Show that the set of bounded real sequences is a metric space with norm defined by $d(x,y)=\sup|x_n-y_n|$.

Question

I am attempting to show that $d=sup|x_n-y_n|$ is a metric when $x_n$ and $y_n$ are real bounded sequences.

I must first demonstrate that $d$ is non-negative, and that $d=0$ iff $x=y$. This step is fairly trivial. Next I must show that $d(x,y)=d(y,x)$, but this clearly follows from the fact that $|x_n-y_n|=|y_n-x_n|$.

I am stuck on the part where I must show that for any $x,y,z$ belonging to the space $X$, it holds that $d(x,z)+d(z,y)\geqslant d(x,y)$. It is very obvious from the picture I drew that this is true, but I'm struggling to form a clear argument.

Answer 1

Note that by the triangle inequality $$ |x_n-z_n|\leq |x_n-y_n|+|y_n-z_n|\leq \sup_n |x_n-y_n|+\sup_n |y_n-z_n| $$ from which the result follows.

Answer 2

Let $(x_n), (y_n), (z_n)$ be three sequences. We have that

$$|x_n-y_n|\le |x_n-z_n|+|y_n-z_n|,\forall n.$$ Now,

$$|x_n-z_n|\le \sup_n |x_n-z_n|<\infty$$ and

$$|y_n-z_n|\le \sup_n |y_n-z_n|<\infty.$$ Thus

$$|x_n-y_n|\le \sup_n |x_n-z_n|+\sup_n |y_n-z_n|,$$ from where

$$\sup_n|x_n-y_n|\le \sup_n |x_n-z_n|+\sup_n |y_n-z_n|,$$ and we are done.