In the formal language of the Zermelo–Fraenkel axioms, the axiom of infinity reads:
there is a set I (the set which is postulated to be infinite), such that the empty set is in I and such that whenever any x is a member of I, the set formed by taking the union of x with its singleton {x} is also a member of I. Such a set is sometimes called an inductive set.
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I'm commenting here on your posts in the wrong order.
ReplyDeleteNote that this formulation of infinity presumes that we have a constant whose name is the usual symbol for the empty set.
It might appear that this enables the existence of the empty set to be proven, but this is an illusion.
Why not?
Well first of all, if our language contains a constant C then we can prove "there exists x such that x = C" in first order logic, without benefit of any set theoretic axioms at all.
However, we can't prove anything about C without some axioms, so
we can't prove that C has no members.
Mentioning C in the pivotal role in the axiom of infinity stated above makes no difference, you still can't prove it has no members from the axiom of infinity, and in fact that axiom works perfectly well whatever set plays that pivotal role.
The axiom can be simplified so that it only states the existence of a non-empty set closed under succession (the function from x to x u {x}).
The axiom of infinity as stated does not prove the existence of an
empty set in a context in which that is not already provable.
Well not in any non-contrived context, it would of course in a context in which there was an axiom (or theorem) asserting that the existence of an infinite set entailed the existence of the empty set.
RBJ