Let S be an ordered set. Let A S be a nonempty finite subset. Then A is bounded.

Furthermore, inf A exists and is in A and sup A exists and is in A. Hint: Use induction.

Printable View

- Oct 17th 2012, 07:47 PMchristianwosProblem with supremum and infimum
Let S be an ordered set. Let A S be a nonempty finite subset. Then A is bounded.

Furthermore, inf A exists and is in A and sup A exists and is in A. Hint: Use induction. - Oct 18th 2012, 12:09 AMchiroRe: Problem with supremum and infimum
Hey christianwos.

Intuitively this seems OK since you have a finite number of elements and if those elements are finite (i.e. not infinity and make sense) then there will be supremum and an infinimum. You don't get this with an empty set and they have (no doubt on purpose) ruled out that exception.

What have you tried? Have you used the fact that the subset will inherit the order from the total set? - Oct 18th 2012, 12:37 AMhollywoodRe: Problem with supremum and infimum
Yes, for a finite set, the inf is just the smallest element and the sup is just the largest element. For a set with one element a, it should follow from the definition that the sup is a and the inf is a. When you add an element, it's either between the inf and the sup, or becomes the inf or becomes the sup.

- Hollywood - Oct 18th 2012, 04:48 AMchristianwosRe: Problem with supremum and infimum
Thank you. This was my original idea for proving it. Start with a one-element set and prove that clearly inf=sup=only element. Then assume it is true for n elements and use induction.