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I'm doing some research on naive set theory and was a little confused over the statement of the axiom of unrestricted comprehension, $\exists$B$\forall$x(x$\in$B$\iff$$\phi$(x)). I am curious as to why this is an iff statement. I was under the assumption that, in naive set theory, a set only exists if it is determined by some property, but this seems to say that a property exists for each set as well. Why doesn't the implication $\exists$B$\forall$x(x$\in$B$\implies$$\phi$(x)) suffice?

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The unrestricted axiom of comprehension says that given any $\phi$, there is a set $B$ s.t. $B$ contains all and only the objects that satisfy $\phi$. The "iff" is taking care of that "all and only" part; it's not saying that for any set $B$, there is a $\phi$ s.t. .... That may be true, but the order of quantifiers matters; the axiom doesn't say that you can swap them. It only says that, again: for any predicate there is a set containing all and only the objects that belong to that predicate's extension.

Peter Smith talked about the alternative being trivial if anything exists, so I won't comment on that.

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Recall, by definition of the empty set, $\forall x\;x \notin \emptyset$.

So trivially, for any $\varphi$, $\forall x(x \in \emptyset \to \varphi(x))$.

So, still trivially, for any $\varphi$, $\exists B\forall x(x \in B \to \varphi(x))$, just so long as the empty set exists. Not enough, then, to give you a set as the extension of $\varphi$, so not what you want!

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  • $\begingroup$ Thank you for the clarification $\endgroup$ – James May 2 '14 at 21:05

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