# Show $d(x,y)=\sup|\alpha^k-\beta^k|$ satisfies triangle inequality

If $X$ is the set of all bounded sequences of real numbers, and $x=(\alpha^k)$ and $y=(\beta^k)$, how do we show that $d$ defined as $$d(x,y)=\sup|\alpha^k-\beta^k|: k\in\mathbb N$$ satisfies the triangle inequality?

I was checking for :

$$\sup|a^i-b^i|+ sup|b^i-c^i|\geq \sup(|a^i-b^i|+|b^i-c^i|)\geq(\sup(|a^i-b^i+b^i-c^i|)$$ ($i\in \mathbb N$)

There is this case where $i=j$ but this can be written as:$$\sup|a^i-b^i|+ \sup|b^i-c^i|\geq (|a^i-b^i|+|b^j-c^j|)\geq(\sup(|a^i-b^i+b^j-c^j|)$$ ($i\in \mathbb N$)

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@ Jason, a^i etc are real number of sequences, so technically the i on the first is not the same as the i on the second they might be but not really so I better change that.. –  Klara Jan 31 '13 at 20:18
did I interpret right? –  uforoboa Jan 31 '13 at 20:20
If it is so the first inequality comes from the fact that in general, if you have two sequences, say $a_k, b_k$, then $\sup_k a_k+\sup_k b_k\geq \sup_k (a_k+b_k)$. For the other, notice that by the triangle inequality you are majorizing termwise. So it's ok. –  uforoboa Jan 31 '13 at 20:22

We need to show that $d(x,z)\leq d(x,y)+d(y,z)$. Let $x=(x_i),y=(y_i),z=(z_i)$. The problem reduces to showing that $\sup |x_i-z_i|\leq \sup |x_i-y_i|+\sup |y_i-z_i|$.
For each $i$, $\sup |x_i-y_i|+\sup |y_i-z_i|\geq |x_i-y_i|+|y_i-z_i|\geq |x_i-z_i|$ by the triangle inequality. Since $\sup |x_i-y_i|+\sup |y_i-z_i|$ is an upper bound of $\{|x_i-z_i|\}_i$, result follows as $\sup |x_i-z_i|$ is the least upper bound of $\{|x_i-z_i|\}_i$.
triangle inequality applied twice for the sup ($L_\infty$ norm) and abs ($L_1$ norm).