Proving $\sup \{|x_j-y_j|: j\in \mathbb{N}\}$ is a distance metric for bounded sequences of reals

Question

So I need to show that $d({\bf x}, {\bf y})= \sup\{|x_j-y_j|:j\in \mathbb{N}\}$ is a distance metric for any ${\bf x},{\bf y}$ that are bounded sequences of reals and ${\bf x}=(x_1, x_2, ...)$. The zero property and symmetry are both easy but I'm a little stuck on, of course, the triangle inequality. Here's what I have so far.

$$d({\bf x}, {\bf z}) = \sup\{|x_j-z_j|:j\in \mathbb{N}\}=\sup\{|x_j-y_j+y_j-z_j|:j\in \mathbb{N}\}$$

For any element out of the set, $|x_j-y_j+y_j-z_j|\leq |x_j-y_j|+|y_j-z_j|$ so

$$\sup\{|x_j-y_j+y_j-z_j|:j\in \mathbb{N}\}\leq \sup\{|x_j-y_j|+|y_j-z_j|:j\in \mathbb{N}\}$$

I'd like to be able to distrubte the sup over this sum, but I'm not sure how to argue that this is legitimate. I tried proving the lemma as follows:

Lemma: Let $E_1, E_2\subseteq \mathbb{R}$, then $\sup\{a+b:a\in E_1, b\in E_2\}\leq \sup E_1+\sup E_2$.

Proof: That $\sup E_1+\sup E_2$ is an upper bound follows from, for any $a\in E_1,b\in E_2$ then $a\leq \sup E_1, b\leq \sup E_2$.

Let $M$ be any other upper bound so that $a+b\leq M$ for all $a\in E_1, b\in E_2$. Then $a\leq M-b$ and so $\sup E_1 \leq M-b$ and likewise $\sup E_2 \leq M-a$.

This is pretty much where I'm stuck. I can't add the inequations, and if I could I'm not sure how that would lead to a desirable result. Maybe I'm approaching it all wrong, I don't know.

Answer 1

You can argue as follows. For any $j$, we have $$|x_j - z_j| \leq |x_j - y_j| + |y_j - z_j|$$ Now, $\displaystyle |x_j - y_j| \leq \sup_j |x_j - y_j|$ and $\displaystyle |y_j - z_j| \leq \sup_j |y_j - z_j|$. Therefore $$|x_j - y_j| + |y_j - z_j| \leq \sup_j |x_j - y_j| + \sup_j |y_j - z_j|$$ Combining this with the first inequality, we obtain $$|x_j - z_j| \leq \sup_j |x_j - y_j| + \sup_j |y_j - z_j|$$ The right hand side does not depend on $j$ (we have taken the supremum of each term over all $j$), so it is an upper bound of the left hand side for every $j$, therefore it is at least as large as the least upper bound of the left hand side: $$\sup_j |x_j - z_j| \leq \sup_j |x_j - y_j| + \sup_j |y_j - z_j|$$

Answer 2

I would simply go by this, $E_2$ may contain negative values and even entirely of negative numbers, therefore any element from $E_2$ added well either result in less than $\sup E_1+\sup E_2$ or the acctual value itself, when we have $\sup |E_1|+\sup |E_2|$ we know it clearly must be less than or equal to it.