# The space of sequences as a complete metric space

Let $s$ be the space of all functions $x:\mathbb{N}\to \mathbb{F}$ where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$. Define addition and scalar multiplication componentwise so that $s$ is vector space. If $x, y\in s$, define $$d(x,y) := \sum_{n=0}^\infty 2^{-n}\frac{|x(n)-y(n)|}{1+|x(n)-y(n)|}.$$ Then $d$ is a metric on $s$, and in fact makes $s$ a topological vector space. Now I am trying to prove that under this metric, $s$ becomes complete. One attempt at a proof was to somehow make $\mathbb{N}$ into a finite measure space so that converging and being Cauchy are the same as being convergent and Cauchy in measure respectively. But I want to also try and prove it without resorting to any measure theory.

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You can prove something more general: Suppose $(X_i,d_i)$ is a complete metric space, then the product $\prod X_i$ is a complete metric space (with the metric which you describe above). –  Asaf Karagila Aug 21 '11 at 5:59
This answer by me to another question might be relevant. –  kahen Aug 21 '11 at 6:28
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## 2 Answers

Suppose that $\langle x_n:n \in \omega \rangle$ is a Cauchy sequence in $s$. The first step is to show that $\langle x_n:n \in \omega \rangle$ converges pointwise, which can be done by showing that each of the coordinate sequences $\langle x_n(k):k\in\omega\rangle$ is Cauchy in $\mathbb{F}$.

For each $\epsilon > 0$ there is an $n_\epsilon \in \omega$ such that $d(x_m,x_n) < \epsilon$ whenever $m,n \ge n_\epsilon$. Then for any $k \in \omega$ we have $$2^{-k}\frac{|x_m(k)-x_n(k)|}{1+|x_m(k)-x_n(k)|} \le d(x_m,x_n) < \epsilon$$ whenever $m,n \ge n_\epsilon$. This inequality can be rewritten $$|x_m(k)-x_n(k)| < 2^k\epsilon (1 + |x_m(k)-x_n(k)|),$$ or $$(1-2^k\epsilon)|x_m(k)-x_n(k)| < 2^k\epsilon.$$

Now fix $k \in \omega$. If $\epsilon$ is chosen to be less than $2^{-k}$, so that $1-2^k\epsilon > 0$, we have $$|x_m(k)-x_n(k)| < \frac{2^k\epsilon}{1-2^k\epsilon} = \frac{1}{1-2^k\epsilon}-1$$ whenever $m,n \ge n_\epsilon$. Clearly $\lim\limits_{\epsilon\to 0^+} \left(\frac{1}{1-2^k\epsilon}-1\right) = 0$, so for each $k\in\omega$, $\langle x_n(k):n\in\omega\rangle$ is a Cauchy sequence in $\mathbb{F}$. $\mathbb{F}$ is complete, so each $\langle x_n(k):n\in\omega\rangle$ converges to some $x(k) \in \mathbb{F}$. Now you need only show that $\langle x_n:n \in \omega \rangle$ converges in $s$ to the sequence $x:\omega\to\mathbb{F}:k \mapsto x(k)$, which is just a matter of keeping track of epsilons. (Stop here if you want to try it on your own.)

We have $$d(x_n,x) = \sum\limits_{k\ge 0}2^{-k}\frac{|x_n(k)-x(k)|}{1+|x_n(k)-x(k)|}.$$ Fix $\epsilon > 0$, and choose $m_0 \in \omega$ big enough so that $2^{-m_0} < \epsilon/2$. Then $$\sum\limits_{k>m_0}2^{-k}\frac{|x_n(k)-x(k)|}{1+|x_n(k)-x(k)|} < \sum\limits_{k>m_0}2^{-k} = 2^{-m_0} < \epsilon/2.$$ Finally, choose $m_1 \in \omega$ big enough so that for each $k \le m_0$, $$|x_n(k)-x(k)|< \frac{\epsilon}{2(m_0+1)}$$ whenever $n \ge m_1$. Then \begin{align*} d(x_n,x) &= \sum\limits_{k=0}^{m_0}\frac{2^{-k}}{1+|x_n(k)-x(k)|}|x_n(k)-x(k)|+\sum\limits_{k>m_0}2^{-k}\frac{|x_n(k)-x(k)|}{1+|x_n(k)-x(k)|}\\ &< \sum\limits_{k=0}^{m_0}|x_n(k)-x(k)|+\epsilon/2\\ &< (m_0+1)\frac{\epsilon}{2(m_0+1)} + \frac{\epsilon}{2}\\ &= \epsilon \end{align*} whenever $n > m_1$, and hence $\langle x_n:n \in \omega \rangle \to x$ in $s$.

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given any $\epsilon > 0$ you can take $\delta>0$ such that $$\frac{\delta}{2^{-j}-\delta}=\epsilon$$ –  leo Aug 21 '11 at 6:01
and then take $N$ such that if $n,m\geq N$, then $d(x_m,x_n)<\delta$. This simplifies the proof of that the coordinate sequences are Cauchy. –  leo Aug 21 '11 at 6:05
And nice at your end. –  leo Aug 21 '11 at 6:11
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Hint: a Cauchy sequence converges pointwise. Show that it converges in your metric.

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