I understand the definition of an isometric embedding, (an injective, distance preserving map) but I don't understand what it means for a metric space to be isometrically embedded in another space. Does it mean that all maps from the first space to the other need to be an isometric embedding? Does just one? I tried searching online but I couldn't find any examples.
[Math] What does it mean for a metric space to be isometrically embedded in another space
metric-spaces
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$(i)$ 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 $X\to C_{b}(X;\mathbb{R})$.
The definition is $d_2(f(x),f(y))=d_1(x,y)$. For this to be a metric we have to assume that $f$ is injective.
For example let $f:\mathbb R \to \mathbb R$ be defined by $f(x)=e^{x}$. the range of this function is $(0,\infty)$ and the induced metric $d_2$ on the range is defined by $d_2(x,y)=|\log \,x -\log \, y |$ (assuming that the domain of $f$ is given the usual metric).
Best Answer
Let $(X,d)$ & $(Y,d_1)$ be a metric spaces. Then $X$ is isometrically embedded to another space means $\exists f:X \to Y$ s.t $d(x,y)=d_1(f(x),f(y)) \forall x,y\in X$