1. ## proving convergence

Let
A be a bounded subset of R with infinitely many elements. If

s
= supA is NOT an element of A, prove that there exists a strictly
increasing sequence (
xn), xn inA such that limit as n approaches infinity of (xn) goes to s.

I know how to prove that the limit of a strictly increasing sequence goes to the supremum, but I'm not sure how to show that there must exist a strictly increasing sequence if the supremum is not an element of the subset. Can someone show me how to do this? I would really really appreciate it.

2. Originally Posted by CindyMichelle
Let
A be a bounded subset of R with infinitely many elements. If

s
= supA is NOT an element of A, prove that there exists a strictly
increasing sequence (
xn), xn inA such that limit as n approaches infinity of (xn) goes to s.

I know how to prove that the limit of a strictly increasing sequence goes to the supremum, but I'm not sure how to show that there must exist a strictly increasing sequence if the supremum is not an element of the subset. Can someone show me how to do this? I would really really appreciate it.

(Note, the set must be infinite because the supremum of a finite set belongs to the set).

Let $\displaystyle \sup A = s$. Choose $\displaystyle \epsilon = 1$ then $\displaystyle s - 1$ cannot be an upper bound so there exists $\displaystyle s_1 \geq s - 1$. Choose $\displaystyle \epsilon = 1/2$ then $\displaystyle s-1/2$ cannot be an upper bound so there exists $\displaystyle s_2 \geq s - 1/2$. Choose $\displaystyle \epsilon = 1/3$ then $\displaystyle s-1/3$ is not an upper bound so $\displaystyle s_3 \geq s - 1/3$. In generel $\displaystyle s_k \geq s - \frac{1}{k}$ where $\displaystyle s_k$ is this constructed sequence. Thus, $\displaystyle |s_k - s| \leq \frac{1}{k}$. Thus this sequence converges to $\displaystyle s$.