Upper bound/Lower Bound

1. May 3, 2005

laminatedevildoll

For the subset M in R (real numbers)

If M={1+1/n : n is an element on N)

then,

- All upper bounds are {x:x an element of R and x > 1}
- Least upper bound is 1
- All lower bounds are {x:x an element of R and x < 0}
- Greatest lower bound is 0

I am not sure if I have the above correct, but for the same problem, how do I find the upper bounds/least upper bound or lower bounds/greatest lower bound if M is a subset in Q (rational numbers)?

Thank you

2. May 3, 2005

kleinwolf

Well, you have to tell which M subset of Q you mean....If you take the one you defined above, then it is by definition a subset of Q..

3. May 4, 2005

honestrosewater

I assume you mean 1 + (1/n) = (n + 1)/n. As kleinwolf said, M is a subset of Q, since (n + 1) and n are integers.
No. First, the least upper bound you've listed isn't in your set of upper bounds. The least upper bound is an upper bound! :) So take another look at your definitions:
Let S be an ordered set and T be a subset of S. If there exists an s in S such that s > t for all t in T, then T is bounded above and s is an upper bound of T. If there exists an s in S such that
1) s is an upper bound of T and
2) if s > x, then x is not an upper bound of T,
then s is the least upper bound of T.

Now, M = {(n + 1)/n : n is in N}. So M is a subset of R and Q, but remember that Q doesn't have the least upper bound property, so if a least upper bound exists, you have a better chance of finding it in R. So plug M and R into your definitions:
Let R be an ordered set and M be a subset of R. M = {(n +1)/n : n is in N}. If there exists an r in R such that r > m for all m in M, then M is bounded above and r is an upper bound of M. If there exists an r in R such that
1) r is an upper bound of M and
2) if r > x, then x is not an upper bound of M,
then r is the least upper bound of M.

Do the same for lower bounds and greatest lower bounds. Since n is in N, you know that (n + 1) > n, and dividing both sides by n you get: (n+1)/n > 1. So m > 1 for every m in M. Look at your definitions, and see if this fits the upper bound, least upper bound, lower bound, or greatest lower bound definitions.
As a hint for the next step, plug in a few small and large values for n and see what happens. Can you find your answers now?

4. May 4, 2005

AKG

You're answers are way off. Let's look at the numbers we'll get in the set:

1 + 1/1 = 2
1 + 1/2 = 1.5
1 + 1/3 = 4/3
.
.
.
1 + 1/9999999999999 = (approximately) 1

In dealing with Q, look at the following:

- All upper bounds are {x:x an element of R and x > 1}
- Least upper bound is 1
- All lower bounds are {x:x an element of R and x < 0}
- Greatest lower bound is 0

Replace the underlined stuff, "R", with "Q". However, that's not all, because the stuff in bold also needs to be replaced, not only for the Q-question but for the original one dealing with the reals, as I hoped the stuff above suggested (the stuff that said that you'll get 2, 1.5, 4/3, ... in the set). I hope it showed up, but I also put ">" and "<" in bold. You have to be careful. There's a difference between saying that all x > 1 are upper bounds, and saying that all x > 1 are upper bounds. In the first case, there is no least upper bound (but there always must be for real numbers) and in the second, there is, namely 1. Of course, 1 is not the LUB in your example.

5. May 4, 2005

laminatedevildoll

M is a subset of R
- All upper bounds are {x:x an element of R and x >= 2}
- Least upper bound is 2
- All lower bounds are {x:x an element of R and x <= 1}
- Greatest lower bound is 1

M is a subset in Q
- All upper bounds are {x:x an element of Q and x >= 2}
- Least upper bound (Does Not Exist?)
- All lower bounds are {x:x an element of Q and x <= 1}
- Greatest lower bound (Does Not Exist?)

I am probably making this even more complicated.

I understand if

M={1/n: n in N} The set of all lower bounds of M is {x:x in R and x<= 0}, and the greatest lower bound is 0. The set of all upper bounds of M is {x:x in E and x>=1} and the least upper bound is 1.

6. May 4, 2005

AKG

I'm not sure what your last sentence is related to. Anyways, with the Q problem, why are you having trouble finding the LUB and GLB? Look at your answers for the real case. 2 and 1. Those are rational numbers, right?

7. May 5, 2005

laminatedevildoll

So, I guess the reals and the q's are the same for LUP and GLB.

8. May 5, 2005

honestrosewater

In this case, yes. Moving from R to Q doesn't change the LUB or GLB of your set; It's just that the LUB or GLB may be an irrational number, so they may not exist in Q. If the LUB or GLB of M were irrational numbers, then M wouldn't have a LUB or GLB in Q.

9. May 5, 2005

laminatedevildoll

In this new case,

M={1/n: n in N} The set of all lower bounds of M is {x:x in R and x<= 0}, and the greatest lower bound is 0. The set of all upper bounds of M is {x:x in E and x>=1} and the least upper bound is 1.

Will the LUB AND GUB be the same for both R and Q?

10. May 5, 2005

honestrosewater

If the LUB (or GLB) is a member of set S, then the LUB (or GLB) exists in S. ;) The values of the LUB and GLB don't change when you move from superset to superset. If the LUB (or GLB) of M exists in two sets, then the LUB (or GLB) of M is the same in both sets. 0 and 1 are members of Q and R.

11. May 5, 2005

laminatedevildoll

I see it now. Thank you for the explanation.