Let $X$ and $Y$ be metric spaces with metrics $d_X$ and $d_Y$ , respectively. Let $f : X \to Y$ have the property that for every pair of points $x_1,x_2$ of $X$, $d_Y (f(x_1),f(x_2)) = d_X(x_1,x_2)$. Show that $f$ is an imbedding. It is called an isometric imbedding of $X$ in $Y$.
My attempt: let $x\in X$ and $\epsilon \gt 0$. Take $\delta =\epsilon \gt 0$ so that $d_X(x,y)\lt \delta ,\forall y\in X \Rightarrow d_Y(f(x),f(y))=d_X(x,y)\lt \epsilon$. Thus $f$ is continuous on $X$. Let $x,x’\in X$ and $f(x)=f(x’)$. Then $d_Y(f(x),f(x’))=d_X(x,x’)=0$, which implies $x=x’$. Hence $f$ is injective. Let $Z=f(X)$. Then clearly $h:X\to Z$ is bijective and $h$ is continuous by theorem 18.2(e). Now we need to show inverse map of $h$ is continuous. Since $h$ is bijective, $f$ is invertible. By definition of invertible, $\exists !g:Z\to X$ such that $h\circ g=\text{id}_Z$ and $g\circ h=\text{id}_X$. We write $g$ as $h^{-1}$. Let $p,q\in Z$. Then $\exists !r,s\in X$ such that $h(r)=f(r)=p$ and $h(s)=f(s)=q$, by definition of bijective map. Since $g\circ h=\text{id}_X$, we have $g(h(r))=r=g(p)$ and $g(h(s))=s=g(q)$. Let $z\in Z$ and $\epsilon \gt 0$ given. Take $\delta =\epsilon \gt 0$ such that, $d_Y(z,m)\lt \delta ,\forall m\in Z \Rightarrow d_Y(z,m)=d_Y(h(x),h(y))=d_X(x,y)=d_X(g(z),g(m))\lt \delta =\epsilon$, our desired result. Hence $h$ is homeomorphism. Note in this case, $p=z$, $q=m$, $r=x$ and $s=y$. Is this proof correct?
Edit: $(Z,d_Z)$ is a metric space. $\mathcal{T}_s$(Subspace topology on $Z$)$=\mathcal{T}_{d_Z}$.
Best Answer
Yes, this is correct. I'd write it more succinctly: $f$ is continuous because we can take $\delta=\varepsilon$ for all $x$ at the same time (so even uniformly continuous). It's injective as $f(x)=f(x')$ implies $0=d_Y(f(x),f(x')) = d_X(x,x')$ so that $x=x'$ by a metric axiom. It follows that $f': X \to f[X]\subseteq Y$ is a bijection (still continuous, as a codomain restriction) and the inverse $g: f[X] \to X$ is clearly also an isometry and so continuous for the same reason as $f$ is. Hence $f:X \simeq f[X]$ and $f$ is an embedding.