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Let $A$ be a non-empty subset of $\mathbb{Z}$. Suppose there exists $n\in\mathbb{Z}$ such that $n\leq a$, for all $a\in A$. Show that $A$ has a minimum. (Suppose there exists $m\in\mathbb{Z}$ such that $m\geq a$, for all $a\in A$. Show that $A$ has a maximum.)

I read about it at: If a nonempty subset of integers is bounded from below, it has a minimum .

I didn't quite get the implementation of the proof by contradiction. Furthermore, minimum-wise, would that also be correct to prove its existence in A for nā‰„0 (Well-Ordering principle) and for n<0. (If so, how would you prove for n<0) Thanks in advance.

John Griffin
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Slavik Egorov
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2 Answers2

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Following your approach, if $n\geq 0$, then $A$ is a nonempty subset of $\mathbb{N}$ and thus has a minimum element by the well ordering of $\mathbb{N}$.

Suppose $n<0$. Then consider the set $A':=\{a-n \mid a\in A\}$. If $a\in A$, then $n\leq a$ implies $0 \leq a-n$. Thus $A'$ is a nonempty subset of $\mathbb{N}$, so there exists a minimum element $a_0'\in A'$. By the definition of $A'$, we have $a_0'=a_0-n$ for some $a_0 \in A$. Show that $a_0$ must be the minimum of $A$.

I'll let you work out the details for the question in parenthesis.

John Griffin
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  • This looks like a great way to put it, though I'm having trouble showing that $a_0$ must be a minimum of $A$. Is there a way to show it without involving the properties of an infimum? ā€“ Slavik Egorov Aug 22 '17 at 12:30
  • @SlavikEgorov It follows by using the fact that $a_0'$ is the minimum of $A'$. Indeed, suppose $a\in A$. Then $a-n\in A'$, so $a_0'\leq a-n$. Hence $a_0-n\leq a-n$, so $a_0\leq a$. ā€“ John Griffin Aug 22 '17 at 14:20
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You can indeed make use of $\mathbb N$ being well-ordered by $\leq$.

Let $B:=\{k\in\mathbb Z\mid k\geq n\}$ so that $A\subseteq B$.

Define $f:B\to\mathbb N$ by $k\mapsto k-n$.

Then $f$ is a bijection that is order preserving, and also its inverse $k\mapsto k+n$ is order preserving.

That means that $\langle B,\leq\rangle$ and $\langle\mathbb N,\leq\rangle$ are isomorphic, and we are allowed to conclude that $\langle B,\leq\rangle$ is a well-order.

Consequently subset $A\subseteq B$ will have a least element.

drhab
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