bounds

[Dr.Peterson]

upper bounds of the set are : 1 and 2
my friend told me that there is no supremum
b= 2 is a upper bound, and b'=1 is another upper bound.
from the def of the supremum, b<=b' so 2<=1 which is not true, and so this set has no supremum.
thats my thought .
btw my friend told me that there is no supremum here and im struggling to find out why.

Clearly you are misunderstanding the definition, so let's clear things up.

You are discussing the set A = {1 − 1/2^n : n ∈ N} = {1/2, 3/4, 7/8, 15/16, ...}. Do you see that its elements never reach 1? That means that any number greater than 1 is an upper bound; and in fact 1 itself is an upper bound. Any of these numbers is greater than or equal to every element of the set A.

Now look at the definition you quoted:

An upper bound b is a least upper bound (or supremum), if b ≤ b’ for any other upper bound b’ — we denote it by sup S​

Think about all those upper bounds I mentioned: 1, and any number greater than 1. Which of those is the least? Clearly 1. So I want to show that 1 is the supremum, and I do that by saying that if b = 1, and b' is any other upper bound, then b ≤ b’. Do you see that this is true? Every other upper bound is greater than 1; any number less than 1 is not an upper bound, because there will be some element of A greater than that. (This is the point at which a proof is needed, but I'm trusting for the moment that you see it is true.)

Now look at what you said:

upper bounds of the set are : 1 and 2
my friend told me that there is no supremum
b= 2 is a upper bound, and b'=1 is another upper bound.
from the def of the supremum, b<=b' so 2<=1 which is not true, and so this set has no supremum.

It is not true that 1 and 2 are the only upper bounds of A. And the fact that 2 ≤ 1 only implies that 2 is not the least upper bound, not that there is no least upper bound. Surely if you are looking for the least, you would consider the one that is less than the other, wouldn't you?