# Why Bourbaki's epsilon-calculus is not suitable for set theory?

Does anybody shed light on what is A. R. D. Mathias' idea that Bourbaki's $$\tau$$-calculus (Logically the same as Hilbert's $$\varepsilon$$-calculus) is not suitable for set theory, especially because of incompleteness of set theory? I have fully read Bourbaki's Theory of Sets and Mathias' two papers on bourbaki's system (1, 2), but still cannot find any "real" logic flaw.

Especially, Mathias mentioned that $$\tau$$-calculus works perfectly on completed system like models of set theory, but works problematically on incomplete system like ZFC. But why? He cited many early publications of Bourbaki members, in which they misused unprovable as false. But at least in 1970 version of Theory of Sets, there is no such misusing.

For my understanding, $$\tau$$-operator works like FIND function in computer programming languages. It will accept property and give a return value. If there is such an element satisfied the property, it will give one correct element. If no such element exists, it will return a certain element (it is still a real element, we cannot say it has no meaning at all.). The quantifier $$\exists xR(x):=R(\tau_x(R))$$ is actually the verification of the returned element.

Also, Mathias says in $$\tau$$-calculus, we cannot discuss the concept of proper classes because they become equal. But, as we known, in ZFC, proper classes are not legal objects and can only be discussed in metalogic.

What is the connection of incompleteness and inproperly use of $$\tau$$-calculus in set theory? Is it only because Mathias' dissatisfaction on Bourbaki members' early misunderstand of incompleteness giving a negative impact on Set Theorist?

• Why call it Bourbaki's if it's due to Hilbert? the choice of Greek letter is unimportant.
– YCor
Nov 17, 2019 at 11:30
• The systems are not absolutely the same, but the principles are the same. Nov 17, 2019 at 11:54
• I'm just trying to understand your question, I'm not the right person to answer. As far as I know, set theory as it stands now has circumvented Bourbaki's foundations and this is rather history of math.
– YCor
Nov 17, 2019 at 12:07
• One reason that Mathias gives -- I do not recall which paper -- is that Bourbaki doesn't honestly use the $\tau$-calculus because it is wildly convoluted and impractical. Ah, here it is: dpmms.cam.ac.uk/~ardm/inefff.pdf The length of the term that defines $1$ is $4,523,659,424,929$. Nov 17, 2019 at 12:23
• @zqfmath But ZFC and Bourbaki's system are not equally cumbersome. In particular, one key feature of ZFC is that it's easy to convince mathematicians that the formalization of mathematics into ZFC can in fact be done (and indeed that they themselves could do it if they wanted to). This does rely on a certain conceptual simplicity at all levels. Nov 17, 2019 at 17:52

One model-theoretic problem with Bourbaki's system is that, for example, the $$V_{\alpha}$$'s, $$H_{\alpha}$$'s and $$L_{\alpha}$$'s are not closed under $$\tau$$ in general - it is not sure even if $$1\in V_{\alpha}$$, for a given infinite ordinal $$\alpha$$. The usual stuff on reflection and models of set theory becomes compromised, because closure under $$\tau$$ must now be taken into account in the satisfaction relation. Bourbaki's $$ZF$$ is also not conservative over $$ZF$$ because the $$\tau$$ symbol is allowed in the comprehension scheme (hence the axiom of choice becomes a theorem).
• I do not really understand this answer, but it seems to me that Bourbaki's axiom scheme UB might be relevant. It says that $\tau$ always chooses inside the smallest possible (Grothendieck) universe. So, if necessary, the problems mentioned here may be overcome in a similar way. Nov 17, 2019 at 20:10
• This technical axiom - which, as I recall, is not given in the théorie des ensembles, maybe you are referring to SGA IV - is given exactly to guarantee that the Grothendieck universes are closed under $\tau$, so that they are models of Bourbaki's set theory. Technically it solves a particular case of the model-theoretic problem that I have mentioned - the case in which $\alpha$ is strongly inaccessible - but it will not work in all cases. Nov 18, 2019 at 0:19