- #1
Anoonumos
- 16
- 0
Hi,
V is a non-empty, upper bounded subset of R. Show that a sequence [tex](v_n)_{n \geq 0}[/tex] in V exists such that: 1)[tex]v_0 \leq v_1 \leq ...[/tex] and 2) the limit of the sequence is sup V. (Hint: use a recursive construction)
V is non-empty and upper bounded so sup V exists.
Suppose [tex]sup V \in V[/tex] 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:
[tex]v_0 \in V[/tex]
[tex] (v_n, supV) \subset (v_{n-1}, supV)[/tex]
But I'm not sure if this is right or how to proof that the limit of this sequence is sup V.
Any ideas?
Homework Statement
V is a non-empty, upper bounded subset of R. Show that a sequence [tex](v_n)_{n \geq 0}[/tex] in V exists such that: 1)[tex]v_0 \leq v_1 \leq ...[/tex] 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 [tex]sup V \in V[/tex] 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:
[tex]v_0 \in V[/tex]
[tex] (v_n, supV) \subset (v_{n-1}, supV)[/tex]
But I'm not sure if this is right or how to proof that the limit of this sequence is sup V.
Any ideas?