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Proposition: If $A \subset \mathbb{Z}$, and if there exists $n \in \mathbb{Z}$ such that $n \leq a$ for every $a \in A$ then $A$ has a minimum. If $m \in \mathbb{Z}$ such that $a \leq m$ for every $a \in A$ then $A$ has a maximum.

Proof: We know that $\mathbb{Z}^+ \cup$ {0}$ = \mathbb{N}$, so we can define a new set L={$|a-n|$ s.t $a \in A$ } , since $L \subset \mathbb{N}$ from well ordering principle L has a minimum, therefore A has a minimum, as desired.(The maximum part is proved exactly the same)

Can someone verify this proof? Is there something missing? The last step, for example, is it ok to derive from the fact that L has a minimum , then A has a minimum? or is it enough to state that? Thanks for the feedback!

(Note:I have proved the well ordering principle before)

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3 Answers 3

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You don’t need the absolute value signs: just let $L=A-n=\{a-n:a\in A\}$, and you’ll automatically have $L\subseteq\Bbb N$. I would add a bit more detail: let $k=\min L$, and show that $k+n=\min A$. At the very least you should say explicitly that $n+\min L=\min A$.

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  • $\begingroup$ @Charles: You’re welcome! $\endgroup$ Oct 13, 2014 at 16:51
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I think you should fill in the gaps. Let $b$ be the minimum of $L$. Use the definition of $L$ along with $b$ to explicitly show that the minimum of $A$ exists. (Also, I'd define $L$ as $\{a - n \,|\, a \in A\}$; no need for the absolute value.)

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Yes, it is correct.

For the second part, probably the clearest to write is to apply the previous statement to $-A:=\{-a\mid a\in A\}$.

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