Question

assume the nested interval property is true and use it to prove the completeness property in R

Answer 1

here is what I know so far. there exists a an element in all these nested closed intervals. a1<a2<a3<..an<x<bn<...b3<b2<b1

Answer 2

What is the definition of completeness you are using? "Every set A bounded above has a l.u.b./sup"?

Answer 3

yes every non empty set of reals that has an upper bound also has a supremum in
R.

I got that the nested inervals provides that the intervals are non empty. do i use the fact that this element in all the sets must be an upper bound of all the a_n?

Answer 4

You can do this constructively by writing down the intervals such that intersection of them all is x = lim bn, and that number is the sup.

Or you can argue by contradiction.

I think I would use the constructive method. It's not hard to see that's possible to write down a strictly monotonically increasing series (an); just be careful with the (bn) so that the intervals are nested.

Answer 5

or rather "in the intersection of them all is lim bn = x and that number is the sup"

Answer 6

actually, no, we need this to be sharper and have lim an = lim bn = x and then show that x = sup A.

Of course in the construction (an) are all in A and the (bn) are all greater than every member of A.

Answer 7

and in that case the (an) is increasing, but not necessarily strictly.

Answer 8

i like the use of the montone cnvg thm. but my prof won't allow that because it is in a future chap. so contradiction would have to be the way to go. Assume that not every non empty set of real that has an upper bound also has a supremum and get a contradiction?

Answer 9

Ok. So yes, take such a set A.

Answer 10

Note first that A must have infinitely many members, other max A would exist and be = sup A

Answer 11

Now let (an) be a strictly increasingly series in A such that ______ and (bn) be a strictly decreasingly series such that bn > a for all a in A such that ______

Answer 12

I'm just trying to figure out exactly what to put in these blanks.

Answer 13

Oh, I see. ...such that lim(an - bn) = 0

Answer 14

but doesn't that mean that we don't have a common element since an and bn are becominmg closer and closer?

Answer 15

I'm not sure exactly what you mean by 'common element' and I'm going to have to chew on this for a couple of minutes to see how to get a sharp proof given where you are in the material.

Let me ask you have any other properties from completeness yet that we can use?

Answer 16

I.e., have you shown Completeness <=> P for some statement P ?

Answer 17

...so we can use not P to help us in our proof by contradiction.

Answer 18

we have got the axioms for reals, the completness property and the nested interval prop. we know that the completeness property implies the nested interval property.

Answer 19

If the intervals are nested we know that they share at least one common element.

Answer 20

By hypothesis yes. The question is how do we construct the intervals so that number is sup A.

Answer 21

Let's try this.

Let B(x,r) be the closed ball of radius r i.e., B(x,r) = [x-r, x+r]

Answer 22

but if lim (an - bn) = 0 our contradiction because as the distace between an and bn shortens it makes that common element impossible.

Answer 23

Yes, but how do we know there are such series (an) and (bn)? So that's what I'm going to try and construct now.

Answer 24

Now for every integer n there is an upper bound for A, call it x_n (not in A of course), such that |x_n - a| < 1/n for some n. Let the (bn) be a decreasing sequence of such x_n.

Answer 25

Then B(b_n,1/n) intersects A by construction. Let a_n = max(a_{n-1}, a) for an arbitrary
\[a \in B(b_n,1/n)\]
and a1 = some arbitrary member of A.

Answer 26

Then the interviews [an,bn] are nested and by construction
\[|b_n -a_n| < 1/n \rightarrow 0\]

Answer 27

I've been waiting 7 minutes for that to post!! lag on this site is terrible.

Answer 28

Now, the intersection of all the interviews in non-empty by hypothesis; call it {x}. Now show x must necessarily be the sup.

Answer 29

So actually, this is a constructive argument after all.

Answer 30

Draw a diagram: this will help a lot.