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I have to show that in any non empty subset of $N$ there is least element. Note: This is not a homework question.

So this is how my incomplete proof looks like. And I tried this by induction:
let $S$ be nonempty subset of $N$. I defined $M = \{m \in N \mid m\le s \text{ for all } s \in S\}.$ Then I showed $1$ is in $M$. I did this using proof by contradiction. Now if $s$ is in $S$ then $s < s^+$. So $s^+$ is not in $M$. Thus $M$ is not equal to $N$.

Thus there exists $z$ in $M$ such that such that $z^+$ is not in $M$. Now I have to show $z$ is the least element of $S$. Since $z$ is in $M$ by the way $M$ is defined $z$ must be least element of $S$. But what if $z$ is not in $S$? What should I do? Help please.

Stefan Hamcke
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Ahmed
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  • Well, you showed $,1\in M,$ under the assumption that $,1\notin S,$ , which is a trivial case. Ok. Your last line in the second parraph is incomprehensible to me, though: couldn't there be a $,s\in S\cap M,$ ? Yes, it could...and this would end your proof! – DonAntonio May 03 '13 at 00:13

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Well, the well ordering of $\,N\,$ is equivalent to the principle of mathematical induction, so you being able to use the latter can easily lead you to prove the former:

Let $\,\emptyset\neq S\subset\Bbb N\,$ and define now $\,K:=\{\;n\in\Bbb N\;;\;n\notin S\;\}\,$ :

(1) if $\,1\in S\,$ then we're done (why?), so we may assume $\,1\in K\,$ .

(2) Suppose $\,1,2,...,n\in K\,$ ; if $\,n+1\notin K\,$ then $\;n+1\in S\,$ is the least element of $\,S\,$ (why?) , so we can assume $\,n+1\in K\,$

Check that (1)+(2) allows you to use the principle of induction and conclude that $\,K=\Bbb N\,$ ...but this is impossible (why?), so this last contradiction proves your question (again, why?)

DonAntonio
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  • There are some similar questions on StackExchange, and I have the same query about your answer. When you say "Suppose 1, 2,, ...n" are you making an assumption that the naturals are totally ordered, and would you therefore need to prove this first ? (one similar question which only uses weak induction is http://math.stackexchange.com/questions/433020/proving-the-equivalency-of-principle-of-mathematical-induction-and-well-ordering) – Tom Collinge May 11 '14 at 15:36
  • When I, or anyone, write "$;1,2,3,...,n;$ , or $;4,7,3,9,78;$ are natural numbers" there is no assumption of order at all, leave alone of well ordering. It is just a way to describe some (finite ammount of) elements in a set. – DonAntonio May 11 '14 at 17:53
  • Thanks for the response: I didn't think you were assuming well-ordering, but aren't you implicitly defining the set {1 , 2...n} as {m in N: 0 < m < n+1} and therefore relying on the total order of N ? – Tom Collinge May 11 '14 at 18:43