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  1. Let $(X,d)$ be a metric space. We denote by $C_b(X;\mathbb{R})$ the space of continuous and bounded functions from $X$ into $\mathbb{R}$, equipped with the sup-norm metric. We define a mapping $O: X \to C_b(X;\mathbb{R})$ as follows. Fix a point $x_0\in X$. Given $x\in X$, $O(x)$ is a function from $X$ into $\mathbb{R}$ such that $O(x)(y) = d(y, x)-d(y, x_0)$: Show the following.

    • $(i)$ $C_b(X;\mathbb{R})$ is a complete metric space.
    • $(ii)$ For each $x \in X$, the corresponding function $O(x): X \to \mathbb{R}$ is continuous and bounded.
    • $(iii)$ Mapping $x \mapsto O(x)$ yields an isometric embedding of $X$ into $C_b(X,\mathbb{R})$. Thus every metric space may be isometrically embedded into a complete metric space.
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Why are you editing the question, its title, and proposing to edit my answer out as well? – T. Eskin Feb 25 '13 at 16:46
@user63929, that is not how this site works. If you made a mistake by posting here, then now you know better. – user58512 Feb 25 '13 at 16:52
@user63929: Maybe if you give a reason, it will be. – Chris Eagle Feb 25 '13 at 16:54
Self-vandalizing your question like this is extremely disrepecful to those who have taken time to answer. People are here and helping without any compensation, and you are showing very bad manners treating their help in this way. – mrf Feb 25 '13 at 16:55
@user58512: What on earth are you talking about? – Chris Eagle Feb 25 '13 at 17:01
  • $(i)$ Take a Cauchy sequence $(f_{n})_{n=1}^{\infty}\subseteq C_{b}(X;\mathbb{R})$ in the sup-norm $\|\cdot\|_{\infty}$ and $\varepsilon>0$. Hence there exists $n_{\varepsilon}\in\mathbb{N}$ so that \begin{equation*} |f_{n}(x)-f_{m}(x)|\leq \|f_{n}-f_{m}\|_{\infty}<\frac{\varepsilon}{3} \end{equation*} for all $n,m\geq n_{\varepsilon}$, which shows that $(f_{n}(x))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}$ for every $x\in X$. Since $\mathbb{R}$ is complete, for every $x\in X$ there exists $f(x)\in \mathbb{R}$ so that $f_{n}(x)\to f(x)$. Since limits are unique in metric spaces, you may define a function $f:X\to\mathbb{R}$ so that $x\mapsto f(x)$. Now $f$ is the point-wise limit of $f_{n}$. We show that $\|f_{n}-f \|_{\infty}\to 0$ and $f\in C_{b}(X;\mathbb{R})$. Fix $x\in X$. Now there exists $n_{0}\in \mathbb{N}$ so that $|f_{n}(x)-f(x)|<\frac{\varepsilon}{3}$ for all $n\geq n_{0}$.. Thus for every $n\geq n_{1}:=\max\{n_{\varepsilon},n_{0}\}$ we have \begin{equation*} |f_{n}(x)-f(x)|\leq |f_{n}(x)-f_{n_{0}}(x)|+|f_{n_{0}}(x)-f(x)|<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\frac{2\varepsilon}{3}. \end{equation*} Hence by taking supremum over all $x\in X$, we have $\|f_{n}-f\|_{\infty}\leq \frac{2\varepsilon}{3}<\varepsilon$ for all $n\geq n_{1}$. Hence $\|f_{n}-f\|_{\infty}\to 0$. You may now either conclude that $f\in C_{b}(X;\mathbb{R})$ since $f$ is the uniform limit of continuous bounded functions, or prove it as follows. Let $x\in X$ and everything else as above remain fixed. Since $f_{n_{1}}$ is continuous, there exists $\delta>0$ so that $f_{n_{1}}B(x,\delta)\subseteq B(f_{n_{1}}(x),\frac{\varepsilon}{3})$. Now for all $y\in B(x,\delta)$ we have \begin{align*} |f(x)-f(y)| &\leq |f(x)-f_{n_{1}}(x)|+|f_{n_{1}}(x)-f_{n_{1}}(y)|+|f_{n_{1}}(y)-f(y)| \\ &\leq 2\|f_{n_{1}}-f\|_{\infty}+|f_{n_{1}}(x)-f_{n_{1}}(y)| \\ &<3\frac{\varepsilon}{3}=\varepsilon, \end{align*} and thus $fB(x,\delta)\subseteq B(f(x),\varepsilon)$. Hence $f$ is continuous. Finally, since \begin{equation*} \|f\|_{\infty}\leq \|f-f_{n_{1}}\|_{\infty}+\|f_{n_{1}}\|_{\infty}<\frac{\varepsilon}{3}+\|f_{n_{1}}\|_{\infty}<\infty, \end{equation*} then $f$ is bounded. Hence $f\in C_{b}(X;\mathbb{R})$, and since $\|f_{n}-f\|_{\infty}\to 0$, we have proven that $C_{b}(X;\mathbb{R})$ is complete.

  • $(ii)$ Note that $|O(x)(y)|=|d(y,x)-d(y,x_{0})|\leq d(x,x_{0})$ for all $y\in X$ by the reverse triangle-inequality. Hence $\|O(x)\|_{\infty}\leq d(x,x_{0})$, which shows that $O(x)$ is bounded for every $x\in X$. The functions $y\mapsto d(y,x)$ and $y\mapsto -d(y,x_{0})$ are continuous, so each $O(x)$ is continuous as a sum of two continuous functions. Hence $O(x)\in C_{b}(X;\mathbb{R})$ for every $x\in X$.

  • $(iii)$ Note at first, that again by the reverse triangle-inequality it follows that for all $x,y\in X$: \begin{align*} \|O(x)-O(y)\|_{\infty}&= \sup_{z\in X}\|O(x)(z)-O(y)(z)\|_{\infty} \\ &=\sup_{z\in X}\|d(x,z)-d(z,x_{0})-d(z,y)+d(z,x_{0})\|_{\infty} \\ &=\sup_{z\in X}\|d(x,z)-d(z,y)\|_{\infty} \\ &\leq d(x,y). \end{align*} On the other hand, \begin{align*} 0\leq d(x,y)&=d(x,y)-d(x,x_{0})+d(x,x_{0})-d(x,x) \\ &=O(y)(x)-O(x)(x) \\ &= |O(y)(x)-O(x)(x)| \\ &\leq \|O(y)-O(x)\|_{\infty}. \end{align*} Hence $\|O(y)-O(x)\|_{\infty}=d(x,y)$ for all $x,y\in X$, which shows that $x\mapsto O(x)$ is an isometric embedding $X\to C_{b}(X;\mathbb{R})$.

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In your first point you show $\lVert f_n- f \rVert < \frac 2 3 \epsilon$ if $n \geq n_1$, which seems strange to me since $n_1$ dependes on a chosen $x \in X$. As I see it, you cannot simply take the supremum if you don't have a uniform $n_0$ which is valid for all $x \in X$. – Andre Jul 19 '15 at 13:16

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