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Many presentations of axiomatic set theory contain an error

[ Content warning: highly technical mathematics ]

[ Addendum 20220223: Also, I was mistaken. ]

I realized recently that there's a small but significant error in many presentations of the Zermelo-Frankel set theory: Many authors omit the axiom of the empty set, claiming that it is omittable. But it is not.

The overarching issue is as follows. Most of the ZF axioms are of this type:

If !!\mathcal A!! is some family of sets, then [something derived from !!\mathcal A!!] is also a set.

The axiom of union is a typical example. It states that if !!\mathcal A!! is some family of sets, then there is also a set !!\bigcup \mathcal A!!, which is the union of the members of !!\mathcal A!!. The other axioms of this type are the axioms of pairing, specification, power set, replacement, and choice.

There is a minor technical problem with this approach: where do you get the elements of !!\mathcal A!! to begin with? If the axioms only tell you how to make new sets out of old ones, how do you get started? The theory is a potentially vacuous one in which there aren't any sets! You can prove that if there were any sets they would have certain properties, but not that there actually are any such things.

This isn't an entirely silly quibble. Prior to the development of axiomatic set theory, mathematicians had been using a model called naïve set theory, and after about thirty years it transpired that the theory was inconsistent. Thirty years of work about a theory of sets, and then it turned out that there was no possible universe of sets that satisfied the requirements of the theory! This precipitated an upheaval in mathematics a bit similar to the quantum revolution in physics: the top-down view is okay, but the most basic underlying theory is just wrong.

If we can't prove that our new theory is consistent, we would at least like to be sure it isn't trivial, so we would like to be sure there are actually some sets. To ensure this, the very least we can get away with is this axiom:

!!A_S!!: There exists a set !!S!!.

This is enough! From !!A_S!! and specification, we can prove that there is an empty subset of !!S!!. Then from extension, we can prove that this empty subset is the unique empty set. This justifies assigning a symbol to it, usually !!\varnothing!! or just !!0!!. Once we have the empty set, pairing gives us !!\{0,0\} = \{0\} = 1!!, then !!\{0, 1\} = 2!! , and so on. Once we have these, the axioms of union and infinity show that !!\omega!! is a set, then from that the axiom of power sets gets us uncountable sets, and the sky is the limit. But we need something like !!A_S!! to get started.

In place of !!A_S!! one can have:

!!A_\varnothing!!: There exists a set !!\varnothing!! with the property that for all !!x!!, !!x\notin\varnothing!!.

Presentations of ZF sometimes include this version of the axiom. It is easily seen to be equivalent to !!A_S!!, in the sense that from either one you can prove the other.

I wanted to see how this was handled in Thomas Jech's Set Theory, which is a standard reference text for axiomatic set theory. Jech includes a different version of !!A_S!!, initially given (page 3) as:

!!A_∞!!: There exists an infinite set.

This is also equivalent to !!A_S!! and !!A_\varnothing!!, if you are willing to tolerate the use of the undefined term “infinite”. Jech of course is perfectly aware that while this is an acceptable intuitive introduction to the axiom of infinity, it's not formally meaningful without a definition of “infinite”. When he's ready to give the formal version of the axiom, he states it like this:

$$\exists S (\varnothing \in S\land (\forall x\in S) x\cup\{x\}\in S).$$

(“There is a set !!S!! that includes !!\varnothing!! and, whenever it includes some !!x!!, also includes !!x\cup\{x\}!!.” (3rd edition, p. 12))

Except, oh no, “!!\varnothing!!” has not yet been defined, and it can't be, because the thing we want it to refer to cannot, at this point, be proved to actually exist.

Maybe you want to ask why we can't use it without proving that it exists. That is exactly what went wrong with naïve set theory, and we don't want to repeat that mistake.

I brought this up on math Stack Exchange and Asaf Karagila, the resident axiomatic set theory expert, seemed to wonder why I complained about !!\varnothing!! but not about !!\{x\}!! and !!\cup!!. But the issue doesn't come up with !!\{x\}!! and !!\cup!!, which can be independently defined using the axioms of pairing and union, and then used to state the axiom of infinity. In contrast, if we're depending on the axiom of infinity to prove the existence of !!\varnothing!!, it's circular for us to assume it exists while writing the statement of the axiom. We can't depend on !!A_∞!! to define !!\varnothing!! if the very meaning of !!A_∞!! depends on !!\varnothing!! itself.

That's the error: the axioms, as stated by Jech, are ill-founded. This is a little hard to see because of the way he prevaricates the actual statement of the axiom of infinity. On page 8 he states !!A_\varnothing!!, which would work if it were included, but he says “we have not included [!!A_\varnothing!!] among the axioms, because it follows from the axiom of infinity.”

But this is wrong. You really do need an explicit axiom like !!A_\varnothing!! or !!A_S!!. As far as I can tell, you cannot get away without it.

This isn't specifically a criticism of Jech or the book; a great many presentations of axiomatic set theory make the same mistake. I used Jech as an example because his book is a well-known authority. (Otherwise people will say “well perhaps, but a more careful writer would have…”. Jech is a careful writer.)

This is also not a criticism of axiomatic set theory, which does not collapse just because we forgot to include the axiom of the empty set.

[ Addendum 20220223: As could perhaps have been predicted, I was mistaken. Details here. Thanks to Math SE user Eike Schulte for explaining my error in a way I could understand. ]

[ Addendum 20220224: A bit more about this. ]

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