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My understanding is that the naturals may be constructed via algebraic terms as per Peano axioms and they may also be constructed based on set theory. I would like to improve my intuitions regarding set theoretical construction approaches. On the algebraic side, I believe I can intuit the successor function adding a multiplicative identity element to an empty set or to a set containing other instances of that element. With set theory, I have difficulty intuiting the meaning of

$$1 := 0 \cup \{0\} = \{0\} = \{\emptyset\}\quad \text{or}\quad 2 := 1 \cup \{1\} = \{0, 1\} = \{\emptyset, \{\emptyset\}\}.$$

I am able to recognize the nesting of prior/lesser sets within those of a given nonzero number but my newb intuition tells me that the constituent elements of those sets are a whole lot of nothing, which makes it difficult for me to intuitively recognize them as representing differing quantities of something. Any words of advice or recommended resources would be much appreciated.

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    $\begingroup$ Each natural number $n$ is represented by a set with $n$ elements; and what easier $n$ elements to choose than the $n$ natural numbers that precede $n$? $\endgroup$ – Lord Shark the Unknown Oct 28 '18 at 14:19
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    $\begingroup$ To add on to @LordSharktheUnknown 's comment, the nice thing about this definition is that $n\leq m \Longleftrightarrow n\subseteq m$. $\endgroup$ – Don Thousand Oct 28 '18 at 14:29
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    $\begingroup$ To add on to @RushabhMehta's comment: $n\subsetneq m \Longleftrightarrow n\in m$. $\endgroup$ – Le Anh Dung Oct 28 '18 at 14:36
  • $\begingroup$ There is a very careful and accessible construction of $\mathbb N$ and from there, the class Ord, in Elements of Set Theory by Herbert B. Enderton. $\endgroup$ – Matematleta Oct 28 '18 at 14:58
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Let me preface this by saying that I am an analyst, and not an expert in set theory or mathematical foundations. The answer that I am about to give is based on an undergraduate class that I took maybe a decade ago, so if anyone with an actual background in foundational set theory wants to correct my errors, I am more than happy to defer to them.

That being said, my feeling about the construction is as follows: the axioms of $\mathsf{ZFC}$ (or any other axiom system) tell us how objects should behave, but it is not obvious that those axioms actually provide enough structure to model the objects that we are interested in. In particular, it is not immediately obvious that $\textsf{ZFC}$ is sufficient to model our intuition about counting. Hence the set theoretic construction of the natural numbers is meant to give a model within the axiom system. Once we know that such a model exists, we can largely get on with our lives.

Depending on precisely how the axioms are stating, $\textsf{ZFC}$ doesn't give us many sets to work with. Indeed, the only set that is required in order to start building things up is the empty set–existence of the empty set can either be taken as an axiom, or follows from the Axiom of Schema Specification. From the empty set, we want to be able to build up the natural numbers.

To do this, we need a starting place (zero), and a way of getting from one set to another (a successor operation). The usual way to do this is as you described, i.e. we define $$ 0 := \varnothing \qquad\text{and}\qquad s(n) := n \cup \{n\}.$$ The definition of zero is, I think, relatively easily understood. The successor operation is a little more opaque. The essential idea is that, given a natural number $n$, we want some kind of set theoretic construction which gives the "next" natural number $s(n)$. To do this, we have to construct a new set out of whole cloth which is distinct from all of the other sets thus far constructed. The way we choose to do this is to consider a set which contains two elements: the set $n$ itself, and the set $\{n\}$, which is a set containing only one element.

The two sets $$ 0 = \varnothing \qquad\text{and}\qquad 1 = \{\varnothing\} $$ look like a whole lot of nothing, but they are actually quite distinct. The set containing the empty set–$\{\varnothing\}$–is an element of 1, but not an element of 0. Perhaps part of the problem here is the distinction between the ideas of "subset" and "is an element of." The empty set is a subset of any set. Hence $$ \varnothing \subseteq 0. $$ On the other hand, the empty set is not an element of 0. That is, $$ \varnothing \not\in \varnothing, \qquad\text{though}\qquad \varnothing \in \{\varnothing\} = 1. $$ The construction of the natural numbers depends on understanding how this nesting works, and noting that a set with no elements is distinct from the set which contains a set which has no elements. Think of $\varnothing$ as an empty grocery bag, and $\{\varnothing\}$ as a grocery bag containing an empty grocery bag. These are two different objects: the bag with a bag in it isn't empty!

In general two sets $n$ and $n \cup \{n\}$ are distinct: we will always have $n \in s(n)$, but $n\notin n$. This also has the advantage of giving us an ordering that matches our intuition: $$ m \le n \iff m \subseteq n \iff m \in n. $$ For example (abbreviating notation so that we don't have to write a whole bunch of empty sets over and over again), $$ 3 = 2 \cup \{ 2 \} = (1 \cup \{ 1 \} ) \cup \{2\} = ((0 \cup \{0\}) \cup \{1\}) \cup \{2\} = \{ 0, 1, 2 \}. $$ Note that $2 \in 3$, and also that $2 = \{0,1\} \subseteq 3$. The advantage of this scheme is that the cardinality of the set $n$ is equal to the number of elements contained in $n$, where "cardinality" has a rigorous mathematical definition, and "the number of elements contained in" matches our intuitive notion.

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  • $\begingroup$ The distinction between zero and one appears to be between the empty set and a set containing only the empty set. I recognize that the empty set contains no elements; it is less apparent to me how the empty set constitutes an element. Don't all sets contain the empty set? If all sets contain the empty set, does that mean that a set _A_containing the empty set and only one other element contains a total of two elements? $\endgroup$ – bblohowiak Oct 29 '18 at 18:29
  • $\begingroup$ Every set contains the empty set as a subset. That is, $\varnothing \subseteq A$ for any set $A$. What is not true is that the empty set is an element of every set. For example, $\varnothing \not\in \{\spadesuit,\heartsuit,\diamondsuit,\clubsuit\}$. What is true is that $\varnothing \in \{\varnothing\} =: 1$. $\endgroup$ – Xander Henderson Oct 29 '18 at 22:19
  • $\begingroup$ I've edited the answer to try to address this point a little more clearly. $\endgroup$ – Xander Henderson Oct 29 '18 at 22:25

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