Let $$X=\left\{(a_n)_{n \in \mathbb N} \in \mathbb R^N : \exists n_0 \in \mathbb N\, \forall n\ge n_0 \big(a_n\le \sqrt{n}\big)\right\}$$ with the metric $$d\big((a_n)_{n \in \mathbb N},(b_n)_{n \in \mathbb N}\big)=\sup_{n \in \mathbb N}\frac{|a_n-b_n|}{n}$$

  1. Determine if $(X,d)$ is separable

  2. Prove that for every Cauchy sequence in $X$ there exists a sequence of real terms such that the Cauchy sequence converges to that one but that $X$ is not complete.

Can anyone give me a hint?


With this definition of $X$, the function $d$ isn’t defined on all of $X\times X$: for example, if $z$ is the zero sequence and $x=\langle -n^2:n\in\Bbb N\rangle$, then $z,x\in X$, but $$\sup_{n\in\Bbb N}\frac{|0-(-n^2)|}n=\sup_{n\in\Bbb N}n$$ doesn’t exist. I’m going to assume that the last condition in the definition of $X$ was supposed to be that $|a_n|\le\sqrt{n}$. Then if $x=\langle x_n:n\in\Bbb N\rangle,y=\langle y_n:n\in\Bbb N\rangle\in X$, then there is an $m\in\Bbb N$ such that $$\frac{|x_n-y_n|}n\le\frac{2\sqrt{n}}n=\frac2{\sqrt{n}}$$ for all $n\ge m$, and $d(x,y)$ is defined.

  1. Consider rational sequences that are eventually $0$.

  2. For $n,k\in\Bbb N$ let $$x_k^{(n)}=\begin{cases}2\sqrt{k},&\text{if }k\le n\\0,&\text{if }k>n\;,\end{cases}$$ and let $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle$. Show that $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ is a Cauchy sequence in $X$ that has no limit in $X$. There is a sequence $x\in\Bbb R^{\Bbb N}\setminus X$ such that $d\big(x^{(n)},x\big)$ is defined for each $n\in\Bbb N$ and converges to $0$ as $n\to\infty$; what is it?

    More generally, show that if $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ is any Cauchy sequence in $X$, where $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle$ for each $n\in\Bbb N$, then $\left\langle x_k^{(n)}:n\in\Bbb N\right\rangle$ is a Cauchy sequence in $\Bbb R$ for each $k\in\Bbb N$. Use this observation to show that there is an $x\in\Bbb R^{\Bbb N}$ such that $d\big(x^{(n)},x\big)$ is defined for each $n\in\Bbb N$ and converges to $0$ as $n\to\infty$.

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  • $\begingroup$ Thanks, I've checked the statement of the exercise and it was without absolute value, but you're right, it makes no sense because then the distance could give infinity. I'm still working on 2., but your suggestions were of big help. $\endgroup$ – Andrew Oct 9 '13 at 4:30
  • $\begingroup$ @Andrew: You’re welcome. $\endgroup$ – Brian M. Scott Oct 9 '13 at 4:36

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