10

I would like to know how from the very basic I can teach some one the above title statement. Here is my plan.

$\textbf{First}$ I will state WOP: Every non-empty set of positive integers contains a least element.

$\textbf{Second}$ I will state PMI:

$(1)$ $0\in T$.

$(2)$ $n\in T\implies n+1\in T$

We show that $T=\Bbb N$.

Suppose I want to prove $\textbf{First}$ statement using The $\textbf{second}$

So I take $T$ be a non-empty subset of natural number, enough to show it has a least element. ( is that right?)

$\textbf{proof}:$ Assume by contradiction that $T$ has no least element, let $$W=\{x:x\notin T\}$$ since $0$ is a lower bound of natural number so $0\in W$, now here can I say like this: let $0,1,\dots,n\in W\Rightarrow n+1\in W$. Then $n+1$ can not be in $T$ since then it would be a lower bound of $T$ and since $0,1,\dots,n\notin T$, so $n+1\notin T$ by induction $W=\mathbb{N}$, so $T=\phi$

$\textbf{Now}$ Suppose every nonempty subset of $\Bbb N$ has a least element. Let $T$ be a subset of $\Bbb N$ with the following properties

$(1)$ $0\in T$.

$(2)$ $n\in T\implies n+1\in T$

We show that $T=\Bbb N$.

Let $T$ be as above. Consider the set of $\Bbb N\setminus T$, and assume by contradiction it is not empty. By the WOP, it has a least element, call it $x$.

I am not able to proceed further, Thank you for help and discussion.

Cameron Buie
  • 102,994
Myshkin
  • 35,974
  • 27
  • 154
  • 332

2 Answers2

3

Induction implies well order: Let $A$ be the set of all elements not in the non-empty $T$. Then $0\in A$ otherwise $0$ is the least element of $T$.

Assume that $k\in A$ for all $k<n$. If $n\in T$ then $n$ is the least element of $T$. Therefore $n\in A$. By induction all elements are in $A$ and then $T$ is empty.

Well order implies induction: Let $P$ be a property such that $P(0)$ and such that for all $n$ if $P(k)$ for all $k<n$ then $P(n)$. Let $T$ be the set of elements not satisfying $P$. If $T$ is non-empty, then it has a first element $a$. Since it is the first element not satisfying $P$, then for all $k<a$ $P(k)$. By assumption $P(a)$. Contradiction. Therefore $T$ is empty, i.e. all elements satisfy $P$.

The answer from a student once: Let $A$ be a non-empty set. Consider $\sum_{x\in A}x$. If $A$ doesn't have a first element we cannot begin to sum. Contradiction.

Advice: First break the mindframe of only using induction when $P(n)$ is an algebraic formula in $n$,by solving problems involving other types of propositions. After a few examples involving disease spread, colors, geometric statements, etc. they are able to imagine the property $n\notin T$ as a candidate to run induction.

OR.
  • 5,941
  • 1
    You might wanna add the student's answer here. – Git Gud Jun 30 '13 at 16:21
  • It's worth noting that the induction you're using here is what is often known as complete induction, and is not the same as PMI as stated above (though they are equivalent). – Cameron Buie Jun 30 '13 at 16:51
  • Strange name. I was always taught that complete was to distinguish from the (incomplete) induction as in: The sun has come out all this days, it will come out tomorrow. So, complete induction was just a name for mathematical induction. Strong is the name I have seen for this. – OR. Jun 30 '13 at 17:40
  • Yeah. I am pretty sure Wikipedia is just perpetuating the mistake some author did in some book (http://logic.stanford.edu/intrologic/chapters/chapter_09.html). – OR. Jun 30 '13 at 17:59
  • That's another common one. Although I've also seen that one used to refer to any induction between standard and complete induction. For example, I've seen the following referred to as "strong induction": If $P(0)$ and $P(1)$ both hold, and if $P(n+2)$ holds whenever $P(n)$ and $P(n+1)$ do, then $P(n)$ holds for all $n$. – Cameron Buie Jun 30 '13 at 20:03
2

In your first part, there are a couple of things you should fix. Suppose by way of contradiction that $T$ is a non-empty set of positive integers with no least element and let $$W=\{x\in\Bbb N:x\notin T\}.$$ (Note the change, there.) Since $T$ is a set of positive integers, then $0\in W$. Since $T$ is non-empty, then there is some positive integer (so some natural number) $m\in T$. Hence, $W\subsetneq\Bbb N,$ but $0\in W,$ so by PMI, there is some $n\in W$ such that $n+1\notin W$ (meaning $n+1\in T$).

