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Prove that every set of natural numbers has a lower bound in $\mathbb{Z}$.

I have no clue how to do this. I would like to prove that one is the min of the set A and conclude since 1 is a min A is bounded below, but I cannot think of what to do if 1 is not in the set.

Approach

let $A=\{k\in \mathbb{Z}:k>0\}$

Then $1 \in A$ since $1>0$. Also $k \geq 0$ for all $k \in A$ so the set is bounded below. By the well ordering principle A has a minimum call it $m$. Then $m\leq1$ and since $m$ is positive $m^2\leq m$.But $m^2 \in A$ since $m^2>0$. But since $m$ is a lower bound this implies $m \leq m^2$. So $m^2 \leq m \leq m^2$. Which is only true if $m=m^2$. Thus $0=m^2-m$

$\implies 0=m(m-1)$ thus $m=1$ so $1$ is a minimum therefore the set is bounded below.

I think my proof is only that 1 is the minimum of all positive integers, is the proof for all sets of natural numbers similar?

Alternate approach: Can I just assume since the set is nonempty and has a minimum that the set has to be bounded below by the well ordering principle?

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    If $A$ is a set of natural numbers, $A\subseteq\mathbb{N}$ and since $1$ is a lower bound for $\mathbb{N}$, it is also a lower bound for $A$. – mxian Jul 07 '19 at 20:21
  • @mxian Does that mean I had to prove that 1 is a lower bound of A or could I just make that assumption? –  Jul 07 '19 at 21:00
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    By proving that $1$ is a lower bound for $A$, you have proved that $A$ has a lower bound in $\mathbb{Z}$ since $1\in\mathbb{Z}$. – mxian Jul 07 '19 at 21:02
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    @Eastcoastdancer This is the well ordering principle. – J.A.G Jul 08 '19 at 19:18
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    https://math.stackexchange.com/questions/358979/proving-the-so-called-well-ordering-principle – J.A.G Jul 08 '19 at 19:22
  • @JaimeGrimalAlves Thank you –  Jul 08 '19 at 19:26
  • @Eastcoastdancer you're welcome!! – J.A.G Jul 08 '19 at 19:28
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    @JaimeGrimalAlves No, this is much simpler than the well-ordering principle, because it asks only for a lower bound, not for a smallest element. It can be deduced from the well-ordering principle, but it can also be proved by just observing that $0$ is a lower bound for any set of natural numbers. – Andreas Blass Aug 15 '19 at 01:08

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