Let M be the set of all summable non-negative sequences $\{x_k\}_{k=1}^\infty$ of real numbers, that is, $x_k \geq 0$ for all k and $\sum_{k=1}^\infty x_k$ converges to a real number. Let $d:M \to \textbf{R}$ be defined by $d(\{x_k\},\{y_k\})=\sum_{k=1}^\infty |x_k-y_k|$. Show that M, d is a metric space. Is there a reason why M can't be all summable sequences - why do we need non-negative?
$\textbf{Show that M,d is a metric space.}$
$\textbf{Proof:}$ For M to be a metric space, it must satisfy the following properties:
- If $\{x_k\},\{y_k\} \in M$ with $d(\{x_k\},\{y_k\})=0$, then $x_k=y_k$ (vice versa).
Let $\{x_k\},\{y_k\} \in M$. Assume $d(\{x_k\},\{y_k\})=\sum_{k=1}^\infty |x_k-y_k|=0$. $$ d(\{x_k\},\{y_k\})=\sum_{k=1}^\infty |x_k-y_k|=0 \iff |x_k-y_k|=0 \iff $$ \begin{cases} x_k-y_k=0 \iff x_k=y_k\\ -(x_k-y_k)=0 \iff x_k=y_k\\ \end{cases} Hence $x_k=y_k$.
- If $\{x_k\},\{y_k\} \in M$, then $d(\{x_k\},\{y_k\})=d(\{y_k\},\{x_k\})$.
Let $\{x_k\},\{y_k\} \in M$. Assume $d(\{x_k\},\{y_k\})=\sum_{k=1}^\infty |x_k-y_k|$. $$d(\{x_k\},\{y_k\})=\sum_{k=1}^\infty |x_k-y_k|=\sum_{k=1}^\infty |-(y_k-x_k)|=\sum_{k=1}^\infty |y_k-x_k|=d(\{y_k\},\{x_k\})$$ Hence $d(\{x_k\},\{y_k\})=d(\{y_k\},\{x_k\})$.
- Triangle Inequality Let $\{x_k\},\{y_k\},\{z_k\} \in M$. Assume $d(\{x_k\},\{z_k\})=\sum_{k=1}^\infty |x_k-z_k|$.
\begin{equation*} \begin{split} d(\{x_k\},\{z_k\}) & =\sum_{k=1}^\infty |x_k-z_k| \\ & =\sum_{k=1}^\infty |x_k-y_k+y_k-z_k|\\ & \leq \sum_{k=1}^\infty |x_k-y_k| +|y_k-z_k|\\ & =\sum_{k=1}^\infty |x_k-y_k| +\sum_{k=1}^\infty |y_k-z_k|\\ & =d(\{x_k\},\{y_k\})+d(\{y_k\},\{z_k\}) \\ \end{split} \end{equation*} Hence $d(\{x_k\},\{z_k\}) \leq d(\{x_k\},\{y_k\})+d(\{y_k\},\{z_k\})$.
Hence M,d is a metric space.
$\textbf{Is there a reason why M can't be all summable sequence?}$
The reason why M can't be all summable sequence is because then the sequence might not converge, particularly when the sequence is negative, and we will get an infinite number when summming it.
Is this explanation correct?