms? $\endgroup$ – mcg256 Dec 26 '13 at 21:43
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    $\begingroup$ Roughly speaking the argument goes as follows. There is a logical axiom $\forall x x=x$. Using the Universal Instantiation rule, we get $x=x$. Now using the Existential Generalization rule, we get $\exists x x=x$. That is, we proved that there exists a set. (See also en.wikipedia.org/wiki/List_of_rules_of_inference) $\endgroup$ – Yury Dec 26 '13 at 22:05
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There are various arguments:

  • $\exists$ in the sense of existence would be wrong in a void domain where nothing exists. Hence the axiom states (among others) that the domain is not void

  • $\exists$ may be introduced as abbreviation for $\neg\forall\neg$. In the case of empty domain all $\forall$ statemens are (vacuously) true, hence all $\exists$ statemens are false. So again, an axiom startiung with $\exists$ rules out void domains.

  • Interpreting $\exists$ as disjunction over the domain of quantification needs to be adjusted both for the case of an infinite domain (where we would illegally produce an infinitely long formula) and the empty domain (where we would produce an empty formula). Thius requires some conventions, such as that the empty formula $\epsilon$ is true (to make $\epsilon \lor a\lor b\lor \ldots\iff a\lor b\lor \ldots$)

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  • $\begingroup$ Perhaps you mean $\epsilon$ is false, since false is the identity for the $\vee$ operation. $\endgroup$ – Ben Millwood Dec 26 '13 at 21:14
  • $\begingroup$ If possible, can you please supply a reference for your statements? I'm working from (have completed) Language, Proof, and Logic by Barwise, et al. and these topics are excluded. $\endgroup$ – mcg256 Dec 26 '13 at 21:40
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An alternative axiom, would be just $$\exists B$$ Then, together with the subset axiom (or the replacement axiom) would guarantee the existence of the empty set.

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  • $\begingroup$ In some developments, that's not a well-formed formula by itself, but it's easy to make an equivalent one (e.g. $\exists B .\top$ or $\exists B. B = B$) $\endgroup$ – Ben Millwood Dec 26 '13 at 21:15

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