# prove A has a largest element

• Sep 17th 2011, 11:09 AM
wopashui
prove 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 AM
Plato
Re: prove A has a largest element
Quote:

Originally Posted by wopashui
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?

Again let $\alpha=\sup(A)$.
If $\alpha\in\mathbb{Z}$ we are done.

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

That means that $\left( {\exists n \in A} \right)\left[ {\left\lfloor \alpha \right\rfloor < n < \alpha } \right]$.
• Sep 17th 2011, 01:26 PM
HallsofIvy
Re: prove A has a largest element
Quote:

Originally Posted by wopashui
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?

You are using words in meaningless ways. A is a set, not a number, so it can't be the "least upper bound" of any set. And Z, the set of all integers, is not bounded and so does not have a least upper bound. I strongly recommend you review the basic definitions.
• Sep 19th 2011, 03:14 PM
wopashui
Re: prove A has a largest element
Quote:

Originally Posted by Plato
Again let $\alpha=\sup(A)$.
If $\alpha\in\mathbb{Z}$ we are done.

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

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

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 $S_0$. in A such that $S<=S_0$. for all S in A.

but why is this has to be in Z though?
• Sep 19th 2011, 03:31 PM
Plato
Re: prove A has a largest element
Quote:

Originally Posted by wopashui
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 $S_0$. in A such that $S<=S_0$. for all S in A.
but why is this has to be in Z though?

@wopashui
You have some serious issues (i.e. basic misunderstandings) with all of these questions.
Please seek a sit-down with a live tutor.
• Sep 19th 2011, 04:10 PM
wopashui
Re: 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 PM
TheChaz
Re: prove A has a largest element
Quote:

Originally Posted by wopashui
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?

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.