Consider two bounded sequences $\{|A_i|\}_{i = 1}^{\infty}$ and $\{|B_i|\}_{i = 1}^{\infty}$, for some $\epsilon > 0$, suppose $\sup_{i \in \mathbb{N}}||A_i| - |B_i|| < \epsilon$ where $\mathbb{N}$ is the set of natural numbers, do we have $|\sup_{i \in \mathbb{N}}|A_i| - \sup_{i \in \mathbb{N}}|B_i|| < \epsilon$?
I can prove the above claim when using max not sup, but I have no idea how to prove rigorously when using sup. Here are my thoughts for the case max. Let $\alpha: = \max_{i \in S}|A_i| = |A_{i_j}|$ and $\beta := \max_{i \in S}|B_i| = |B_{i_k}|$ where $S$ is a finite set, and suppose now we have $|\alpha - \beta| \geq \epsilon$, i.e., $||A_{i_j}| - |B_{i_k}|| \geq \epsilon$, and without loss of generality assume $|A_{i_j}| \geq |B_{i_k}|$, then we have $|A_{i_j}| - |B_{i_j}| \geq |A_{i_j}| - |B_{i_k}| \geq \epsilon$, contradicting the fact that $\max_{i \in S}||A_i| - |B_i|| < \epsilon$.
The $\max$ here is not necessarily well-defined.
Let $x(i) = |A_i|$ and $y(i) = |B_i|$, then we have two sequences of real numbers. Since the two sequences are non decreasing and bounded then $\sup_{i \in \mathbb N} (x(i)) < \infty$ and $\sup_{i \in \mathbb N} (y(i)) < \infty$.
Let $z:=\sup_{i \in \mathbb N}(z(i))$. Then from triangle inequality we have $|x +y | \leq |x| +|y|.$
Since $|x| \leq |x-y|+ (|-y|=|y|) \Rightarrow |x|-|y| \leq |x-y|$.
Also,
Since $|y| \leq |y-x|+ (|-x|=|x|) \Rightarrow |y|-|x| \leq |x-y|$.
Then we get $$||x|-|y|| \leq |x-y|.$$
Thus, the given information is equivalent means exactly expected.