View Single Post
honestrosewater
honestrosewater is offline
#3
May4-05, 02:03 AM
PF Gold
honestrosewater's Avatar
P: 2,330
Quote Quote by laminatedevildoll
For the subset M in R (real numbers)

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