# Thread: Nested Interval Property (NIP) implies Least Upper Bound propery (LUB)

1. ## Nested Interval Property (NIP) implies Least Upper Bound propery (LUB)

I'm stuck in proving this. I get that we can set up the sequence of nested intervals by starting with an element a of nonempty set A with an upper bound b and create interval I_1 = [a,b]. Then create interval I_2 nested in I_1 by letting:

I_2 = [a, (a+b)/2] if (a+b)/2 is a upper bound

or

I_2 = [(a+b)/2, b] if (a+b)/2 is a lower bound

We continue this interval having process and generate our sequence of nested intervals. Now the typical argument states that since lim(1/2)^n , n->infinity = 0, this leads to us having only one point of intersection in Intersection of all I_i, i = 1 ... infinity. Finally with a few additional arguments we can show this point must be the least upper bound. My question is how do we argue that lim(1/2)^n , n->inf = 0. Or put another way: for every e, there exists a N such that (1/2)^n < e, when n >= N. I don't see how this can be proved without the Monotone Convergence Theorem which I only know how to prove using the LUB property.

2. You shouldn't need the LUB property to show that 1/2^n → 0 as n→∞. It only requires the archimedean property. First prove (by induction) that 2^n>n. Then given ε>0, choose N with N>1/ε (that's where the archimedean property comes in). It will follow that (1/2)^n<ε whenever n≥N.

3. But isn't the Archimedean Property also proved as a result of LUB?

4. Originally Posted by wellfed
But isn't the Archimedean Property also proved as a result of LUB?
The LUB axiom implies the archimedean property, but the converse is not true. So the AP is a weaker axiom than the LUB.

5. Originally Posted by Opalg
The LUB axiom implies the archimedean property, but the converse is not true. So the AP is a weaker axiom than the LUB.
Yes, but what I think the poster means is that you need to justify the Archimedean Property (and not just quote it) without referring to the LUB Property.