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In the thread Set theories without "junk" theorems?, Blass describes the theory T in which mathematicians generally reason as follows:

Mathematicians generally reason in a theory T which (up to possible minor variations between individual mathematicians) can be described as follows. It is a many-sorted first-order theory. The sorts include numbers (natural, real, complex), sets, ordered pairs and other tuples, functions, manifolds, projective spaces, Hilbert spaces, and whatnot. There are axioms asserting the basic properties of these and the relations between them. For example, there are axioms saying that the real numbers form a complete ordered field, that any formula determines the set of those reals that satisfy it (and similarly with other sorts in place of the reals), that two tuples are equal iff they have the same length and equal components in all positions, etc.

There are no axioms that attempt to reduce one sort to another. In particular, nothing says, for example, that natural numbers or real numbers are sets of any kind. (Different mathematicians may disagree as to whether, say, the real numbers are a subset of the complex ones or whether they are a separate sort with a canonical embedding into the complex numbers. Such issues will not affect the general idea that I'm trying to explain.) So mathematicians usually do not say that the reals are Dedekind cuts (or any other kind of sets), unless they're teaching a course in foundations and therefore feel compelled (by outside forces?) to say such things.

Question: If set theorists just want to do set theory and not worry about foundations (and encodings of mathematical objects as sets), do they also work in the theory T? Or are they always regarding every object as a set?

Also, do I understand it correctly that it's hard to actually formalize the syntax of the theory T, because of the many types and connotations of natural language involved? But then, what's "first-order" about T, if T is communicated through natural language?

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    $\begingroup$ No, but they usually have some while discussing GCH (worst math joke ever?) $\endgroup$
    – guest
    Commented Oct 27, 2016 at 0:46
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    $\begingroup$ Another joke -- yes they work in T, the modal logic with axiom $\Box A\rightarrow A $, since forcing implies truth... $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Oct 27, 2016 at 7:45
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    $\begingroup$ I'd say very little of what set theorists do is actually about foundations, unless you understand "foundations" in a very broad sense (for example, when a set theorist proves that certain open problem from another area of mathematics is undecidable in $\mathsf{ZFC}$, that doesn't count as "foundations" to me)... $\endgroup$
    – David Fernandez-Breton
    Commented Oct 27, 2016 at 17:23
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    $\begingroup$ @DavidFernandezBreton Really? My view is that such a result would be a very welcome result in foundations. What counts as foundations for you, if not that? $\endgroup$
    – Joel David Hamkins
    Commented Oct 27, 2016 at 21:22
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    $\begingroup$ @JoelDavidHamkins , for me "foundations" denotes more the purely "philosophical" reflection about whether, for example, $\mathsf{ZFC}$ is an appropriate system to work as the foundations of math, or whether our definition of "natural number" agrees with some pre-existent intuitive notion of number, and so on. The moment you prove a theorem about something like abelian groups (even if such a theorem is an independence result), you're no longer doing foundations of math: you're doing math. (this is just part 1 of the comment) $\endgroup$
    – David Fernandez-Breton
    Commented Oct 28, 2016 at 14:00

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Caveat number 1: strictly speaking, no one actually works in the theory $T$, just as no one actually works in the theory $\mathsf{ZFC}$. Mathematicians work by means of carefully used natural language and not within a formal system. Formal systems are formulated as approximations that try to model what mathematicians actually do while at work. Now to address the question, with the above caveat in mind, are we always regarding every object as a set? Not necessarily always, just sometimes. The point is that $\mathsf{ZFC}$ and $T$ are bi-interpretable, so you can switch between both viewpoints at will without that changing the stuff that you can prove (and even better: both $T$ and $\mathsf{ZFC}$ are just approximations to what we actually do, so we can just do math as usual, and not worry about these nuances, and whatever it is that we're doing can in theory be translated to the formal system of your choice).

