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Show that $\mathbb N $ with the usual order is well ordered.

My attempt

Let $A\neq\emptyset$ a subset of $\mathbb N$. Since $A\subset \mathbb R$ is under-bounded by $0$, there is an infimum. Let denote $a=\inf A$. By property of the infimum, , there is a $x\in A$ such that $$a\leq x\leq a+1.$$ Since $a, x\in \mathbb N$, we necessarily have $x=a$, and thus $a\in\mathbb N$.

Is it correct ? It look to have a problem, the fact that I consider $\mathbb N$ as a subset of $\mathbb R$.

Rick
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  • You don't need to know anything about the reals to prove this. – Thomas Andrews Oct 27 '15 at 13:19
  • http://math.stackexchange.com/questions/49555/proving-that-the-set-of-natural-numbers-is-well-ordered – Jean-Claude Arbaut Oct 27 '15 at 13:19
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    Also, you haven't shown that $a\in\mathbb N$. You've just assumed it. "Since $a,x\in\mathbb N$, we necessarily have $x=a$ and thus $a\in\mathbb N$." is blatantly circuluar. – Thomas Andrews Oct 27 '15 at 13:21
  • @ThomasAndrews: ok, I see. And like this : Let $\emptyset \neq A\subset \mathbb N$ and suppose there is no minimal element. Since $\mathbb N$ is under bounded by $0$, $A$ is also underbounded. So $$\exists M\in\mathbb N: \forall n\in A, n\geq M.$$ If $M\in A$ we are finish. Suppose $M+k\notin A$ for all $k\geq 1$. In particular, $A\subset [M,\infty [$ and $A\cap [M,+\infty [=\emptyset$, therefore $A=\emptyset$ which is a contradiction with $A\neq \emptyset$. Therefore, there is a $k$ such that $M+k\in A$ and $M+k\leq n$ for all $n\in A$ what prove the claim. Is it correct ? – Rick Oct 27 '15 at 13:32

1 Answers1

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I think this is easiest: pick an $a\in A$. The set $$A'=\{a'\in A\mid a'\leq a\}$$ is finite and non-empty, and therefore has a smallest element $a_0$. Then $a_0$ will also be the smallest element of $A$.

Arthur
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