----- > [!proposition] Proposition. ([[metrizable compact spaces are second-countable]]) > Any [[metrizable]] [[compact]] [[topological space]] is [[second-countable space|second-countable]]. > [!proof]- Proof. ([[metrizable compact spaces are second-countable]]) > Let $(X,d)$ be a [[metrizable]], [[compact]], [[topological space]]. The topology on $X$ is [[topology generated by a basis|generated by]] [[basis for a topology|basis elements]] of the form $B_{d}(x,r) \in \mathscr{B}$, where $x \in X \text{ and }r>0$. > Given $n \in \mathbb{N}$, put an open ball of radius $\frac{1}{n}$ around each $x \in X$. This is an [[cover|open covering]] of $X$, by [[compact|compactness]] it admits a finite subcover $\mathscr{B}_{*}^{(n)}$. Now $\mathscr{B}_{*}:=\bigcup_{n \in \mathbb{N}}^{}\mathscr{B}_{*}^{(n)}$ is [[countably infinite|countable]] as a countable union of finite sets. It is also a [[basis for a topology]]: by construction, it covers $X$, and if we let $B_{*_{1}}=B_{d}\left( x_{1}, \frac{1}{n_{1}} \right)$, $B_{*_{2}}=B_{d}\left( x_{2}, \frac{1}{n_{2}} \right)\in \mathscr{B_{*}}$, then since $B_{d}\left( x_{1}, \frac{1}{n_{1}} \right) \cap B_{d}\left( x_{2}, \frac{1}{n_{2}} \right)$ is open in $X$ (as an intersection of open sets), given $x \in X$ there exists $\varepsilon>0$ such that $B_d(x, \varepsilon) \subset B_{d}\left( x_{1}, \frac{1}{n_{1}} \right) \cap B_{d}\left( x_{2}, \frac{1}{n_{2}} \right)$. Now choose $n_{3}$ such that $\frac{1}{n_{3}} < \varepsilon$. Since $\mathscr{B}^{(2n_{3})}$ covers $X$, there exists $x_{3} \in X$ such that $x \in B_{d}\left( x_{3}, \frac{1}{2n_{3}} \right)$. We claim that $B_{d}\left( x_{3}, \frac{1}{2n_{3}} \right) \subset B_{d}\left( x_{1}, \frac{1}{n_{1}} \right) \cap B_{d}\left( x_{2}, \frac{1}{n_{2}} \right)$. Indeed, letting $x_{0} \in B_{d}\left( x_{3}, \frac{1}{2n_{3}} \right)$, we have $d(x,y) \leq d(x,x_{3}) + d(x_{3}, x_{0})< \frac{1}{2n_{3}}+\frac{1}{2n_{3}}=\frac{1}{n_{3}}<\varepsilon$ and this completes that part of the proof. > Note that each $\mathscr{B}^{(n)}$ open-covers $X$. We claim that $\mathscr{B}_{*}$ generates the same [[metric topology]] on $X$ as $d$. To show this, we will use the [[basis-nestling characterization of comparing topologies]]. Let $x_{0} \in X$ with $x_{0} \in B \in \mathscr{B}$. We want to find $B_{*} \in \mathscr{B}_{*}$ s.t. $x \in B_{*} \subset B$. $B=B_{d}(x_{0}, \varepsilon)$ for some $\varepsilon>0$; by the [[archimedian property]] we can pick $n_{0}$ large enough that $\frac{1}{n_{0}}< \varepsilon$. Recall that $\mathscr{B}_{*}^{(n)}$, the open balls of the form $B_{d}\left( x, \frac{1}{n_{0}} \right)$, for finitely many $x$ is a finite open cover of $X$. Take the subset $\{B_{d}\left( x_{1}, \frac{1}{n_{0}} \right), B_{d}\left( x_{2}, \frac{1}{n_{0}} \right),\dots,B_{d}\left( x_{m}, \frac{1}{n_{0}} \right)\} \subset \mathscr{B}^{(n_{0})}$ of balls which contain $x_{0}$, and then consider $B_{d}\left( x_{1}, \frac{1}{n_{0}} \right) \cap B_{d}\left( x_{2}, \frac{1}{n_{0}} \right) \cap \dots \cap B_{d}\left( x_{m}, \frac{1}{n_{0}} \right).$ Since the collection $\mathscr{B}_{*}$ is a [[basis for a topology]], this intersection contains an basic open set $B_{*} \in \mathscr{B}_{*}$ containing $x$. Since each open ball comprising the intersection has radius less than epsilon, $B_{*}$ does too. Hence $x \in B_{*} \subset B=B_{d}(x_{0}, \varepsilon)$ and one direction is settled. > The other direction is obvious. Given $x_{0} \in X$ and $B_{*} \in \mathscr{B}_{*}$ containing $x_{0}$, we know that $B_{*}=B_{d}\left( x_{0}, \frac{1}{n_{0}} \right)$ for some $n_{0} \in \mathbb{N}$ and then any choice $\varepsilon < \frac{1}{n_{0}}$ will witness $x_{0} \in B \subset B_{*}$. ^e2c3a7 ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```