Let $X_{n}$ and $Y_{n}$ are two bounded sequence. I need to prove $\sup_{n\in\mathbb N}|X_{n}-Y_{n}|=0$ iff $X_{n}=Y_{n}$ for all $n$

Question

Let $X_{n}$ and $Y_{n}$ are two bounded sequence. I need to prove $\sup_{n\in\mathbb N}|X_{n}-Y_{n}|=0$ iff $X_{n}=Y_{n}$ for all $n$.

I want this proof. Because if you see this in metric space then this is come from $l_{\infty}$ metric space. First of all if $ X_{n}=Y_{n}$ then the condition holds. But difficulty arise for reverse case. I first suppose if $X_{n} \neq Y_{n}$ and we have $\sup_{n\in \mathbb N}|X_{n}-Y_{n}|=0$. There exist atleast one $n$ for which $X_{n}\neq Y_{n}$ for this $n$, $\sup_{n\in \mathbb N}|X_{n}-Y_{n}| > 0$ then here is a contradiction. I now prove this way. Is there any property of Supremum for which I say directly?

Answer 1

A more direct argument would be that for any $n\in\mathbb{N}$,

$$0\leq\lvert X_n-Y_n\rvert\leq\sup_{m\in\mathbb{N}}\lvert X_m-Y_m\rvert=0,$$

so $\lvert X_n-Y_n\rvert=0$, i.e. $X_n=Y_n$.