Now, clearly, $1\in W,$ for otherwise, $1$ would be the least element of $T$. But then $2\in W,$ for otherwise $2$ would be the least element of $T$. We can continue in this fashion through finitely-many steps (so PMI isn't brought into play in this part), and see that also $3,4,...,n-1\in W$. Hence, $0,1,2,...,n\in W,$ but $n+1\in T,$ so $T$ has a least element. Contradiction.

It's worth noting that I rewrote the proof in this part largely to avoid the use of what is known as complete induction, and use only PMI as stated in your post.

To continue your second part: We know that $x$ can't be $0$, since $0\in T$. Put $n=x-1$. Since $x$ is a positive integer, then $n\in\Bbb N$, and since $x$ is the least element of $\Bbb N\setminus T,$ then we know $n\in T$ (since $n<x$). But then by assumption $x=n+1\in T,$ a contradiction.

Cameron Buie
  • 102,994
  • your contradiction seems very strange, what we need to prove that is a contradiction? I did not get the point – Myshkin Jun 30 '13 at 16:49
  • Which contradiction? One contradiction came from the assumption that $T$ was an inductive set of naturals, so since $n\in T$ we know $n+1\in T$. But by definition of $n:=x-1,$ we have $x\in T$, and we chose $x$ to be the least element that wasn't in $T,$ so we have a contradiction. In the other, we showed that $T$ has a least element, but we assumed that it didn't have a least element, so we have a contradiction. – Cameron Buie Jun 30 '13 at 16:55
  • yeah, I am asking about the 2nd contradiction in your comment, we needed to show that $T$ has a least element right? and you finally proved it, then why did you assume that it does not have a least element? – Myshkin Jun 30 '13 at 17:16
  • The assumption that $T$ is non-empty allowed us to bring PMI (as you stated it) into play and get $n\in W$ such that $n+1\notin W$. The assumption that $T$ had no least element was then used to show that each of $1,...,n-1\in W$. We could go another way, though: (a) Prove as a Lemma that any finite non-empty set of positive integers has a least element. (PMI will be used here.) (b) Assume that $T$ is a non-empty set of positive integers. (c) Find $n\in W$ such that $n+1\notin W$ (as above). (d) Conclude that if $n+1$ isn't the least element of $T$, then ... – Cameron Buie Jun 30 '13 at 17:33
  • ... the set of elements of $T$ strictly between $0$ and $n$ has a least element by our Lemma, and that least element will be the least element of $T.$ – Cameron Buie Jun 30 '13 at 17:33
  • Oops! Just realized I didn't answer your question. The reason I assumed that $T$ didn't have a least element is because it was easier to proceed by contradiction than to prove the Lemma I mentioned above (it is trivial *if* one knows some early cardinality results, but you said you were looking for a "basic" approach, which I took to mean "requiring as little background as possible"). – Cameron Buie Jul 01 '13 at 00:32
  • @Cameron Buie. I am particularly interested in the proof that PMI (avoiding complete induction) -> the naturals being well ordered. A question I have on your proof is that it seems to rely on the Naturals being totally ordered so that you can work through the elements 0,1,2, ...n. So in a comprehensive proof would you first need to prove total ordering ? – Tom Collinge May 11 '14 at 14:44
  • @TomCollinge: That's an excellent observation. I would say that one probably should, but it might depend on which formulation of "the natural numbers" one is working with. Which are you working with? – Cameron Buie May 12 '14 at 12:54
  • @Cameron Buie. I'd like to be able to show the proof in ZF (excluding choice) working closely from the axioms, so I guess I'd start from the natural numbers as the minimal successor (inductive) set and try to avoid arithmetic. – Tom Collinge May 12 '14 at 14:10
  • @TomCollinge: So (if I'm understanding you correctly), for example, you have $0:=\emptyset,$ $1:={0},$ $2:={0,1},$ and in general, the successor to $n$ is given by $s(n):=n\cup{n}$? If so, then that makes our job fairly easy. We define the order relation by $m<n$ iff $m\ne n$ and $m\cap n=m.$ This is immediately irreflexive by definition, and readily transitive, so is a strict partial order relation among the naturals. Moreover, it is clear that $m<s(m)$ for any $m$ by definition of the successor function. (cont'd) – Cameron Buie May 12 '14 at 15:56
