Show that the supreme of successions that converge to $0$ is a metric

Let $X=\{\{x_{n}\}_{n=1}^{\infty}\subset \mathbb{R}:\lim_{n\rightarrow \infty}x_{n}=0\}$ be the set of all successions that converge to $0$ and let $d$ be a function defined by $d(x,y)=sup_{n\in \mathbb{N}}\{|x_{n}-y_{n}|\}$, with $x,y\in X$, prove that $d$ is a metric in $X$.

Now, as far as I know i think that it isn't because of the second property, we can't say that $d(x,y)=0 \Leftrightarrow x=y$ because since $x,y\in X$ it means that $\lim_{n\rightarrow \infty}x_{n}=0$ and $\lim_{n\rightarrow \infty}y_{n}=0$, there is a point in which both go to $0$ and the supreme is cero but they are not equal, am I right? Or what am I doing wrong?, What can I do with the triangle inequality Thanks for your help

Solution 1:

You are wrong:\begin{align}d(x,y)=0&\iff\sup_{n\in\Bbb N}|x_n-y_n|=0\\&\iff(\forall n\in\Bbb N):|x_n-y_n|=0\\&\iff(\forall n\in\Bbb N):x_n=y_n\\&\iff x=y.\end{align}

Solution 2:

Since $|x_n-z_n|=|x_n-y_n+y_n-z_n|\leq|x_n-y_n|+|y_n-z_n|$,$\forall n\in\mathbb{N}$
we have that $$sup_n|x_n-y_n|\leq sup_n|x_n-y_n|+sup_n|y_n-z_n|$$ which means that $d(x,z)\leq d(x,y)+d(y,z)$ for every $x,y,z\in X.$
Actually, this space is often denoted as $\ell_0$ and this metric as $d_{\infty}$.
Also it is known that the metric space $(\ell_0,d_{\infty})$ is a complete metric space ( i.e. every Cauchy sequence is convergnet). I recommend you to give it a try and prove that.