Suppose \(f:X\rightarrow Y\) is an isometry. If \(x,x'\in X\) and \(x\neq x'\text{,}\) then \(d(x,x')>0\text{.}\) So \(f(x)\neq f(x')\text{.}\)
Definition4.112.Isometric Isomorphism.
An isometric isomorphism between metric spaces \(X\) and \(Y\) is a surjective isometry \(f:X\rightarrow Y\text{.}\) In this case, \(f\) is bijective and \(f^{-1}:Y\rightarrow X\) is also an isometry. \(X\) and \(Y\) are isometrically isomorphic if exists an isometric isomorphism \(f:X\rightarrow Y\text{.}\)
Definition4.113.Completion.
A completion of a metric space \(X\) is a pair \((\hat{X},j)\) where \(\hat{X}\) is a complete metric space and \(j:X\rightarrow \hat{X}\) is an isometry such that \(j(X)\) is dense in \(\hat{X}\text{.}\)
Example4.114.
If \(X=\mathbb{Q}\text{,}\) then \((\mathbb{R},j:\mathbb{Q}\hookrightarrow\mathbb{R})\) is a completion of \(\mathbb{Q}\text{.}\)
If \(X=(a,b)\subseteq \mathbb{R}\text{,}\) then \(([a,b], j:(a,b)\hookrightarrow[a,b])\) is a completion of \((a,b)\text{.}\)
\(X=(a,b)\subseteq \mathbb{R}\text{,}\) then \(([a+1,b+1], j)\) where \(j(x)=x+1\) is a completion of \((a,b)\text{.}\)
If \(X\) is a complete metric space, then \((X,\text{id}_X)\) is a completion of \(X\text{.}\)
Theorem4.115.Completion Existence.
Let \(X\) be a metric space.
\(X\) has a completion \((\hat{X},j)\)
Let \((\hat{X}_1,j_1)\text{,}\)\((\hat{X}_2,j_2)\) be completions of \(X\text{.}\) Then there is a unique isometric isomorphism \(f:\hat{X}_1\rightarrow\hat{X}_2\) such that \(f(j_1)=j_2\text{.}\)
SubsectionSeparability
Definition4.116.Separable.
A metric space \(X\) is seperable if there is a countable dense (recall dense means \(\overline{S}=X\)) set \(S\subseteq X\text{.}\)
Example4.117.
\(\mathbb{R}\) is separable since \(\mathbb{Q}\subseteq \mathbb{R}\) is countable and dense in \(\mathbb{R}\text{.}\)
\(\mathbb{C}\) is separable. \(\mathbb{Q}[i]=\{{a+ib}\,:\,{a,b\in\mathbb{Q}}\}\subseteq \mathbb{C}\) is dense in \(\mathbb{C}\text{,}\) and \(\mathbb{Q}[i]\) is countable since \(\mathbb{Q}[i]\sim\mathbb{Q}\times \mathbb{Q}\) by \((a,b)\mapsto a+ib\text{.}\)
\(\mathbb{R}^n,\mathbb{C}^n\) are separable.
If \(X\) is an uncountable set with the discrete metric, then \(X\) is not separable. In fact, \(X\) is the only dense subset of \(X\text{,}\) and \(X\) is not countable.
Theorem4.118.Compact Implies Separable.
Every compact metric space is separable.
Lemma4.119.Separable Lemma.
A metric space \(X\) is separable iff there exists a countable set of open sets \(\mathcal{S}\) in \(X\) such that for all \(x\in X\text{,}\) and all \(U\subseteq X\) open with \(x\in U\text{,}\) there exists \(V\in \mathcal{S}\) so that \(x\in V\subseteq U\text{.}\)
(\(\impliedby\)) Suppose \(\mathcal{S}\) is given. We may assume \(\emptyset\notin \mathcal{S}\) by replacing \(\mathcal{S}\) with \(\mathcal{S}\setminus\{{\emptyset}\}\) if needed. For each \(V\in \mathcal{S}\text{,}\) let \(x_V\in V\text{,}\) and let
The function \(\mathcal{S}\rightarrow S:V\mapsto x_V\) is surjective, and \(\mathcal{S}\) is countable, so \(S\) is also countable. I claim that \(S\) is also dense in \(X\text{.}\) Suppose \(\overline{S}\neq X\text{.}\) Fix \(x\in X\setminus\overline{S}\text{.}\) Then by property of \(\mathcal{S},\) taking \(U=X\setminus\overline{S}\text{,}\) we can find \(V\in \mathcal{S}\) with \(x\in V\subseteq X\setminus\overline{S}\text{.}\) Then \(x_V\in S, x_V\in V,\) and \(S\cap V=\emptyset\text{,}\) a contradiction.
(\(\implies\)) Let \(S\subseteq X\) be a countable set with \(\overline{S}=X\text{.}\) Let \(\mathcal{S}=\{{B_\varepsilon(x)}\,:\,{x\in S,\varepsilon\in \mathbb{Q},\varepsilon>0}\}.\) Then there is a surjective function \(S\times\mathbb{Q}_{\geq 0}\rightarrow\mathcal{S}\) by \((x,\varepsilon)\mapsto B_\varepsilon(x)\text{.}\) Since \(S\) and \(Q_{\geq 0}\) are countable, \(\mathcal{S}\) is also countable. Suppose now that \(x\in X,U\subseteq X\) is open, and \(x\in U\text{.}\) We need to find \(V\in\mathcal{S}\) with \(x\in V\subseteq U\text{.}\) Fix \(\varepsilon_0>0\) such that \(B_{\varepsilon_0}(x)\subseteq U\text{.}\) Now let \(\varepsilon\in(0,\varepsilon_0]\cap \mathbb{Q}\) Since \(\overline{S}=X\text{,}\) we have \(x\in\overline{S}\text{.}\) So exists \(y\in S\) with \(d(x,y)<\frac{\varepsilon}{2}\text{.}\) Now
Then \(\frac{\varepsilon}{2}\in\mathbb{Q}\) and \(y\in S\text{,}\) so \(B_{\frac{\varepsilon}{2}}(y)\in\mathcal{S}\text{,}\) set \(V=B_{\frac{\varepsilon}{2}}(y)\text{.}\)
Remark4.120.
In the language of topology, this says a metric space is separable iff it is second countable.
Theorem4.121.Separable Subsets.
If \(X\) is a separable metric space and \(Y\subseteq X\text{,}\) then \(Y\) is separable.
Suppose \(X\) is separable, and let \(\mathcal{S}\) be given as in the lemma above. Let \(\mathcal{S}'=\{{V\cap Y}\,:\,{V\in \mathcal{S}}\}\text{.}\) Then \(\mathcal{S}'\) is a collection of open subsets of \(Y\) and \(\mathcal{S}'\) is countable since \(\mathcal{S}\) is countable and the function \(\mathcal{S}\rightarrow\mathcal{S}':V\rightarrow V\cap Y\) is surjective. Let \(Y'\subseteq Y\) be an open set and let \(y\in Y\) with \(y\in U'\text{.}\) We need a \(V'\in\mathcal{S}'\) with \(y\in V'\subseteq U'\text{.}\) Since \(U'\) is open in \(Y\text{,}\) there is an open set \(U\subseteq X\) such that \(U'=U\cap Y\text{.}\) Since \(y\in U\text{,}\) there exists \(V\in\mathcal{S}\) such that \(y\in V\subseteq U\text{.}\) Let \(V'=V\cap Y\in \mathcal{S}'\) and note \(y\in V'\subseteq U'\text{.}\) Apply the lemma again to \(Y\text{,}\) and \(Y\) is separable.
