A homeomorphism preserves topological structure, but an isometry preserves exact distances. The focus keyword isometry marks this stronger notion of sameness. Isometries preserve Cauchy sequences and completeness in a way homeomorphisms need not. This lesson studies onto isometries, sequence-space shifts, and isometric embeddings into spaces of bounded continuous functions. :::definition[Isometry] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A mapping $f:X\to Y$ is called an isometry if $d_Y(f(x),f(y))=d_X(x,y)$ for all $x,y\in X$. ::: :::definition[Isometric Metric Spaces] Two metric spaces $X$ and $Y$ are called isometric if there exists an onto isometry from $X$ to $Y$. ::: :::theorem[Onto Isometries are Homeomorphisms] Every onto isometry between metric spaces is a homeomorphism. ::: :::proof Given that $f:X\to Y$ is an onto isometry. To prove that $f$ is a homeomorphism. If $f(x)=f(y)$, then $d_X(x,y)=d_Y(f(x),f(y))=0$, so $x=y$. Thus $f$ is bijective. If $x_n\to x$, then $d_Y(f(x_n),f(x))=d_X(x_n,x)\to0$, so $f$ is continuous. The inverse also preserves distances because $d_X(f^{-1}(u),f^{-1}(v))=d_Y(u,v)$. Hence $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. $\square$ ::: :::solved-problem Let $f$ be an isometry of a complete metric space $(X_1,d_1)$ onto $(X_2,d_2)$. Prove that $X_2$ is complete. ::: :::solution Let $\{y_n\}$ be Cauchy in $X_2$. Choose $x_n\in X_1$ with $f(x_n)=y_n$. Since $f$ is an isometry, $d_1(x_n,x_m)=d_2(y_n,y_m)$, so $\{x_n\}$ is Cauchy. Completeness of $X_1$ gives $x_n\to x$. Then $y_n=f(x_n)\to f(x)$, so $X_2$ is complete. ::: :::solved-problem Let $X=\ell^p$, $p\geq1$, and define $\omega(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$. Prove that $\omega$ is an isometry. ::: :::solution For $x,y\in\ell^p$, $d(\omega(x),\omega(y))=(|0-0|^p+\sum_{i=1}^{\infty}|x_i-y_i|^p)^{1/p}=d(x,y)$. Hence, $\omega$ is an isometry. ::: :::solved-problem Let $X$ be a metric space and fix $x_0\in X$. Define $f_x(y)=d(y,x)-d(y,x_0)$. Show that $x\mapsto f_x$ is an isometric embedding into $C_b(X,\mathbb R)$ with the uniform metric. ::: :::solution For every $y$, $|f_x(y)|\leq d(x,x_0)$, so $f_x$ is bounded. Also $|f_{x_1}(y)-f_{x_2}(y)|\leq d(x_1,x_2)$, so $d_u(f_{x_1},f_{x_2})\leq d(x_1,x_2)$. Taking $y=x_2$ gives equality. Hence, the embedding is isometric. ::: :::remark This isometric embedding gives a way to construct completions: take the closure of the isometric image inside a complete function space. ::: :::exercise 1. Prove every isometry is one-to-one. 2. Give a homeomorphism that is not an isometry. 3. State a property preserved by onto isometry. ::: :::answer 1. If images are equal, distance in the domain is zero. 2. $x\mapsto x/(1+|x|)$ from $\mathbb R$ to $(-1,1)$. 3. Completeness. :::