How can I show this inequality??

Question

$$\text{Let}\ {\{a_n\}}\ \text{and}\ {\{b_m\}}\ \text{is bounded}$$

How can I show that $$|\sup{\{a_n\}}-\sup{\{b_m\}}|$$ is equal or smaller than $$\sup\{{|a_n-b_n|\}}$$?

Answer 1

Given $\epsilon>0$, an $n$ is such that $\sup a_{n}-\epsilon<a_{n}$, so $\sup a_{n}-\sup b_{m}-\epsilon<a_{n}-b_{n}\leq|a_{n}-b_{n}|\leq\sup|a_{n}-b_{n}|$, this gives $\sup a_{n}-\sup b_{n}\leq\sup|a_{n}-b_{n}|$. Symmetry gives $\sup b_{n}-\sup a_{n}\leq\sup|b_{n}-a_{n}|=\sup|a_{n}-b_{n}|$, the result follows.

Answer 2

This follows from the fact that $\sup_n (a_n+b_n) \le \sup_n a_n +\sup_n b_n$.

Hence $\sup_n a_n = \sup_n ((a_n-b_n) + b_n) \le \sup_n (a_n-b_n) +\sup_n b_n$ and so $\sup_n a_n - \sup_n b_n \le \sup_n (a_n-b_n) \le \sup_n |a_n-b_n|$.

Reversing the roles of $a,b$ gives the desired result.