Prove $d(x,y)=\sup _{n} \left| \sum _{k=1}^{n}(x_k-y_k)\right |$ is a metric Let $\gamma$ be the set of convergent series.$$\gamma = \{x=(x_k), x_k \in \mathbb{R} : \sum x_k <\infty\}$$
Prove that $(\gamma , d)$ is  a metric space, with $$d(x,y)=\sup _{n} \left| \sum _{k=1}^{n}(x_k-y_k)\right |\,\,\,(x, y \in \gamma).$$ 
 A: Everything except the triangle inequality is obvious, I'll leave them for you. For the triangle inequality, it is a matter of applying $\sup_n$ in the right order. We have: $$\left|\sum_{k = 1}^n (x_k - y_k) \right|\leq \left|\sum_{k = 1}^n (x_k - z_k) \right|+\left|\sum_{k = 1}^n (y_k - z_k) \right| \leq \sup_n\left|\sum_{k = 1}^n (x_k - z_k) \right|+\sup_n\left|\sum_{k = 1}^n (y_k - z_k) \right|.$$
Then, the right side of the inequality is an upper bound for $\left|\sum_{k = 1}^n (x_k - y_k) \right|$. Then, we apply $\sup_n$ and obtain:
$$\sup_n\left|\sum_{k = 1}^n (x_k - y_k) \right|\leq \sup_n\left|\sum_{k = 1}^n (x_k - z_k) \right|+\sup_n\left|\sum_{k = 1}^n (y_k - z_k) \right|.$$
I can apply $\sup_n$ like this because the previous inequality holds for all $n \geq 1$.
A: The only nontrivial thing to prove here is the triangle inequality.
Hint:
\begin{align*}
d\left(x,z\right) & =\sup_{n}\left|\sum_{k=1}^{n}x_{k}-z_{k}\right|\\
 & =\sup_{n}\left|\sum_{k=1}^{n}x_{k}-y_{k}+y_{k}-z_{k}\right|\\
 & =\sup_{n}\left|\sum_{k=1}^{n}x_{k}-y_{k}+\sum_{k=1}^{n}y_{k}-z_{k}\right|
\end{align*}
