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Thu, 24 Feb 2022
More about the axioms of infinity and the empty set
[ Content warning: inconclusive nattering ] Yesterday I discussed how one can remove the symbol !!\varnothing!! from the statement of the axiom of infinity (“!!A_\infty!!”). Normally, !!A_\infty!! looks like this: $$\exists S (\color{darkblue}{\varnothing \in S}\land (\forall x\in S) x\cup\{x\}\in S).$$ But the “!!\varnothing!!” is just an abbreviation for “some set !!Z!! with the property !!\forall y. y\notin Z!!”, so one should understand the statement above as shorthand for: $$\exists S (\color{darkblue}{(\exists Z.(\forall y. y\notin Z)\land (Z \in S))} \\ \land (\forall x\in S) x\cup\{x\}\in S).\tag{$\heartsuit$}$$ (The !!\cup!! and !!\{x\}!! signs should be expanded analogously, as abbreviations for longer formulas, but we will ignore them today.) Thinking on it a little more, I realized that you could conceivably get into big trouble by doing this, for a reason very much like the one that concerned me initially. Suppose that, in !!(\heartsuit)!!, in place of $$\exists Z.(\color{darkgreen}{\forall y. y\notin Z})\land (Z \in S),$$ we had $$\exists Z.(\color{darkred}{\forall y. y\in Z\iff y\notin y})\land (Z \in S).$$ Now instead of !!Z!! having the empty-set property, it is required to have the Russell set property, and we demand that the infinite set !!S!! include an element with that property. But there is provably no such !!Z!!, which makes the axioms inconsistent. My request that !!\varnothing!! be proved to exist before it be used in the construction of !!S!! was not entirely silly. My original objection partook of a few things:
I believe that expanding abbreviations as we did above addresses this issue adequately.
“Meaningful” is the wrong word here. I'm willing to agree that the defined symbol necessarily has a sense. But without an existence proof the symbol may not refer to anything. This is still a live issue, because if the symbol doesn't denote anything, your axiom has a big problem and ruins the whole theory. The axiom has a sense, but if it asserts the existence of some derived object, as this one does, the theory is inconsistent, and if it asserts the universality of some derived property, the theory is vacuous. I think embedding the existence claim inside another axiom, as is done in !!(\heartsuit)!!, makes it easier to overlook the existence issue. Why use complicated axioms when you could use simpler ones? But technically this is not a big deal: if !!Z!! doesn't exist, then neither does !!S!!, and the axioms are inconsistent, regardless of whether we chose to embed the definition of !!Z!! in !!A_\infty!! or leave it separate. One reason to prefer simpler axioms is that we hope it will be easier to detect that something is wrong with them. But set theorists do spend a lot of time thinking about the consistency of their theories, and understand the consistency issues much better than I do. If they think it's not a problem to embed the axiom of the empty set into !!A_\infty!!, who am I to disagree? [Other articles in category /math] permanent link |