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?

  • $\begingroup$ $\Bbb R$ is complete, so every Cauchy sequence in $\Bbb R$ converges. Let $b_k=\lim_nb_{n,k}$. The sequence $x=\langle b_1,b_2,\ldots\rangle$ is a natural candidate for the limit of the $x_n$. $\endgroup$ – Brian M. Scott Nov 13 '16 at 23:56
  • $\begingroup$ How can I show this? $\endgroup$ – lithium123 Nov 14 '16 at 0:21

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 $|x_m| < \epsilon$.


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