Exercise 21.2

Question

Let $X$ and $Y$ be metric spaces with metrics $d_X$ and $d_Y$, respectively.
Let $f \colon X 	o Y$ have the property that for every pair of points $x_1,x_2$ of $X$,
\begin{equation*}
  d_Y(f(x_1),f(x_2)) = d_X(x_1,x_2).
\end{equation*}
Show that $f$ is an imbedding. It is called an \emph{	extbf{isometric imbedding}} of $X$ in $Y$.

Amit Awasthi · Accepted Answer

\begin{proof} We first show $f$ is injective: \begin{equation*} f(x_1) = f(x_2) \implies d_Y(f(x_1),f(x_2)) = 0 \implies d_X(x_1,x_2) = 0 \implies x_1 = x_2 \end{equation*} by properties of metrics, and so we have an injective map. \par Now we consider the map $f'\colon X o \Im(X) \subseteq Y$, which is a bijection; it suffices to show that $f',f'^{-1}$ are continuous to show $f$ is an imbedding. Let $x \in X$ and $\epsilon > 0$ be given; then, letting $\delta = \epsilon$, we have \begin{equation*} d_X(x,y) < \delta \implies d_Y(f'(x),f'(y)) < \epsilon, \end{equation*} and so $f$ is continuous. Given $y \in Y$ and $\epsilon > 0$, letting $\delta = \epsilon$ gives \begin{equation*} d_Y(x,y) = d_Y(f(a),f(b)) < \delta \implies d_X(f'^{-1}(x),f'^{-1}(y)) = d_X(a,b) < \epsilon, \end{equation*} where $a,b$ exist by the bijectivity of $f$, and so $f'^{-1}$ is continuous. \end{proof}

Dan Whitman · Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} First, let $Z = f(X)$ and let $f'$ denote the function $f$ with the range restricted to $Z$ so that clearly $f'$ is surjective. First, we must show that $f$ and therefore also $f'$ is injective, from which it clearly follows that $f'$ is a bijection. So suppose that $x_1,x_2 \in X$ where $f(x_1) = f(x_2)$. Then we have that \begin{gather*} d_X(x_1, x_2) = d_Y(f(x_1), f(x_2)) = d_Y(f(x_1), f(x_1)) = 0 \end{gather*} by property (1) of the metric $d_Y$, so that it must be that $x_1 = x_2$ since $d_X$ is a metric and $d_X(x_1, x_2) = 0$. This of course means that $f$ and $f'$ are injective and hence $f'$ is bijective. We show that $f'$ and ${f^{\prime}}^{-1}$ are both continuous using Theorem~21.1 since $X$ and $Z$ are both metric spaces, noting that $Z \subset Y$ is a metric space with metric $d_Z = d_Y estriction Z imes Z$ by the previous exercise. So consider any $x \in X$ and any $\e > 0$, and let $\d = \e$. Then for any $x_1,x_2 \in X$ where $d_X(x_1,x_2) < \d$ we have \begin{gather*} d_Z(f'(x_1), f'(x_2)) = d_Y(f(x_1), f(x_2)) = d_X(x_1,x_2) < \d = \e \,, \end{gather*} which suffices to show that $f'$ is continuous. A similar argument shows that ${f^{\prime}}^{-1}: Z o X$ is continuous. For $y_1,y_2 \in Z$ and $\e > 0$, again let $\d = \e$ and suppose that $d_Z(y_1,y_2) <\d$. Let $x_1 = {f^{\prime}}^{-1}(y_1)$ and $x_2 = {f^{\prime}}^{-1}(y_2)$. Then we have \begin{align*} d_X({f^{\prime}}^{-1}(y_1), {f^{\prime}}^{-1}(y_2)) &= d_X(x_1, x_2) = d_Y(f(x_1), f(x_2)) = d_Y(f'(x_1), f'(x_2)) \ &= d_Y(f'({f^{\prime}}^{-1}(y_1)), f'({f^{\prime}}^{-1}(y_2))) = d_Y(y_1, y_2) \ &= d_Z(y_1, y_2) < \d = \e \,, \end{align*} which of course shows that ${f^{\prime}}^{-1}$ is also continuous. This shows that $f'$ is a homeomorphism so that $f$ is an imbedding as desired.

Exercise 21.2

Answers

Comments

Comments

Exercise 21.2

Answers

Comments

Comments

Add answer