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I understand the difference between the supremum and the greatest element and between the supremum and the maximal elements. But I'm not sure about the difference between the supremum (least upper bound) and the minimal upper bound. They both seem fairly similar to me.

Thanks

EDIT:

To address some of the comments, the reason I think there is a difference is because this site says so http://en.wikipedia.org/wiki/Supremum#Minimal_upper_bounds. My definition of minimal upper bound would have been the exact same as that for least upper bound.

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Perhaps you should first give your definition of minimal upper bound! –  Mercy Aug 20 '12 at 12:20
    
Why do you think there is a difference? –  Did Aug 20 '12 at 12:22
    
I suggest you have a look at this: math.stackexchange.com/questions/178773/… –  M Turgeon Aug 20 '12 at 12:38

3 Answers 3

up vote 5 down vote accepted

This is a very dumb example, but suppose we augment the real line by adding two additional elements, $\infty$ and $\widetilde{\infty}$. We order this set by declaring $x < \infty$ and $x < \widetilde{\infty}$ for all real numbers $x$. (But we declare no order between these two new elements, so, in particular, the ordering is no longer total).

Note that both $\infty$ and $\widetilde{\infty}$ are upper bounds of $\mathbb{R}$ (the real reals) in this augmented order.

  • Since there is no upper bound of $\mathbb{R}$ strictly below either, they are actually minimal upper bounds.
  • However neither $\infty$ nor $\widetilde{\infty}$ is a least upper bound, because neither $\infty < \widetilde{\infty}$ nor $\widetilde{\infty} < \infty$ holds. (A least upper bound must be strictly below any other upper bound.)

This is the difference between the adjectives minimal and least.

  • To be minimal with respect to some property an object must have that property, and no object strictly below it can share that same property.
  • To be least with respect to some property an object must have that property, and it must be strictly below any other object having that property.

In terms of partially ordered sets, every least object will be a minimal object, but the converse may not hold. The two concepts do coincide in linear (total) orders.

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Another example would be the complex numbers ordered by their real part. Then any purely imaginary number would be a minimal upper bound to the negative numbers (because there's no complex number with lower real part which is above all negative numbers) but not a least upper bound (because all purely imaginary numbers have the same real part). –  celtschk Aug 20 '12 at 13:23
    
Thank you very much for that! cleared it up! –  BYS2 Aug 21 '12 at 12:53

When you use a set, you may have a order defined on it such that there won't be only one minimal upper bound. This may occur when the set is not totally ordered.

For example, let's define a set: $A=\mathbb{N}\cup\{a,b\}$. And we will extend the normal order on $\mathbb{N}$ so that $a>n$ and $b>n$ for every $n\in\mathbb{N}$ but we won't define any relation between $a$ and $b$.

In this case, the minimal upper bounds of A is $\{a,b\}$ but there is no least upper bound.

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Thanks for that! –  BYS2 Aug 21 '12 at 12:53

In linearly (i.e. totally) ordered sets, there is no difference. In partially ordered sets, there is a difference. An element $x$ is a minimal upper bound if $x$ is an upper bound and nothing less than $x$ is an upper bound. There may be more than one minimal upper bound, all of which are mutually incomparable. A least upper bound, on the other hand, is an upper bound that is less than or equal to all other upper bounds. There can be only one of those.

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Alright, thank you for that! –  BYS2 Aug 21 '12 at 12:54

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