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This is how the topology textbook I'm reading (Munkres) defines integers:

A subset of the real numbers is "inductive" if it contains 1 and $1+x$ for all $x$ in the subset. The intersection of all inductive subsets of the reals is the set of positive integers.

Why take this route involving the intersection of so many sets? I could define the positive integers given reals as $1$ along with any sum of positive integers and get the same set much more easily.

Joe
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    Along with any sum of ... what? "of positive integers"? Isn't that circular? – bof Mar 17 '22 at 22:31
  • Depends on your axioms. A classic definition of the reals in pure set theory is in terms of inductive sets (as a subset of the entire universe, not just the reals). Depending on how Munkres defines reals, defining the integers in terms of them could be circular. It's probably just done to give you tangential exposure to inductive sets as a concept. But I've not read the book. – preferred_anon Mar 17 '22 at 22:32
  • @bof If that's circular, wouldn't Peano's axioms (zero is a natural number and the successor to a natural number is a natural number) be circular too? – Retracted Mar 17 '22 at 22:34
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    Circular definitions are bad. I don't know about "circular axioms" – bof Mar 17 '22 at 22:36
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    @bof If you could explain in greater detail how generator-style "x along with all results of using a certain operator on x any number of times" definitions are impermissible, that would make a great answer. Maybe because although it defines things that are positive integers, it doesn't mention what numbers aren't. – Retracted Mar 17 '22 at 22:39
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    First, you’ve only defined positive integers here. But this is a usual way of defining the integers from the reals, if you are going in that direction. More traditional to define the reals in terms of the integers. – Thomas Andrews Mar 17 '22 at 22:48
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    I wouldn't call it "impermissible". If you're working "semi-formally", whether you can "just do it" or whether some justification is needed depends on what "logical basis" you're starting from. I don't have your textbook (which you didn't identify) but from a certain starting point, the argument in your textbook is the natural way of justifying your "generator-style" definition. (I won't try to post an answer, that is better left to a logician. My understanding of logic is rudimentary.) – bof Mar 17 '22 at 22:49
  • But your definition is very hard to make formal - it requires an inductive definition of the set, but if you start with the real numbers, you also might not start knowing about induction. We can define the positive integers using the intersection using just basic set theory. It is true, it is a bit of a long haul, but it is a common way to find the minimum set that has a property, if the property is closed under intersections. (That is, if $A_i$ has the property for all $i\in I,$ then $\bigcap_{i\in I}A_i$ has the property.) – Thomas Andrews Mar 17 '22 at 22:53
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    Likely that method is used because it is a special case of a general method of defining a subalgebra generated by certain elements (as the intersection of all subalgebras containing those elements). This will be explained in a course on (universal) algebra. See also this answer. – Bill Dubuque Mar 17 '22 at 23:03
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    This is an approach which is fairly typical among analysts: we tend to view the reals as fundamental---they can be defined axiomatically. From the real numbers, it is then possible to extract the positive integers in the manner described. – Xander Henderson Mar 17 '22 at 23:04
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    Here's another example: we can define the Gaussian integers $,\Bbb Z[i] = { a + b,i\ :\ a,b\in \Bbb Z},$ to be the intersection of all subrings of $\Bbb C$ containing both $\Bbb Z$ and $i$. This is the smallest subring of $\Bbb C$ that contains $\Bbb Z$ and $i.,$ Equational algebras like groups, rings, fields etc that are defined by operations with $\rm\color{#c00}{universal}$ axioms like $,\color{#c00}{ \forall}:! x,y!:\ xy = yx,$ are always closed under intersection, so such intersections give a universal way to define the "smallest" subalgebra generated by certain elements. – Bill Dubuque Mar 17 '22 at 23:26
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    In your case this method allows us to define $\Bbb N$ as the sub-Peano algebra of $\Bbb R$ with no generators, where a Peano algebra is a set containing $1$ and closed under the successor operation $,s(x) = x+1.,$ That's equivalent to what Munkres does. – Bill Dubuque Mar 17 '22 at 23:26
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    @Xander I don't see any answer at all to the OP's question in the linked dupe. See my prior comments for one way to answer. – Bill Dubuque Mar 17 '22 at 23:30
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    Btw, the claims that your idea is circular are incorrect. Your idea amounts to defining $\Bbb N$ as the additive semigroup of $\Bbb R$ generated by $1$, i.e. the smallest subset of $\Bbb R$ that contains $1$ and is closed under addition. As above, this is the intersection of all additive sub-semigroups of $\Bbb R$ containing $1.\ \ $ – Bill Dubuque Mar 17 '22 at 23:44
  • Comments are not for extended discussion; this conversation has been moved to chat. – Xander Henderson Mar 20 '22 at 17:00