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    $\begingroup$ A small detail: the people who are typing things into Coq are not using ZFC, they are using type theory. As for you not caring whether $\mathbb{Q}$ is a subset of $\mathbb{R}$, at some level you're right that you shouldn't have to care, but on another you're running the danger of an inconsistency. They used to not care whether every function can be written as a power series, they just did it. So, if you really do not want to care, you should change your preferred foundation from ZFC to one that actually lets you not care about this sort of thing. That would be Univalent Foundations. $\endgroup$
    – Andrej Bauer
    Commented Oct 27, 2016 at 6:43
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    $\begingroup$ @AndrejBauer Why do you think ZFC requires you to care about whether $\mathbb{Q}\subset\mathbb{R}$? It seems to me that ZFC is fully compatible with a structuralist perspective about such things. My impression is that set theorists generally are as structuralist about such matters as other kinds of mathematicians. $\endgroup$
    – Joel David Hamkins
    Commented Oct 27, 2016 at 12:14
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    $\begingroup$ I am serious about ZFC, and if I am in a structuralist mood about $\mathbb{R}$, then my theorems will state: in every complete ordered field, blah blah blah; or every complete separable dense linear order, etc., depending on what $\mathbb{R}$ means to you. There is no need to have a particular copy of $\mathbb{R}$ identified as the official reals, and this is what structuralism amounts to. The particular constructions are used to show existence only, and it is a trivial matter to build instances showing that $\mathbb{Q}\subset\mathbb{R}$ is possible, if you want it. $\endgroup$
    – Joel David Hamkins
    Commented Oct 27, 2016 at 14:34
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    $\begingroup$ Usually, though, the structuralism goes without saying, and so one asserts a theorem about $\mathbb{R}$, with the understanding that it doesn't matter which copy of $\mathbb{R}$ you use. (In some very rare cases, it may matter---for example, if one is paying attention to interaction with the cumulative hiearchy---and in those cases one will say a little more about the requirements of the situation.) This happens also in category theory, for example, where one may want to use a copy of $\mathbb{R}$ that is available in the current universe that has been fixed, rather than arising above it. $\endgroup$
    – Joel David Hamkins
    Commented Oct 27, 2016 at 14:42
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    $\begingroup$ I view the structuralist part of the foundation as the easy part, in that it can be implemented in a high level language, simply by preferring certains kinds of theorems. Meanwhile, the value of ZFC as a foundation is the simplicity and clarity of the most fundamental existence assumptions, free of the meta-theory and mass of defined terms that often seem to complicate alternative foundations. In my opinion, this is why set-theory has led to such a successful and clarifying meta-foundational program, via forcing and large cardinals. $\endgroup$
    – Joel David Hamkins
    Commented Oct 27, 2016 at 20:45
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You address your question to set theorists, and so let me answer as a set theorist that, yes, when I think purely as a set theorist, then indeed the idea that every object is a set just goes without saying — it is a basic elementary part of the ZFC conceptual framework. One needn't ever even remark on this in an argument made to other set theorists.

For example, the cumulative $V_\alpha$ hierarchy provides an extremely rich structural background for the set-theoretic universe, which is especially informative and helpful in the analysis of various set-theoretic axioms, especially the large cardinal axioms that reach very high into this hierarchy. The picture of all objects existing as sets in the cumulative hierarchy is basically pervasive in set-theoretic arguments, and is definitely a fundamental part of the set-theoretic understanding of the mathematical universe. One may freely carry out an argument by $\in$-induction, for example, concluding that every object $x$ has a certain property, although really what one has proved is that every well-founded set has the property. In this sense, I would say that when set theorists are operating as set theorists amongst set theorists, they are not usually operating in the theory T described by Andreas, but rather in something much closer to ZFC, in a language expanded by concepts that have been defined in set theory. (As David mentioned, esentially no mathematician, including set theorists, works in a purely formal system.)

Meanwhile, however, this doesn't mean that set-theorists don't make use of type-theoretic concepts. For example, set theorists have diverse concepts of what counts as a real number, and one can commonly find various real-number concepts used in set theory, including: an element of Cantor space $r\in 2^\omega$; a subset of the natural numbers $r\subseteq\omega$; an element of Baire space $r\in \omega^\omega$; and so on. Often the algebraic field-theoretic structure of the reals is less important or relevant to set-theoretic concerns, and set theorists typically care about real numbers as: a package containing countably much information. In many set-theoretic contexts, it doesn't matter which particular real-number concept one is using, and in this sense talk about the reals reduces essentially to the use of a real-number type.

Finally, however, when set theorists communicate with other mathematicians, then of course they are naturally able to communicate in something much closer to the theory $T$ that you mentioned.

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    $\begingroup$ I'd say that the set theorist's theory $T$ has a sort for classes because classes arise as mathematical concepts in set-theoretic arguments. $\endgroup$
    – Andrej Bauer
    Commented Oct 27, 2016 at 6:46
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    $\begingroup$ To add a concrete example: one workhorse method which relies fundamentally on the “everything is built out of sets” framework is the presentation of forcing using names (or Boolean-valued sets, etc.). One can do forcing-type arguments in a more $T$-like setting, using sheaves; but I’ve never seen that approach taken in mainstream set theory. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Oct 27, 2016 at 9:28
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    $\begingroup$ @AndrejBauer Yes, set theorists do often work in theories with a sort for classes, and this is completely explicit, for example, in Gödel-Bernays set theory and in Kelley-Morse set theory. In this sense, the class sort is not really a part of the informal theory $T$, but present explicitly in the formal theories GBC and KM and the various intermediate theories such as ETR. Set theorists are usually quite explicit about their background theory regarding classes, and one could replace ZFC in my answer with GBC or KM in the two-sorted second-order language of set theory. $\endgroup$
    – Joel David Hamkins
    Commented Oct 27, 2016 at 11:44
  • $\begingroup$ Joel, What is ETR? $\endgroup$
    – Bob Solovay
    Commented Nov 2, 2016 at 1:54
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    $\begingroup$ Regarding the edit by user218210, I think my original ${}^\omega 2$ and ${}^\omega\omega$ is standard notation for the function spaces. The right-side exponentation $2^\omega$ and $\omega^\omega$ can be ambiguous because they might also refer to the cardinal or ordinal, instead of the space of sequences. But I didn't roll it back... $\endgroup$
    – Joel David Hamkins
    Commented Apr 21, 2021 at 10:07

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