---- Let $X$ be a [[topological space]] and let $A \subset X$. > [!proposition] Proposition. ([[closure is interior together with boundary]]) > $\text{Int } A$ and $\text{Bd }A$ are disjoint, and $\overline{A}= \text{Int }A \sqcup \text{Bd }A$. > [!proposition] Corollary. > By set theory, this means that $\text{Bd }A = \overline{A}- \text{Int }A$ and that $\text{Int }A=\overline{A}-\text{Bd }A$. Compare to [[open iff boundary is closure minus set]]. > [!proof] Proof. ([[closure is interior together with boundary]]) > *Disjointness.* Since $\text{Int }A \subset \overline{A}$, $(\text{Int } A) \cap \overline{A} \cap \overline{X-A}=\text{Int }A \cap \overline{X-A}$. If $x \in \overline{X-A}$, then all [[neighborhood]]s $U \ni x$ intersect ${X-A}$ nontrivially. So no [[neighborhood]] of $x$ is fully contained in $A$, hence $x \notin \text{Int }A$. If $x \in \text{Int }A$, then because $\text{Int }A$ is [[open set|open in]] $A$ there exists a [[neighborhood]] $U \ni x$ for which $x \in U \subset \text{Int } A$. Now since $\text{Int }A \subset A$ and $(X-A) \cap A = \emptyset$, $U \cap (X-A)=\emptyset$. So $x \notin \overline{X-A}$. \ *Equality.* Let $x \in \overline{A}$. We want to show $x \in \overline{X-A}$ or $x \in \text{Int }A$. Let $U$ be a [[neighborhood]] of $x$. By [[neighborhood-basis characterization of set closure]], $U \cap A \neq \emptyset$. If $U \subset A$, then $U$ (and hence $x$) belongs to $\text{Int }A$ by definition. If $U \not \subset A$, then we have $U \cap (X-A) \neq \emptyset$, in which case $x \in \overline{X-A}$ by [[neighborhood-basis characterization of set closure]]. Thus $\overline{A} \subset \text{Int }A \sqcup \text{Bd }A$. \ Let $x \subset \text{Int }A \sqcup \text{Bd }A$; proceed by cases. If $x \in \text{Int }A$, then $x \in A$ and thus $x \in \overline{A}$. If $x \in \text{Bd }A$, then by definition $x \in \overline{A}$. So, $\overline{A} \supset \text{Int }A \sqcup \text{Bd }A$. ---- #### ---- #### Refer. nces > [!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 > ```