- #1
amcavoy
- 665
- 0
Let [tex]\left\{x_{n}\right\}[/tex] be a nonempty sequence of monotonically increasing rational numbers bounded from above. Prove that [tex]\left\{x_{n}\right\}[/tex] has a least upper bound in [tex]\mathbb{R}[/tex].
If we choose a monotonically decreasing sequence of upper bounds [tex]\left\{b_{n}\right\}[/tex] with the property that [tex]2^{n-1}\left(b_{n}-x_{n}\right)=b_{1}-x_{1}[/tex], can we show that there exists an accumulation point and conclude that it is the least upper bound? Given the choice of [tex]\left\{b_{n}\right\}[/tex], we can say that [tex]\left(x_{n+1},b_{n+1}\right)\subset\left(x_{n},b_{n}\right)[/tex], but how do we show from here that there is a least upper bound?
Thank you.
If we choose a monotonically decreasing sequence of upper bounds [tex]\left\{b_{n}\right\}[/tex] with the property that [tex]2^{n-1}\left(b_{n}-x_{n}\right)=b_{1}-x_{1}[/tex], can we show that there exists an accumulation point and conclude that it is the least upper bound? Given the choice of [tex]\left\{b_{n}\right\}[/tex], we can say that [tex]\left(x_{n+1},b_{n+1}\right)\subset\left(x_{n},b_{n}\right)[/tex], but how do we show from here that there is a least upper bound?
Thank you.