5

Math people:

The title is the question. I am not a logician, and to me, it seems self-evident that any nonempty set of natural numbers has a smallest element. I am reading an analysis book that uses the Completeness Axiom of the real numbers to prove this "fact" (I put "fact" in quotes because a logician might not accept something so obvious as a fact but might require that I assume some axiom). To me, this really seems like overkill, and that you should be able to prove it using something much weaker.

EDIT: I am not asking specifically about axioms of ZFC. My question does not even mention ZFC. I highly doubt that Choice is necessary here and I suspect it may not even be helpful. I would like (i) confirmation that using the Completeness Axiom of the real numbers to prove that every nonempty set of natural numbers has a least element is massive overkill and (ii) some weaker axioms that give me the same conclusion, the weaker the better.

Stefan Smith
  • 8,192
  • 3
    You can definitely prove it from something much weaker. For example, it's enough to use the single axiom Every non-empty set of natural numbers has a smallest element. –  Aug 26 '13 at 20:13
  • The natural numbers consist of a set theoretic object. The empty set is a natural number. "Smallest element" means smallest in terms of cardinality. So, there would have to exist a natural number with the least cardinality. Cardinality gets defined in terms of functions between sets. The empty set cannot get mapped to anything, as there does not exist any elements of it such that you can map a member of the empty set to something else. Thus, the empty set comes as unanalyzable in terms of cardinality. You can't compare its cardinality to anything else. So, there is no way to say... – Doug Spoonwood Aug 27 '13 at 02:04
  • that there exists a set smallest in terms of its cardinality in terms of set theory. We can't tell if the empty set has greater cardinality or smaller cardinality than the set of the empty set usually denoted by "1". Again, the cardinality of the empty set is NOT analyzable, since it does not have any members, and you have to have a member of a set in order to have a mapping between sets. So, you can't prove this in set theory. And you can't prove it for natural numbers, since they consist of a set-theoretic object. – Doug Spoonwood Aug 27 '13 at 02:09
  • Every finite non-empty set of integers has a smallest element, but many infinite subsets don't. Completeness of the reals is sometimes introduced axiomatically (complete ordered field) but even if we do we need to verify there exists such a thing. The system does not like long strings of comments. I will delete most of mine, and suggest you also delete. That will make room for any new questions you have! – André Nicolas Aug 27 '13 at 14:32
  • @StefanSmith How does cardinality get defined? – Doug Spoonwood Aug 28 '13 at 01:46
  • 1
    With finite sets we can fully trust our intuition. Although things get tricky when the issue of Dedekind finite sets comes up. – André Nicolas Aug 29 '13 at 17:00
  • @StefanSmith No. The cardinality of a finite set consists of a measure equivalent to what you would get if you (could) count the number of members of a finite set, with each element only permitted to get counted once. The empty set does not have any elements to count. Consequently, it does not have cardinality. Otherwise, cardinality is NOT equivalent to a counting measure... you can't count zero objects. You can postulate the empty set as having cardinality zero, but you have no demonstration of such, because you need either to count the objects or show how the objects can get counted. – Doug Spoonwood Aug 29 '13 at 23:47
  • @Stefan Smith And you simply can't show how the empty set's lack of objects could get counted. So, you've only provided a definition which implies the empty set as having zero cardinality. You have no proof of such, and you can't given that one can invoke the equivalence of cardinality of finite sets with a counting measure. – Doug Spoonwood Aug 29 '13 at 23:48
  • @AndreNicolas : are you saying that if you define reals using Dedekind cuts it is difficult to prove that a finite nonempty set of reals has a smallest element? Do you need any additional axioms to prove it, or can you use arithmetic and the assumption/fact/axiom/whatever that the naturals are well-ordered? – Stefan Smith Aug 31 '13 at 15:45
  • 1
    No additional axioms needed, but we need to define an ordering on Dedekind cuts, and verify it is an order relation. Every step is quite short, but there are several. – André Nicolas Aug 31 '13 at 15:56
  • @DougSpoonwood : if you want to be formal, you can view $\varnothing$ as an injection from $\varnothing$ to any other set. So you can compare the cardinality of $\varnothing$ to that of other sets. Of course, the cardinality of $\varnothing$ is zero because $\varnothing$ has no elements. I trust Andre Nicolas when he writes that with finite sets we can fully trust our intuition. – Stefan Smith Aug 31 '13 at 17:18
  • @StefanSmith If "I" comes as an injection, then it qualifies as some function F. If F denotes a function from X to Y, then it qualifies as a subset of the Cartesian product (X x Y) := C. If C denotes the Cartesian product of sets X and Y, then C consists of the set of all ordered pairs (x, y) where x belongs to X and y belongs to Y. But, there does NOT exist any ordered pair (x, y), where X := ∅, because no x belongs to ∅. So, no Cartesian product C exists, nor does any function F exist, and consequently, no injections I from the empty set exist. Why should you blindly trust Andre Nicolas? – Doug Spoonwood Aug 31 '13 at 20:55
  • @DougSpoonwood Andre has displayed his wisdom in countless posts. Remember your basic set theory. $\varnothing$ is a subset of any set, hence you can interpret it as a function (viewing a function as a subset of the Cartesian product of two sets) between any two sets, etc. This is my last comment on this, I give up. – Stefan Smith Aug 31 '13 at 21:26
  • @StefanSmith I find it interesting that you tell me to remember my basic set theory when in fact you've shown evidence that you haven't remembered it. The Cartesian product of two sets X and Y consists of the set of all ordered pairs (x, y), where x belongs to X and y belongs to Y. But, for ∅ and Y there do not exist any ordered pairs (x, y), because no x belongs to ∅. So, NO, you can't just interpret the empty set as a function since it qualifies as a subset of a set. {(a, a), (a, b)} := A is a subset of [{a, b}x{a, b}], but A is NOT a function. There does not exist any injection from ∅. – Doug Spoonwood Sep 01 '13 at 23:33

