Proving that $\sup\limits_{n\geq 1}\{a_n+b_n\}\neq \sup\limits_{n\geq 1}a_n+\sup\limits_{n\geq 1}b_n$

Question

Let $\{a_n\}\;\;\text{and}\;\;\{b_n\},\;\;n=1,2,3,...$ be bounded subsets of real numbers, then

$$\sup\limits_{n\geq 1}\{a_n+b_n\}\leq \sup\limits_{n\geq 1}\{a_n\}+\sup\limits_{n\geq 1}\{b_n\}$$

The proof of the above is trivial. However if $A=\{a_n\},\;\;B=\{b_n\}$ and $A+B=\{a_n+b_n\}$ then, I wish to prove that

$$\sup\limits_{n\geq 1}\{a_n+b_n\}\neq \sup\limits_{n\geq 1}a_n+\sup\limits_{n\geq 1}b_n$$

Please, can anyone help me out?

Answer 1

Take $$a_n:=(-1)^{n} \ \ \text{ and }\ \ b_n:=(-1)^{n-1}$$ so that $$\sup_n \{a_n+b_n\}=0$$ and $$\sup_n \{a_n\}+\sup_n\{b_n\}=2$$

Answer 2

$$ \{ a_n \} = \{ 1,0,1,0,1,0,.....\}\implies \sup \{a_n\} = 1$$

$$\{ b_n\} = \{ 0,1,0,1,0,1,.....\}\implies \sup\{ b_n\} = 1 $$

$$\{a_n +b_n\} =\{ 1,1,1,1,1,.....\}\implies \sup\{ a_n + b_n\} = 1 $$

$$ 1\ne 1+1 $$