Constructing a Sequence to Show the Existence of a Limit

[Anoonumos]

Hi,

Homework Statement

V is a non-empty, upper bounded subset of R. Show that a sequence (v_n)_{n \geq 0} in V exists such that: 1)v_0 \leq v_1 \leq ... and 2) the limit of the sequence is sup V. (Hint: use a recursive construction)

Homework Equations

The Attempt at a Solution

V is non-empty and upper bounded so sup V exists.
Suppose sup V \in V Construct the sequence: sup V = v0 = v1...
And I know how to go from there.

I'm having problems with the case: sup V is not in V.
I was thinking of constructing the sequence:
v_0 \in V
(v_n, supV) \subset (v_{n-1}, supV)

But I'm not sure if this is right or how to proof that the limit of this sequence is sup V.
Any ideas?

 

[HallsofIvy]

Let v_0 be any member of V. We can do this because V is non-empty.

The recursion step: for any v_i, (sup V+ v_i)/2< sup(V) and so is not an upper bound on V/. There must exist some v_{i+1} in V such that (v_i+ sup(V))/2< v_{i+1}. That cuts the distance from the previous term to sup(V) in half on each step.

 

[Anoonumos]

Thanks!

 

[Anoonumos]

Sorry for the double post, but I came across another problem when writing it down. v(i+1) doesn't have to exist right? And there doesn't have to be a smallest v(i+1). So how would one formulate the sequence?

v_{i+1} \in [v_1, sup V) perhaps?

Or is it enough to state v(i+1) exists and v(i+1) ? vi and v(i+1) is in V, to formulate a sequence?

 

[HallsofIvy]

No, v_{i+1} by this method must exist: Since in this case we are assuming that sup(V) is not in V itself, our (v_i+ sup(V))/2 is NOT sup(V) and so is not an upper bound for V. There must exist at least one member of V larger than (v_i+ sup(V))/2. I said nothing about v_{i+1} being the smallest such member of V- choose any of them.

But it is not sufficient just to say that v_{i+1} is some number in V larger than v_i. That sequence would not necessarily converge to sup(V).

 

[Anoonumos]

I understand. I have one final question.
Nevermind, thanks :)