  • Then, PMI should let you show that for all naturals $m,n$ with $m\ne n$ and $m\not\lt n,$ we must have $n<m.$ Hence, the relation is total on the naturals. (Since you're trying to show that PMI implies that the naturals are well-ordered, then it's okay to use it, here.) Now, if you're using a different definition of the order relation, then the best way to proceed may be by first showing that your definition is equivalent to the one I gave above, but it may also be that your definition makes things even easier. Let me know. – Cameron Buie May 12 '14 at 16:00
  • @Cameron Buie. Much appreciate your feedback. Definition I have for m $\le$ n is m $\in$ n, which then relies on N being a transitive set (m $\in $ n is equivalent to m $\subset$ n) to show reflexivity and transitivity. I have Halmos's book where he goes on to prove totality (inductively) and then leaves well-ordering as an exercise. So, my main question here was whether you had found a short-cut round proving totality: it seems not. There's a related question here http://math.stackexchange.com/questions/49555/proving-that-the-set-of-natural-numbers-is-well-ordered ..... – Tom Collinge May 13 '14 at 08:28
  • .....which attempts to prove PMI -> WO with minimum assumptions, but doesn't seem to get a proper answer. If you have the time I'd be interested to read your comments / answer on that. – Tom Collinge May 13 '14 at 08:29
  • @Tom: Apologies for the delay. One can actually show totality without induction. Let me outline an approach that gets you there. (1) Every element of $\Bbb N$ is a transitive set. Indeed, $n\in\Bbb N$ and $k,m$ are such that $k\in m$ and $m\in n,$ then since $\Bbb N$ is transitive and $m\in n\in\Bbb N,$ we have $m\in\Bbb N,$ so since $k\in m$ we similarly have $k\in\Bbb N.$ Now, we have $k,m,n\in\Bbb N$ and $k\in m\in n,$ so since $\in$ is a transitive relation on $\Bbb N,$ we have $k\in n.$ Thus, $n$ is a transitive set, as desired. (cont'd) – Cameron Buie May 29 '14 at 14:34
  • (2) If $A$ is a transitive set whose elements are transitive sets, then $\in$ is a total relation on $A.$ Indeed, suppose $a,b\in A$ are distinct, so without loss of generality, we have $a\not\subseteq b,$ whence $a\setminus b\ne\emptyset.$ By Axiom of Foundation (also known as Regularity), there is some $x\in a\setminus b$ such that $x\cap(a\setminus b)=\emptyset.$ Since $x\in a\setminus b\subseteq a$ and $a$ is transitive, then $x\subseteq a,$ and since $x\cap(a\setminus b)=\emptyset,$ it follows that $x\subseteq b.$ Since $b\in A$ and $A$ is transitive, then $x\in A.$ (cont'd) – Cameron Buie May 29 '14 at 14:56
  • Since $x,b\in A$ with $x\subseteq b$ and $A$ is a transitive set, then $x\in b$ or $x=b.$ Since $x\in a\setminus b,$ then $x\notin b,$ and so $x=b.$ But $x\in a,$ so $b\in a.$ Thus, $\in$ is total on $A,$ as desired. – Cameron Buie May 29 '14 at 14:58
  • That's one way we can prove totality of $\in$ on $\Bbb N,$ but if you're unfamiliar with Foundation/Regularity, that may not be an approach you'd favor. Let me know if you have any questions. – Cameron Buie May 29 '14 at 14:59
  • @CameronBuie Thanks for your further thoughts on this. I have a concern in your proof. As I understand foundation it is a constraint that a set must comply with rather than a result that can be assumed for any "set". The proof I have that the natural numbers are well-founded relies on guess what - that they are totally ordered. I think I have made some progress with induction which I posted for comments http://math.stackexchange.com/questions/798992/is-this-a-correct-proof-of-the-principle-of-complete-induction-for-natural-numbe . Not much feedback so far, I'd value yours. – Tom Collinge May 30 '14 at 14:20
  • @CameronBuie - Proof that the N is well-founded. Given that natural numbers are totally ordered by set inclusion (corresponding to their natural ≤ ordering). Any non-empty subset M of N has a minimum element defined by Mmin = ∩ m ∊ M. If Mmin = 0, then Mmin = {} and so Mmin ∩ M = {}. Otherwise the only elements of Mmin are all less than Mmin (by total order) and so are not elements of M and again Mmin ∩ M = {}. So, the smallest element of M always has an empty intersection with M (and in fact is the only element to do so). – Tom Collinge May 30 '14 at 14:28
  • @Tom: It turns out that, under the axioms of ZF minus Foundation, the Axiom of Foundation holds if and only if $\in$ is a well-founded relation on every set. The proof (as I've seen it) relies on quite a few preliminary results, so I won't attempt to outline it in the comments here at the moment. You've piqued my curiosity, though; I suspect that it can be proved without relying on PMI or anything about the natural numbers specifically, but I'm not yet certain. I'll update you as developments are made, and will probably make a self-answered question at some point for reference. – Cameron Buie May 30 '14 at 17:53