Let A be a nonempty bounded above subset of Z. Prove that A has a largest element.

To prove A has a largest element does it mean to prove A is the least upper bound of Z, if so how do I do this?

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- Sep 17th 2011, 11:09 AMwopashuiprove A has a largest element
Let A be a nonempty bounded above subset of Z. Prove that A has a largest element.

To prove A has a largest element does it mean to prove A is the least upper bound of Z, if so how do I do this? - Sep 17th 2011, 11:42 AMPlatoRe: prove A has a largest element
Again let $\displaystyle \alpha=\sup(A)$.

If $\displaystyle \alpha\in\mathbb{Z}$ we are done.

If $\displaystyle \alpha\notin\mathbb{Z}$ then $\displaystyle \left\lfloor \alpha \right\rfloor < \alpha $.

That means that $\displaystyle \left( {\exists n \in A} \right)\left[ {\left\lfloor \alpha \right\rfloor < n < \alpha } \right]$.

**That is a contradiction. WHY?** - Sep 17th 2011, 01:26 PMHallsofIvyRe: prove A has a largest element
- Sep 19th 2011, 03:14 PMwopashuiRe: prove A has a largest element
it's a contradction because it's bounded above, i.e there is an upper bound for A, A will converge to some number eventually, however, we don't need to show A has a least upper bound here, we just need to show that there is an element $\displaystyle S_0$. in A such that $\displaystyle S<=S_0$. for all S in A.

but why is this has to be in Z though? - Sep 19th 2011, 03:31 PMPlatoRe: prove A has a largest element
- Sep 19th 2011, 04:10 PMwopashuiRe: prove A has a largest element
No, I understand this question and why is a contridiction, I'm just not sure why it has to be in Z, if A is a nonempty bounded above sunset of R, will this still work?

- Sep 19th 2011, 04:20 PMTheChazRe: prove A has a largest element
That you understand is definitely arguable (and may indeed be argued by those already in the discussion).

Now, it appears that you are introducing a new question.

Consider the intervals [0, 1) and [0, 1] as subsets of ℝ

Both are bounded above. Both have the same least upper-bound (supremum).

One of these intervals, however, does not have a largest element. Indeed - to prove this, you could take and arbitary element and find one that is larger, and still in the interval.