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 :)