# Show that metric space is complete.

What I have so far:

Let $(a_n)_{n\in \mathbb{N}}$ be a Cauchy Sequence in E. This means $$\forall \epsilon > 0, \exists N \in \mathbb{N} : \forall j, k \geq N, d(a_j, a_k) < \epsilon$$ In this metric space, this means that $$d(\left( (a_n^{(j)})_{n\in \mathbb{N}}\right)_{j\in \mathbb{N}}, \left( (a_n^{(k)})_{n\in \mathbb{N}}\right)_{k\in \mathbb{N}}) < \epsilon$$ $$j, k \geq N \Rightarrow \sup_{n\in \mathbb{N}} |(x_n^{(j)})_{n\in \mathbb{N}} - (x_n^{(k)})_{n\in \mathbb{N}}| < \epsilon$$ For each $j, k \in \mathbb{N}$, set: $$Q_{j, k} = \{m \in E : |x_m^{(j)} - x_m^{(k)}| > \sup_{n\in \mathbb{N}} |(x_n^{(j)})_{n\in \mathbb{N}} - (x_n^{(k)})_{n\in \mathbb{N}}| \}$$

Now I'm not sure what to do with this set. There's more information on this metric in terms of functions and norms, but my class has only learn about this in terms of convergent sequences.

HINT: The crucial observation is this:

Suppose that $x=\langle x_n:n\in\Bbb N\rangle,y=\langle y_n:n\in\Bbb N\rangle\in E$; then $|x_n-y_n|\le d(x,y)$ for each $n\in\Bbb N$.

Now for $n\in\Bbb N$ let

$$x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle\in E\;,$$

and suppose that $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ is a Cauchy sequence. Then for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d\left(x^{(n)},x^{(\ell)}\right)<\epsilon$ whenever $n,\ell\ge m_\epsilon$. Now use the observation at the beginning: for each $k\in\Bbb N$ we have $\left|x_k^{(n)}-x_k^{(\ell)}\right|<\epsilon$ whenever $n,\ell\ge m_\epsilon$, so for each $k\in\Bbb N$ the sequence $\left\langle x_k^{(n)}:n\in\Bbb N\right\rangle$ is a Cauchy sequence in $\Bbb R$ and therefore has a limit $y_n$. Now consider the sequence $\langle y_n:n\in\Bbb N\rangle$.

Suppose $a_n$ is Cauchy with this distance. This means that for all $\epsilon>0$ there is some $N$ such that if $n,m \ge N$ then $d(a_n,a_m)= \sup_k |a_n(k)-a_m(k)| < \epsilon.$

First you need to come up with a candidate sequence.

Since $a_n(k)$ is Cauchy for any fixed $k$, we have $a_n(k) \to a(k)$ for some $a(k)$ (since $\mathbb{R}$ is complete). This $a$ is our candidate sequence, and we wish to show that (i) $a \in E$ and (ii) $d(a_n,a) \to 0$.

To show (i), note that if we choose $\epsilon=1$ there is some $N$ such that $|a_n(k)-a_m(k)| < 1$ for all $k$ and for all $n,m \ge N$. Since $a_N \in E$ there is some $B$ such that $|a_N(k)| \le B$ for all $k$. Hence $|a_n(k)| \le |a_N(k)| + |a_n(k)-a_N(k)| \le B+1$ for all $k$ from which it follows that $|a(k)| \le B+1$ for all $k$ and hence $a \in E$.

To show (ii), choose $\epsilon>0$ and choose $N$ such that $d(a_n,a_m) < {1 \over 2 } \epsilon$ for all $n,m \ge N$. Choose some $k$ and $n \ge N$, then $|a(k)-a_n(k)| \le |a(k)-a_m(k)|+|a_m(k)-a_n(k)|< |a(k)-a_m(k)|+{1 \over 2 } \epsilon$ for any $m \ge N$. Now choose $m$ large enough (depends on $k$) such that $|a(k)-a_m(k)| < {1 \over 2 } \epsilon$, from which we get $|a(k)-a_n(k)| < \epsilon$ for all $n \ge N$. Hence $d(a,a_n) \le \epsilon$ for all $n \ge N$.