5 Answers5

6

Most people, logicians included, will indeed agree that intuitively the claim that every non-empty subset of the natural numbers has a least element (which is knows as the well-ordering principle of the naturals) is self-evident. Another such self-evident property of the natural numbers is the principle of induction.

As is often the case with self-evident things, actually proving them can be tricky. It turns out (and it's not hard to prove directly) that the well-ordering of the naturals is equivalent to the principle of induction (which is a second order principle).

Now things become more complicated when one realizes that there exist first order models of arithmetic in which all of the Peano axioms hold as well as all the induction scheme axioms (that is, one axiom per first order formula) in which the second order principle of induction fails. That means that in such models of the naturals the well-ordering principle fails too. So things are not so simple, or self-evident, after all.

I do agree that using the completeness of the reals in order to prove the well-ordering of the naturals is a bit of an over kill as the former is less intuitively clear than the latter. But, it is instructive to see that imposing completeness on the reals (a second order property again) forces the naturals to obey the principle of induction.

Ittay Weiss
  • 79,840
  • 7
  • 141
  • 236
  • 2
    models of the natural numbers that are not well-ordered are known as nonstandard models of arithmetic, and they are quite important. And yes, if a theorem crucially used the well-ordering of the naturals then the proof (and perhaps the claim) will not be valid in a nonstandard model. – Ittay Weiss Aug 26 '13 at 22:11
  • @StefanSmith: Dear Stefan and Ittay, As one example of a result that holds in the standard model of the natural numbers, but whose status in non-standard models is currently unknown, one could take Fermat's Last Theorem. Regards, – Matt E Aug 26 '13 at 23:33
  • 1
    @StefanSmith: Dear Stefan, When people speak of FLT being true, they mean in the standard model (which is certainly the model that Fermat had in mind, and that all number theorists normally have in mind). But some theorems about the standard model are visibly proved using just the first order Peano axioms, and so actually hold true in all the non-standard models too. But FLT, although proved, is (at least not obviously) proved just using the Peano axioms. (The proof uses a huge amount of machinery and technique from various different areas of math, and as far as I know no-one is really ... – Matt E Aug 27 '13 at 00:00
  • 1
    ... sure whether those different inputs can be reworked to just rely on the Peano axioms, or whether they rely on other aspects of the standard model that (perhaps) don't hold in non-standard models.) Regards, – Matt E Aug 27 '13 at 00:01
  • @Stefan: To complement Matt E's comments, http://math.stackexchange.com/questions/213253/what-does-it-mean-for-something-to-be-true-but-not-provable-in-peano-arithmetic/ here is a thread discussing exactly the meaning of "true" in the context of arithmetics. – Asaf Karagila Aug 27 '13 at 01:20
2

The standard ordering on $\mathbb{N}$ is a well-ordering (the well-ordering principle); this is equivalent to the axiom of induction, which is one of the axioms of Peano arithmetic.

Proving that every bounded set of real numbers has a greatest lower bound does require the completeness axiom.

