Elementary Analysis proof, sequences: Cauchy and Monotone sequences; limit theorems

141 Views Asked by At

problem

My TA gave us a hint, that I should first deduce that if $\displaystyle s_n$ converges to some $\displaystyle s$ that is an element of real numbers, then I can see that $\displaystyle m \leq s \leq M$.

I'm still pretty bad at analysis so far, but if I can get the convergence of $\displaystyle s_n$ to equal that $\displaystyle s$, then I can see that the Sup and Inf bind the $\displaystyle s$. Then I can substitute into the inequality?..

Help is appreciated! Thanks. EDIT: we can disregard the sequence $\displaystyle t_n$ here.

1

There are 1 best solutions below

0
On

Let $S$ be the set of all subsequential limits of $s_n$. Since $\limsup s_n = \sup S$ and $\liminf s_n = \inf S$, we have $\limsup s_n \ge \liminf s_n$. Let $s_{n_k}$ be a subsequence of $s_n$ such that $s_{n_k} \to \limsup s_n$. Since $m \le s_{n_k} \le M$ for all $k$, we have $m \le \limsup s_n \le M$. By a similar argument, $m \le \liminf s_n \le M$. Thus $m \le \liminf s_n \le \limsup s_n \le M$.