I'm self studying analysis right now. There is a problem in the textbook I cannot solve.
Problem: Let $X$ be the metric space consisting of real valued sequences $\{x_n\}$that converge to $0$, with the metric $d(\{x_n\},\{y_n\}) =$ sup$_n |x_n - y_n|$. Prove that $X$ is complete.
I've done this so far, but I do not know how to progress further: Sps. $\{x_n\}$ is a Cauchy sequence in $(X,d)$. Let $x_n = \{b_{n,1},b_{n,2} ,b_{n,3} ... \}$ where $b_{n,k} \in \mathbb{R}$. Then, given $\epsilon >0$, there exists natural $N$ so for all $m,n > N, d(x_n, x_m) < \epsilon$ . This means $d(\{b_{n,k}\},\{b_{m,k}\}) < \epsilon$, so $|{b_{n,k} - b_{m,k}}| < \epsilon$ for all natural $k$.
Fix $k$. Then for all $m, n > N$, $|{b_{n,k} - b_{m,k}}| < \epsilon$. Thus $ \{ b_{1,k}, b_{2,k}, b_{3,k} ... \} $ is Cauchy.
I now not know how to carry on. Can someone help me with this?
The usual answer to these type of questions can be put into three steps. Our space of consideration here is $c_0$, the set of real sequences that converge to $0$. We are considering the sup norm in sequence space.
(i) Find a possible limiting value, $x$.
If $(x_n)_k \in c_0$ is a sequence of points in $c_0$ that is Cauchy (note that a point in $c_0$ is also a sequence), then exists $N$ such that for all $k \ge N$, we have $ d((x_n)_k, (x_n)_l) \le \epsilon $. Thus, $$ \sup_{n \in \mathbb{N}} |x_{n,k} -x_{n,l} | \le \epsilon \quad \quad (1)$$ Hence, for each $i \in \mathbb{N}$, the sequence $\{x_{i,k} \}_{k \in \mathbb{N}}$ is a Cauchy sequence. So limits to a value, $x_i \in \mathbb{R}$ by the completness of reals.
Define $x := (x_1, x_2, \ldots) $.
(ii) Show that the sequence of points in space limits to $x$.
From previous inequality $(1)$ by limiting $l$ to infinity, we obtain that for all $k \ge N$ $$ \sup_{n \in \mathbb{N}} | x_{n,k} - x_n | \le \epsilon \quad \quad (2)$$
Hence, as $\epsilon$ was arbitrary $\lim_{k \rightarrow \infty} (x_n)_k = x$. To clarify the limit,
$$ |x_{n} - x_{n,k} | \le |x_{n} - x_{n,l} | + |x_{n,l} - x_{n,k} | \le |x_{n} - x_{n,l}| + \epsilon $$ for all $l,k \ge N$. So by limiting the RHS, noting that absolute value map is continuous, we obtain our desired inequality.
(iii) Lastly, we need to show that $x $ is indeed in the space.
From $(2)$, note that exists $M$ such that for all $m \ge M$, $|x_{m,k}| < \epsilon$, so $$|x_{m}| \le |x_{m,k} - x_{m} | +|x_{m,k}| \le 2 \epsilon$$ Again, as $\epsilon$ was arbitrary, we can always find a corresponding sequence $(x_{n})_{k_\epsilon}$ and $M_{\epsilon}$ such that for all $ m \ge M_{\epsilon}$ we have $|x_m| < \epsilon$.