1. ## supremum

For each of these sets for their, supremum and infimum (if they exist), maximum and minimum (if they exist), interior (explain your reasoning), boundary (explain your reasoning), set of accumilation points (explain your reasoning), closure
and classify each of these sets as open, closed, neither open nor closed, or both open and
closed.

a) [3,5)
b) N --- natural numbers
c) { x is an element of Q: o <= x <= sqrt(2) } = Q and [0,sqrt(2)]

the and means the upside down U

2. Originally Posted by luckyc1423
For each of these sets for their, supremum and infimum (if they exist), maximum and minimum (if they exist), interior (explain your reasoning), boundary (explain your reasoning), set of accumilation points (explain your reasoning), closure
and classify each of these sets as open, closed, neither open nor closed, or both open and
closed.

a) [3,5)
b) N --- natural numbers
c) { x is an element of Q: o <= x <= sqrt(2) } = Q and [0,sqrt(2)]

the and means the upside down U
I can only help you with supremum and infimum, maximum and minimum, boundary, and the classifications of open, closed etc.

I have no idea what "interior" or "closure" means, and i only have a very vague idea of what "accumilaation" means

a) [3,5)
Let S be the set [3,5)

supremum = 5
infimum = 3
maximum = doesnt exist
minimum = 3
boundary (explain): bounded above and below. Below by 3, since if s is an element of S, then 3<= s for all s in S. Above by 5, since if s is an element of S, 5>= s for all s in S
classification: Both open and closed. Closed at the lower bound, open at the upper bound

b) N --- natural numbers

supremum = doesnt exist (i guess you could say + infinity, dont hold me to this though)
infimum = 1
maximum = doesnt exist
minimum = 1
boundary (explain): Bounded below by 1. Since if n is an element of N, then 1<= n for all n in N. It is unbounded above
classification: closed at lower bound open at upper bound

c) { x is an element of Q: o <= x <= sqrt(2) } = Q and [0,sqrt(2)]

Let R be the set { x is an element of Q: o <= x <= sqrt(2) }

supremum = sqrt(2)
infimum = 0
maximum = doesnt exist
minimum = doesnt exist
boundary (explain): bounded below by 0, since if r is an element of R, 0 <= r for all r in R. bounded above by sqrt(2), since if r is an element of R, sqrt(2) >= r for all r in R
classification: open

I guess someone will come along soon who can help you with the other parts

3. The boundary of a set, denoted as Bdry(S) is the set of points that are in the set and limits points of its complement or points in its complement that are limits points of the set.
Thus Bdry((3,5])={3,5}; Bdry{N}={1} [here be careful some text say that 0 is a natural number!];
Bdry({ x is an element of Q: o <= x <= sqrt(2) }=[0,sqrt(2)]. That is the whole set is a subset of its own boundary.

4. Thank you guys, it really helps when you guys explain things..when I go back to look at how everything is done it really helps out having these examples explained from you.