Is $(A,\rho)$ complete? A is a set of all null sequences $x=(\alpha_n)$ of complex numbers C.
$\rho=sup_i|\alpha_i-\beta_i|$ is a metric on A such that $x=(\alpha_n)$ and $y=(\beta_n)$. Assume C to be complete under the usual metric.
Is $(A,\rho)$ complete?
I know that if $(x_n)$ is Cauchy in A where $x_n$=(_)_j then for each (_)_n is a Cauchy sequence in C. But how should I proceed? Also, apologies for the kind of weird mathematical notation. I didn't know how to add double subscripts.
 A: Let $(\alpha^n)_{n=1}^\infty$ be a Cauchy sequence. For each $n$ write $\alpha^n=(\alpha_i^n)_{i=1}^\infty$. Let $\epsilon>0$. By definition of Cauchy sequence there is some $n_0\in\mathbb{N}$ such that for all $m,n\geq n_0$ we have $||\alpha^n-\alpha^m||<\epsilon$. From the definition of norm it follows that for each $i\in\mathbb{N}$:
$|\alpha_i^n-\alpha_i^m|\leq ||\alpha^n-\alpha^m||<\epsilon$
And so for each $i$, the sequence $(\alpha_i^n)_{n=1}^\infty$ is a Cauchy sequence of complex numbers. Since $\mathbb{C}$ is complete it follows that there is some $\alpha_i$ such that $\alpha_i^n\to\alpha_i$ when $n\to\infty$. Finally, let $\alpha=(\alpha_i)_{i=1}^\infty$. We have to show that $\alpha$ is indeed a null sequence and that $\alpha^n\to\alpha$ when $n\to\infty$.
First we show that $\alpha$ is null. Let $\epsilon>0$. Again, there is some $n_0\in\mathbb{N}$ such that for all $m,n\geq n_0$ and for all $i\in\mathbb{N}$ we have:
$|\alpha_i^m-\alpha_i^n|\leq\frac{\epsilon}{2}$
In particular $|\alpha_i^m-\alpha_i^{n_0}|<\frac{\epsilon}{2}$ for all $m\geq n_0$ and all $i$. Taking $m\to\infty$ we obtain $|\alpha_i-\alpha_i^{n_0}|\leq\frac{\epsilon}{2}$. Also, the sequence $(\alpha_i^{n_0})_{i=1}^\infty$ is null and so there is some $i_0$ such that $|\alpha_i^{n_0}|\leq\frac{\epsilon}{2}$ for all $i\geq i_0$. And then, for all $i\geq i_0$ we get:
$|\alpha_i|\leq |\alpha_i-\alpha_i^{n_0}|+|\alpha_i^{n_0}|\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
And so $\alpha$ is indeed null.
Finally, we will show that $\alpha^n\to\alpha$ when $n\to\infty$. Again, let $\epsilon>0$. Again, there is some $n_0$ such that if $m,n\geq n_0$ and if $i\in\mathbb{N}$ then:
$|\alpha_i^m-\alpha_i^n|<\epsilon$
So now take a constant $n\geq n_0$. For all $m\geq n_0$ we have $|\alpha_i^m-\alpha_i^n|<\epsilon$. Taking $m\to\infty$ we obtain $|\alpha_i-\alpha_i^n|\leq\epsilon$. This is true for all $i$, and so this means $||\alpha-\alpha^n||\leq\epsilon$. And this is true for all $n\geq n_0$, so by definition we indeed get $\alpha^n\to\alpha$.