Kris
  • 1,829
  • This doesn't answer the question. What axioms do you need to prove the induction principle? – Git Gud Aug 26 '13 at 19:48
  • 1
    It's one of the axioms of Peano arithmetic... – Kris Aug 26 '13 at 19:49
  • PA can be built from the $\sf ZFC$ axioms. – Git Gud Aug 26 '13 at 19:50
  • 1
    What are 'the axioms'? Not everything needs to be reduced to ZFC. – Kris Aug 26 '13 at 19:50
  • Isn't the OP asking exactly for the $\sf ZFC$ axioms? After all he did tag it set theory. – Git Gud Aug 26 '13 at 19:51
  • He tagged it 'elementary set theory', which is something rather different. – Kris Aug 26 '13 at 19:51
  • 1
    @GitGud ZFC is only one theory of sets, of course. – Thomas Andrews Aug 26 '13 at 19:52
  • @ThomasAndrews It's the one that matters here. – Git Gud Aug 26 '13 at 19:52
  • @Kris He tagged it set theory, I retagged it elementary set theory and that's the proper tag for this kind of questions. – Git Gud Aug 26 '13 at 19:53
  • 1
    @GitGud: ZFC is (massive) overkill to prove that every set of naturals has a least element. In fact, ZFC is massive overkill for most things. Beware assuming that ZFC is the default framework for mathematics. – Kris Aug 26 '13 at 19:53
  • @Kris And that's a good reason to ask exactly what axioms one needs to prove the well ordering principle in $\Bbb N$. – Git Gud Aug 26 '13 at 19:55
  • 1
    to which the answer is: the principle of induction for $\mathbb{N}$. i feel like i've just walked in a big circle. – Kris Aug 26 '13 at 19:56
  • @Kris That's not an axiom. – Git Gud Aug 26 '13 at 19:56
  • 2
    @GitGud I can never tell when you are being deliberately unhelpful or just incidentally so. – Thomas Andrews Aug 26 '13 at 19:57
  • @ThomasAndrews How am I being unhelpful? If I were to ask this question, this wouldn't be an answer to me. There are nine axioms in $\sf ZFC$. Exactly which ones are needed (or at least what is a minimal subset of these axioms) to show the well-ordering principle in $\Bbb N$? This hasn't been answered and in my opinion this is the question. – Git Gud Aug 26 '13 at 19:59
  • 3
    Because you are not in ZFC, because in second order number theory, which includes sets, induction is an axiom, because you simply won't say why you consider only ZFC the way to approach this problem. You have hung a lot of bad assumptions on a questioner's tagging, which is obtuse. Again, perhaps deliberate, perhaps not. @GitGud – Thomas Andrews Aug 26 '13 at 20:01
  • @Kris : I like your answer and comments. Is Peano Arithmetic weaker than ZF (you know what I mean and I trust you not to nitpick)? – Stefan Smith Aug 26 '13 at 20:09
  • Dear Kris, Induction, as it is encoded in the Peano axioms, is just induction on first order formulas, which is (much) weaker than the full well ordering of the naturals, or than induction for arbitrary subsets of $\mathbb N$. (See Ittay Weiss's answer.) Regards, – Matt E Aug 26 '13 at 20:22
2

One reason this question is a bit tricky to answer is the following: either you are working with the natural numbers directly, and then induction, or equivalently well-ordering, are basic properties that you might well take as axioms. Otherwise, the natural numbers are appearing as a subset of some other set, and then it's not so clear what other contexts there are that are simpler than $\mathbb N$ considered as a subset of $\mathbb R$.

That being said, here is one axiom that might suit you (although depending on your view-point it might be circular): there are only finitely many natural numbers less than any given natural number. (The circularity might enter if you adopt certain definitions of finite.)

This implies well-ordering, i.e. the existence of a least element for any non-empty subset of $\mathbb N$: if $S \subset \mathbb N$ is non-empty, we can choose $n \in S$. If $n$ is a least element, we are done; otherwise, there are only finitely many elements of $S$ that are less than $n$, and since the ordering on $\mathbb N$ is total, we can find a least element of $S$ among them.

Matt E
  • 123,735
1

One can use the Induction Principle in order to prove the well ordering principle (that states that every non empty set of natural numbers has a smallest element). For example, you can see T. M. Apostol, Calculus (Volume I), theorem I.37 (well ordering principle) and paragraph I.45 (proof, starting from induction principle). The arguement is by contradiction.

user91126
  • 2,326
-1

Assume that for all natural numbers "a", (a+0)=a. With both assumptions in place, it follows that (0+0)=0. No other natural number has the property that (x+x)=x. Thus, 0 comes as unique. Addition comes as monotonically increasing. That is,

For all x, for all y belonging to the set of natural numbers

 If x<y, then (c+x)<(c+y), where c indicates any constant.

 Also, if (c+x)<(c+y), then x<y.

Now suppose that x=0. By the second statement, it follows that

 if (c+0)<(c+y), then 0<y.

"y" qualifies as an arbitrary natural number, so 0 comes as the least natural number. That gives us a "hint".

Now consider any arbitrary set of natural numbers. Define a monotonicity relation for a function $ on that arbitrary set such that

 if (c $ x)<(c $ y), then x<y.  And if x<y, then (c $ x) < (c $ y)

Define another function # such that there exists some element j belonging to the arbitrary set of natural numbers such that for all x, (j # x)=j. It follows that (j # j)=j. Now consider the statement

 if (c $ j)<(c $ y), then j<y.

Since y came as arbitrary, j comes as the least element of the set. Since the set came as arbitrary it holds for all subsets of natural numbers.

Doug Spoonwood
  • 11,211