Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thank-you.

share|improve this question
2  
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
2  
@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
add comment

1 Answer

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.

share|improve this answer
    
It did. Thank you very much! –  Bryan Urízar Apr 17 '13 at 12:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.