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Let $(X,d)$ be a metric space.

Let $C(X)$ be the set of all Cauchy seq. on $X$ and define for $(x_n)_{n \in \mathbb{N}}$ , $(y_n)_{n \in \mathbb{N}}$ the following relation

$$ (x_n)_{n \in \mathbb{N}} \sim (y_n)_{n \in \mathbb{N}} \Leftrightarrow d(x_n,y_n) \to 0 \ (n \to \infty) $$

Let $\hat{X} = C(X) / \sim $ and define the metric $\hat{d}$

$$ \hat{d}\left([(x_n)_{n \in \mathbb{N}}], [(y_n)_{n \in \mathbb{N}}]\right) = \lim_{n \to \infty} d(x_n, y_n) \\ ([(x_n)_{n \in \mathbb{N}}, [(y_n)_{n \in \mathbb{N}}] \in \hat{X} $$

For every $x \in X$ the seq. $(x,x,x, \dots) \in C(X)$. Define

$$ \tilde{X} = \{\tilde{x} : x\in X, \ \tilde{x} = [(x,x,x,\dots)] \} $$

Show that $\tilde{X}$ is dense in $\hat{X}$. I have already proved that $X$ and $\tilde{X}$ are isometric

After the answer from Brian I came up with the following.

So $x \in \hat{X}$ thus $x= [(x_n)_{n \in \mathbb{N}}]$ where $(x_n)_{n \in \mathbb{N}}$ is Cauchy. Since $(x_n)_{n \in \mathbb{N}}$ is Cauchy, then for every $\epsilon > 0 $, $\exists N \in \mathbb{N}$ s.t $\forall n,m > N$ $d(x_n.x_m) < \epsilon$. Then for $\hat{d}(x, \tilde{x}_j)$ we have that $\hat{d}(x, \tilde{x}_j) = \hat{d}([x_1,x_2,x_3, \dots],[x_j,x_j,x_j, \dots ]) = \lim_{n \to \infty} d(x_n, x_j) < \epsilon$ if $j > N$

Olba12
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1 Answers1

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HINT: It suffices to show that if $x=\langle x_n:n\in\Bbb N\rangle\in\hat X$, there is a sequence in $\tilde X$ converging to $x$. The sequence $\langle\widetilde{x_n}:n\in\Bbb N\rangle$ in $\tilde X$ is a good candidate to consider.

Brian M. Scott
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  • I'm sorry, but I realised that I didnt include that I have proved that $X$ and $\tilde{X}$ are isometric (it's included now). I dont know if that is of use. anyways, I will Think about your hint now. – Olba12 Nov 25 '16 at 22:02
  • @Olba12: It’s not needed here, since you’ll be concerned only with $\hat d(x,\widetilde{x_n})$. – Brian M. Scott Nov 25 '16 at 22:05
  • So $x \in \hat{X}$ thus $x= [(x_n){n \in \mathbb{N}}]$ where $(x_n){n \in \mathbb{N}}$ is Cauchy. Since $(x_n){n \in \mathbb{N}}$ is Cauchy, then for every $\epsilon > 0 $, $\exists N \in \mathbb{N}$ s.t $\forall n,m > N$ $d(x_n.x_m) < \epsilon$. Then for $\hat{d}(x, \tilde{x}_j)$ we have that $\hat{d}(x, \tilde{x}_j) = \hat{d}([x_1,x_2,x_3, \dots],[x_j,x_j,x_j, \dots ]) = \lim{n \to \infty} d(x_n, x_j) < \epsilon$ if $j > N$ – Olba12 Nov 25 '16 at 22:34
  • If I understand you right, this proves that $\hat{X} \subset cl(\tilde{X})$, the other part comes from that $\tilde{X} \subset \hat{X} \Rightarrow cl(\tilde{X}) = cl(\hat{X}) = \hat{X}$? – Olba12 Nov 25 '16 at 22:42
  • @Olba12: You actually just get $\lim_nd(x_n,x_j)\le\epsilon$, but that’s good enough. And yes, you then have $\hat X\subseteq\operatorname{cl}\tilde X\subseteq\hat X$, so $\operatorname{cl}\tilde X=\hat X$. – Brian M. Scott Nov 25 '16 at 22:54
  • why $\lim_{n \to \infty} d(x_n,x_j) \leq \epsilon$ i.e why $\leq$ and not $<$? – Olba12 Nov 25 '16 at 23:13
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    @Olba12: Consider the case $x_n=1-\frac1n$ for $n\ge 1$. Then $\lim_n|x_n-x_0|=1$ even though $|x_n-x_0|<1$ for every $n>1$. It’s a good general rule: when you take limits, strict inequalities generally become non-strict inequalities. – Brian M. Scott Nov 25 '16 at 23:16