1 Answers1

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You appear to be asking why we can't simply define $\mathbb N$ as an inductive set, i.e. a subset of $\mathbb R$ satisfying the following two axioms:

  1. $1\in\mathbb N$
  2. $(\forall x)(x\in\mathbb N\to x+1\in\mathbb N)$

The issue is that there are many sets which satisfy these two axioms. While the usual natural numbers are indeed an inductive set, so is:

  • The set of positive real numbers.
  • The set of positive real numbers unequal to $1/2$.
  • The real numbers themselves!

Therefore, we need to add a third axiom to make our definition of $\mathbb N$ workable. One possibility, which is very similar in spirit to Munkres' definition, is the following:

  1. $\mathbb N$ is the "smallest" set satisfying (1) and (2), i.e. if $X\subseteq\mathbb R$ is inductive, then $\mathbb N\subseteq X$.

However, we might find it difficult to rigorously prove that there is a subset $\mathbb N$ of $\mathbb R$ which satisfies these three axioms. To get around this issue, we could instead use Munkres' definition of $\mathbb N$ as the intersection of all inductive sets, before proving that $1\in \mathbb N$ and $(\forall x)(x\in\mathbb N\to x+1\in\mathbb N)$. Then, it immediately follows from the definition of intersections that $\mathbb N$ satisfies (3).

This is in fact exercise 25 of chapter 2 of Spivak's Calculus (4th edition), page 34.

Joe
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  • This is indeed Munkres' spirit to the extent that I understand correctly. In fact, after defining N as this intersection he goes on to prove that if A is an inductive (axioms 1 and 2) subset of N then A=N, in other words that N has no proper inductive subset. However, I a bit bothered by the fact that nowhere does Munkres prove that N is not empty. Sure the intersection is well defined but N may be empty under that definition. N might also be empty under the alternative definition you suggested. – Marcus Junius Brutus Mar 15 '24 at 18:30
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    @MarcusJuniusBrutus: If $\mathcal A$ is the collection of all inductive sets, then for every $X\in\mathcal A$, we have $1\in X$, and so $1\in\bigcap_{X\in\mathcal A}X$. Thus, the intersection is nonempty. – Joe Mar 15 '24 at 18:47
  • Silly of me. Yes, so $\mathbb{N} $ is not empty as it contains at least 1. Whether, at this point given the theorems, postulates, and definition in our disposal, we can further prove that it contains any arbitrarily high number of elements, or indeed an infinite number of elements is not clear to me. I shall think about this when I am back at the office. – Marcus Junius Brutus Mar 16 '24 at 19:36
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    @MarcusJuniusBrutus: There is a bit of difficulty here; when discussing such "basic" objects such as the natural numbers we can often run into problems of circularity. In set theory, probably the most common definition of "finite" is: a set $X$ is finite if it can be put into bijection with a proper initial segment of the natural numbers! On the other hand, in set theory, the natural numbers tend to be defined differently to how they are defined here (although there are similarities behind the approaches, and they both involve inductive sets). – Joe Mar 16 '24 at 19:51
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    @MarcusJuniusBrutus: In other words, while I'm sure you will be able to verify that $\mathbb N$ contains infinitely many elements according to the naive notion of "infinite", I think you will have a hard time proving it. If you want to be totally rigorous, then you would probably need to descend down to the level of axiomatic set theory. – Joe Mar 16 '24 at 19:53
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    You can find the details in Paul Halmos' book on set theory, for instance. (Although even Halmos omits to mention some of the logical issues that arise–see here for some indication of the issues.) – Joe Mar 16 '24 at 19:55
  • Thank you! Yes trying to be as rigorous as I can endure but I now appreciate this is about where my limits are. OTOH I think if one wanted to go full rigorous one might then opt for the alternative construction of integers (such as in Tao's analysis). Axiomatically wishing the reals into existence (as Munkres does) and then fretting about the natural numbers might be perverse... – Marcus Junius Brutus Mar 16 '24 at 21:25