Elementary Classical Analysis, Marsden, 2nd ed, Chapter 1, Problem 28

Problem

Let $x_n$ be a Cauchy sequence in $\mathbb{R}$ and let $A_n=\sup\{x_n,x_{n+1},\cdots\}$ and $B_n=\inf\{x_n,x_{n+1},\cdots\}$. Prove $A_n$ converges to the same limit as $B_n$, which in turn is the same as the limit of $x_n$.

Answer

Since $\mathbb{R}$ is complete, $x_n \rightarrow x \in \mathbb{R}$. 

For $\epsilon>0$, there exists $N$ such that $|x_n - x|< \epsilon$ for all $n \geq N$. Since $A_n=\sup\{x_n,x_{n+1},\cdots\}$, $A_n \geq x_n >x - \epsilon$ for all $n \geq N$. On the other hand, note that $x_n < x+\epsilon$ for all $n \geq N$ $\Rightarrow$  $A_n \leq x + \epsilon$ for all $n \geq N$  $\Rightarrow$  $|A_n - x| \leq \epsilon$ for all $n \geq N$. This means $A_n \rightarrow x$.

Since $B_n=\inf\{x_n,x_{n+1},\cdots\}$, $B_n \leq x_n < x + \epsilon$ for all $n \geq N$. On the other hand, note that $x_n > x-\epsilon$ for all $n \geq N$ $\Rightarrow$  $B_n \leq x - \epsilon$ for all $n \geq N$  $\Rightarrow$  $|B_n - x| \leq \epsilon$ for all $n \geq N$. This means $B_n \rightarrow x$.



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