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This a very elementary question, but I am quite new to set theory. I came across this quote on the Wikipedia page for "Hereditary Set".

In formulations of set theory that are intended to be interpreted in the von Neumann universe or to express the content of Zermelo–Fraenkel set theory, all sets are hereditary, because the only sort of object that is even a candidate to be an element of a set is another set.

Is this quote saying that all sets in ZFC are hereditary sets? How does one then construct a set such as ℝ?

Thank you

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    $\begingroup$ I think this question has been asked before, but: yes, in $\mathsf{ZFC}$ everything is a "pure set." Things which we don't think of as pure sets (like $\mathbb{R}$) are "implemented" in this framework, similarly to how an algorithm is implemented in a programming language. For instance, we usually implement the natural numbers as the "finite ordinals," the rationals as equivalence classes of ordered pairs of those, and the reals as particular subsets of those. We may also prove an appropriate "implementation-independence" result: $\endgroup$
    – Noah Schweber
    Sep 30, 2020 at 17:21
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    $\begingroup$ for example, in the case of $\mathbb{R}$ that any two complete Archimedean ordered fields are isomorphic, in the case of $\mathbb{C}$ that any two algebraically closed fields of characteristic $0$ and transcendence dimension $2^{\aleph_0}$ are isomorphic, and so forth. (On more subtle issues around implementation-dependence, see the discussion here.) $\endgroup$
    – Noah Schweber
    Sep 30, 2020 at 17:22
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    $\begingroup$ In the formal language used in ZFC there is no word "set". There are merely assertions about things that do or do not exist. $\endgroup$
    – DanielWainfleet
    Sep 30, 2020 at 17:24
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    $\begingroup$ @DanielWainfleet That's a good point - "everything is a set" is something we say from outside the theory, to informally distinguish $\mathsf{ZFC}$ from other reasonable ways of providing a "background theory" for mathematics. $\endgroup$
    – Noah Schweber
    Sep 30, 2020 at 20:07
  • $\begingroup$ @NoahSchweber. I remember having some confusion about it. In a graduate-level course in Set Theory, in a discussion with another student, he said "Set. Set. Set. Everything's a set." $\endgroup$
    – DanielWainfleet
    Oct 1, 2020 at 14:10

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To answer your first question, I would refer you to DanielWainfleet and Noah Schweber's comments to your question, or Andrés E. Caicedo's comment to this answer.

As to the second, "how do you make $\mathbb{R}$ in ZFC?", you repeatedly take equivalence classes of smaller sets.

First, you begin with the empty set. You define the successor operator as $S(x) = x \cup \{x\}$. So, $0 = \varnothing$, $1 = S(0)$, $2 = S(1)$, etc. This gives you $\mathbb{N}$. The next step then is to define $\mathbb{Z}$, which will be "equivalence classes of $\mathbb{N} \times \mathbb{N}$" where $(a, b) \sim (c, d)$ if $a + d = c + b$. After that, we define $\mathbb{Q}$ by equivalence classes of $\mathbb{Z} \times \mathbb{Z}$, where $(a, b) \sim (c, d)$ if $ad = bc$.

Constructing $\mathbb{R}$ now that we have $\mathbb{Q}$ is more complex, and is typically done by considering equivalence classes of Cauchy sequences, or by considering Dedekind cuts. You can view these constructions in all of their gory details by reading the appendix to chapter 1 of Rudin's "Principles of Mathematical analysis" and by doing the final exercises in chapter 3 of the same book.

I have skipped a lot of details along the way of course, most notably defining what, exactly, addition and multiplication are in $\mathbb{N}$ and $\mathbb{Z}$, and that these satisfy the typical axioms we expect of addition and multiplication.

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    $\begingroup$ You can construct $\mathbb R$ directly from naturals via quasi-homomorphisms, makes it maybe a little easier path for set definition. $\endgroup$
    – zwim
    Sep 30, 2020 at 17:37
  • $\begingroup$ @zwim That sounds very interesting! Could you give me a reference that I can explore? $\endgroup$
    – Duncan Ramage
    Sep 30, 2020 at 18:04
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    $\begingroup$ This one has proofs of most properties maths.mq.edu.au/~street/EffR.pdf $\endgroup$
    – zwim
    Sep 30, 2020 at 18:16
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    $\begingroup$ That all sets are hereditary has nothing to do with regularity. It just means that every element of a set is a set, which follows from extensionality. $\endgroup$
    – Andrés E. Caicedo
    Sep 30, 2020 at 20:03
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    $\begingroup$ If you think that was embarrassing:.. Paul Cohen, in a lecture presenting an over-view of his thoughts & his research in ZFC, forgot that the existence of $\omega_1$ does not require the Axiom of Choice. $\endgroup$
    – DanielWainfleet
    Oct 1, 2020 at 14:22

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