Definition 1: For any set $a$ , its successor $a^+=a\cup \{a\}$.

Informally , we want to construct natural numbers such that :

Definition 2: A natural number is a set that belongs to every inductive set.

Then we can construct a set $\omega$ whose members are exactly the natrual numbers . $\{x | x \text{ belongs to every inductive set } \}$
The discussion above was in Herbert B.Enderton's book . However , I did not see how to make a connection between $\{0,1,2,3,... \}$ and the natural number we define above .

My attempt :
$(1)$ $\omega$ is inductive , and is a subset of every other inductive set .
$(2)$ $\omega$ is the smallest inductive set . Every inductive subset of $\omega$ is concides with $\omega$ .
$(3)$ If we can prove $N=\{\emptyset,\emptyset^+,\emptyset^{++},\emptyset^{+++}, ... , ... \}$ is actually a set , then $N=\omega$ since $N$ is inductive and every member $x \in N$ also belongs to $\omega$ .

My question :
I want to define $N$ as $$\{x\in \omega | x \text{ is finite times successor of }\emptyset \}$$ However , the question happens when we want to define "finite times" . Although we can define finite by members belong to $\omega$ , how can we define finite times while we did not define even one number in our usual sense ?

  • $\begingroup$ Try to define recursively using the properties of an inductive set. The set of all natural number is unique up to a morphism. So I do not think Definition 2 is well defined without mentionning some kind of maps. $\endgroup$ – Zongxiang Yi Jun 20 at 6:08
  • $\begingroup$ For further investigations into philosophy, the OP can study and compare three ways (including the one found in Enderton's book) of constructing ω; see sections $\quad$ Extracting the natural numbers from the infinite set $\quad$ and $\quad$ An apparently weaker version $\quad$ in the wikipedia article Axiom of infinity. $\quad$ en.wikipedia.org/wiki/Axiom_of_infinity (+1) for philosophical inquiries! $\endgroup$ – CopyPasteIt Jun 20 at 11:19
  • $\begingroup$ In Whitehead and Russell's PM, inductive numbers are defined as ancestors of $0$ with respect to the relation $+1$; ancestral relation is defined in terms of hereditary class; $0$ is defined as the cardinality of an empty set. See Section E of Principia Mathematica, 1910. $\endgroup$ – George Chen Jun 22 at 15:10

You cannot prove $N=\{0,0^+, 0^{++}, \ldots\}$ is a set because the right hand side is not even a definition, merely suggestive notation. Your attempt of defining a natural number as something that takes the form of a finite number of successors applied to $0$ fails for the reasons you are beginning to suspect: you haven't defined what it means to be a finite number yet, let alone what it means to have said form, and such a thing seems rather hopeless without first having a definition of what it means to be a natural number.

What we might be tempted to do is "write down" an infinite disjunction: $$ \mathbb N = \{x: x =0\lor x=0^+\lor x=0^{++}\lor\ldots\}.$$ This would be a more logic-oriented approach to the intuitive "definition". However, infinite formulas are not allowed in the first order logic that underlies set theory. There are good reasons why we stick to first order logic, but I won't argue it here... I'll just note that making this definition requires we reason about completed infinities, which might be uncomfortably circular to a person approaching with a foundationalist mindset.

So we need to be somewhat less direct in our definition, and Enderton gives the most common approach. We define the notion of an inductive set as one that contains $0$ and is closed under the successor function, and then, crucially, we assume that an inductive set exists (this is the axiom of infinity, which, as its name suggests, is required for there to be any infinite sets at all). Then we define a natural number as a set belonging to every inductive set, and hence the set of natural numbers is the intersection of all the inductive sets.

(If there are no inductive sets, this definition does not work as intended since every set is vacuously a member of every inductive set. The set of natural numbers cannot be defined. However, the property of being a natural number can still be defined, but one needs to use a different definition that can be most succinctly phrased in terms of ordinals: a natural number is an ordinal that is not greater than or equal to any limit ordinal. CopyPasteIt's link has something similar that will work as well.)

The fact that we take the smallest possible inductive set is what corresponds to the idea that the set only contains zero and its successors, i.e. the only things that we need to be there in order to have an inductive set. However, we cannot hope to prove something of the form $\forall x\in\mathbb N(x=0\lor x=1\lor\ldots)$... as I remarked above, we cannot even express this notion in our language... if we could have we would have probably defined it this way.

So there's a reason Enderton defines the set of natural numbers as the intersection of all inductive sets rather than as $"\{0,1,2,\ldots\}"$ or as the set of all sets that can be obtained from $0$ by a finite number of applications of the successor function: this definition works in the desired framework and the latter two don't.

An earlier version of this answer made some remarks about non-standard models that may have misled you into thinking that somehow these naturals we define in set theory are not the "real" natural numbers. Make no mistake: what Enderton is doing here is giving a rigorous definition of the natural numbers within the framework of set theory. (And we can also define all the usual structure, arithmetic, etc.) The intention is to make precise the intuitive notion and also to unify with any number of other mathematical concepts that can also be encoded in ZF. So this set is intended to be the natural numbers for all intents and purposes.

(This is not the only way of looking at this: nobody says we have to use set theoretical foundations. Moreover, the concept of the natural numbers also has its own effective axiom system (PA or second order variants thereof) that we can use to study arithmetic and analysis in isolation. What is the 'real' natural numbers isn't really a sharp or in my opinion meaningful question.)

  • $\begingroup$ Very appreciate for your help . There still has one thing confused me : It seems that throughout Enderton's book , "natural numbers" , "integers" and "real numbers " are not the same as those concept in calculus or analysis. So how to apply the theorem which we proved in set theory such as Induction principle to other field ? $\endgroup$ – J.Guo Jun 20 at 7:12
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    $\begingroup$ I strongly suggest that you explicitly state that we can define $\mathbb{N}$ as the intersection of all inductive sets in ZFC only because we have at least one inductive set to begin with, which is given by the axiom of infinity. Indeed, the sole purpose of that axiom of ZFC is to enable to construction of the natural numbers in this manner. $\endgroup$ – user21820 Jun 20 at 14:07
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    $\begingroup$ @J.Guo What is “the set $\mathbb N$ which we usually use in analysis?” $\endgroup$ – spaceisdarkgreen Jun 20 at 15:21
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    $\begingroup$ @J.Guo: The point is that the natural numbers $\mathbb{N}$ is merely an abstract structure that as a whole together with arithmetic operations on it satisfy certain properties. Did you follow the link I provided you? It gives all the properties that we expect it to satisfy, whether in real analysis or in any other mathematical field. Whether or not such a structure exists is another question. Either you assume it exists (as in an axiomatic treatment of real analysis), or you make other assumptions (such as ZFC axioms) that allow you to prove that such a structure exists. $\endgroup$ – user21820 Jun 20 at 15:24
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    $\begingroup$ @J.Guo But that is no definition at all. You say it “means” $\{0,1,2,\ldots\}$ but what does that expression mean? Those ellipses are pretty vague. Of course we all know what it’s supposed to mean intuitively, but we need a precise definition based on the rules of our framework. $\endgroup$ – spaceisdarkgreen Jun 20 at 16:26

You can't define the set $\{\emptyset, \emptyset^+, \emptyset^{++}, \ldots\}$. If we can construct a set $X$ with the property $$\emptyset^{(n)}\in X \text{ for every natural number(external) } n,$$ then the compactness theorem provides that there might exist an element of $X$, which is not $\emptyset^{(n)}$ for every $n$.


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