--- tags: definition Created: 2023-06-20 Modified: 2023-06-20 mathLink: auto #definition #geometry-topology/point-set-topology --- \ Properties:: [[characterization of closure in subspace]], [[neighborhood-basis characterization of set closure]], [[closure is set together with limit points]] Sufficiencies:: *[[Sufficiencies]]* Equivalences:: [[neighborhood-basis characterization of set closure]] ---- > [!definition] Definition. ([[closure]]) > Let $A$ be a subset of [[topological space]] $X$. The **closure of $A$**, denoted $\text{Cl }A$ or $\overline{A}$, is defined as the intersection of all [[closed set|closed sets]] containing $A$, that is, as the smallest closed set containing $A$. > \ > If $A$ is [[closed set|closed]], $\overline{A}=A$. In general, $\text{Int \ } A \subset A \subset \overline{A}$. > [!equivalence] Equivalences. > - [[neighborhood-basis characterization of set closure]] > [!basicproperties] >- ![[closure and union#^48a094]] > >- ![[closure and intersection#^a21f89]] > - (Preservation under inclusion) If $A \subset B$, then $\overline{A} \subset \overline{B}$. Let $a \in \overline{A}$. By [[neighborhood-basis characterization of set closure]] we know that for any open [[neighborhood]] $U$ of $a$, we have $U \cap A \neq \emptyset$. Since $A \subset B$, this implies $U \cap B \neq \emptyset$. Thus $a \in \overline{B}$. - $A \subset X$ is closed in $X$ iff $A = \overline{A}$. $\to.$ Suppose $A \subset X$ is closed in $X$. That $A \subset \overline{A}$ is immediate from the definition of [[closure]] (even when $A$ is not closed). Let $a \in \overline{A}$. Then $a \in A$ immediately, or $a$ is a [[limit point]] of $A$; since $A$ is closed the latter case still entails $a \in A$ ([[closure is set together with limit points|closed iff contains all limit points]]). $\leftarrow.$ Suppose $A = \overline{A}$. Then $A$ is closed as the [[closed sets behave complementarily to open sets|arbitrary intersection]] of [[closed set]]s. - (Idempotency) $\overline{\overline{A}}=\overline{A}$. From [[closed sets behave complementarily to open sets]] we know that the [[closed set|closed sets]] in $X$ are stable under arbitrary intersection. Hence, $\overline{A}$ is closed in $X$ by definition. [[closure is set together with limit points|When we form]] $\overline{\overline{A}}$ by adjoining to $\overline{A}$ all of its [[limit point|limit points]], we are just left with $\overline{A}$ again because $\overline{A}$ [[closure is set together with limit points|already contains all of its limit points]]. ---- #### If $Z \subset X$ is [[closed set|closed in]] $X$ and $U \subset Z$ is dense in $Z$ (i.e., the closure of $U$ in $Z$ equals $Z$), then the closure of $U$ in $X$ equals $Z$. ---- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag # Legacy: math 395 - Main Theorem If $A$ is a subset of a [[metric space]] $X$, then the set $\overline{A}$ consisting of $A$ and all its [[limit point]]s is a [[closed set]] of $X$. A subset of $X$ is closed iff it *contains ALL its [[limit point]]s*. The set $\overline A$ is called the [[closure]] of $A$.