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I've been trying to prove a slightly different statement to the definition of the well ordering principle given by Lang in Undergraduate Algebra and given here for reference:

Every non-empty set of integers $\geq 0$ has a least element.

I've been trying to prove that: Any non-empty set $A$ of integers which is bounded above has a largest element.

However, I'm left wondering what does it mean for a set of integers to be bounded? Does this bound have to be an integer also or can it be a real number? What is the general definition?

Assuming it is an integer I think I got a proof of my statement:

Since $A$ is bounded above there is a $q \in Z$ such that $m\leq q$ for all $m \in A$. Now consider the set $A' = \{n \mid n = q - m$ for some $ m \in A\}$. $A'$ is obviously not empty as $A$ isn't empty. Note that for any $n \in A'$ $n \geq 0$ and therefore the well ordering principle gives a $s \in A'$ such that for all $n \in A'$, $s \leq n$. As $s \in A'$ this implies there is an $m' \in A$ such that $s = q - m'$. Therefore, for all $n \in A'$, $q - m' \leq n = q - m$ and as a result for all $m \in A$, $m \leq m'$ and so $m'$ is our largest element.

If someone could verify it or give me any criticism I'd appreciate it as I'm working on my proof technique.


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The argument is fine. You’re working in $\Bbb Z$, so bounded above does mean bounded above by an element of $\Bbb Z$. – Brian M. Scott Apr 17 '13 at 11:38
@Bryan Urizar you can always assume your upper bound to be integer: if your bound $x$ is a real number, take the smallest integer greater than $x$ – Federica Maggioni Apr 17 '13 at 11:46
up vote 1 down vote accepted

Your proof is completely correct: nice!

In a general ordered set $S$ (like that of the integers $(\Bbb Z, \le)$), we have that "$A \subseteq S$ is bounded above" takes the following meaning:

There is an $s \in S$ such that for all $a \in A: a \le s$

An alternative definition, popular in analysis, arises when we regard the set of integers $\Bbb Z$ as a subset of the reals $\Bbb R$. In that case, "$S \subseteq \Bbb Z$ is bounded above" yields a real number $r$ such that for all $s \in S: s \le r$.

But then we can take an integer larger than $r$ (for example, the ceiling of $r$) and be in the case we were before. (Here we use the intuitively obvious fact that there is no real number $r$ greater than all integers.)

I hope that this enhances your understanding of the "bounded above" notion.

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It did. Thank you very much! – BU982T Apr 17 '13 at 12:25

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