---- > [!definition] Definition. ([[precompact]]) > A subset $A$ of a [[topological space]] $X$ is said to be **precompact** if its [[closure]] $\overline{A}$ is [[compact]]. ^definition > [!equivalence] Equivalence for metrizable spaces. > If $X$ is [[metrizable]], then $A \subset X$ is precompact if and only if every [[sequence]] $(a_{n}) \subset A$ has a [[converge|convergent]] [[subsequence]] in $X$. ^equivalence > [!proof] > Recall that [[for metrizable spaces, compact iff limit point compact iff sequentially compact]]. > > Let $X$ be [[metrizable]]. Suppose $A \subset X$ is precompact, and let $(a_{n})$ be a [[subsequence]] of $A$. Then $\overline{A}$ is [[compact]], hence [[sequentially compact]]. So $(a_{n}) \subset A \subset \overline{A}$ has a [[subsequence]] converging to some $a \in \overline{A} \subset X$. > > Conversely, suppose every $(a_{n}) \subset A$ has a convergent subsequence in $X$. It's enough to show $\overline{A}$ is [[sequentially compact]], i.e. that every $(\overline{a}_{n}) \subset \overline{A}$ has a convergent subsequence in $\overline{A}$. For each $\overline{a}_{n}$, [[neighborhood-basis characterization of set closure|define]] a sequence $(a_{n})$ in $A$ by taking $a_{n} \in A$ such that $d(\overline{a}_{n}, a_{n})<1/n$. Extract a [[subsequence]] $(a_{n_{k}})$ converging to some $\overline{a} \in \overline{A}$. Now $d(\overline{a}_{n_{k}}, a_{n_{k}}) \to 0$, hence $\overline{a}_{n_{k}} \to \overline{a}$. ---- #### ---- #### 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 > ```