---- > [!definition] > Let $X$ be a [[topological space]] and $A \subset X$. The **interior** of $A$ is defined to be the union of all open sets in $X$ contained in $A$, i.e., the $\subset$-largest open set contained by $A$. \ It is denoted $\text{Int } A$. Compare to [[exterior]] and [[boundary]]. > [!basicproperties] > > > 1. (preservation under inclusion) If $A \subset B$, then $\text{int }A \subset \text{int }B$. > > Let $a \in \text{int }A$. Then $a$ belongs to an open set $U$ of $X$ with $U \subset A$. Since $U \subset B$ (because $A \subset B$) we have that $a$ belongs to an open set $U$ of $X$ with $U \subset B$. Therefore, because $U$ is an open set containing $a$ and $U \subset B$, $a \in \text{ int }B$. > > 2. $U$ is open in $X$ iff $\text{int }U = U$. > > $\to.$ Suppose $U$ is open in $X$; we will show $\text{int }U \subset U$ and $U \subset \text{int }U$. The direction $\text{int }U \subset U$ is true by definition of interior (even when $U$ is not open in $X$). To show the reverse inclusion, let $x \in U$. $x$ belongs to an open set in $X$ contained in $U$, namely, $U$ itself— thus $X$ obviously belongs to the union of all open sets in $X$ contained in $U$, i.e., $x \in \text{int }U$. So $U \subset \text{int }U$. > > $\leftarrow.$ Suppose $\text{int }U = U$. Then $U$ is the union of open sets in $X$ and therefore is open in $X$. > > 4. ([[idempotent|idempotency]]) $\text{int}(\text{int }A)=\text{int }A$. > > We know that for any open set $U$ in $X$, $\text{int }U = U$. As a union of open sets in $X$, $\text{int }A$ is open in $X$, and hence $\text{int }(\text{int }A)=\text{int }A$. > > 5. $\text{int}(A \cup B) \supset \text{int }A \cup \text{int }B$. > > > > > $\text{int}(A \cap B) = \text{int }A \cap \text{int }B$. > > Let $x \in \text{int}(A \cap B)$. Then $x$ lives in an open set $U \subset X$ for which we have both $U \subset A$ and $U \subset B$. Because $U$ is open in $X$, by definition of [[topological interior]] we have $U \subset \text{int }A$ and $U \subset \text{int }B$, thus $U \subset \text{int }A \cap \text{int }B$. So, $x \in U \subset \text{int }A \cap \text{int }B$ and hence $x \in \text{int }A \cap \text{int }B$. > > Conversely, let $x \in \text{int }A \cap \text{int }B$. Then there exist open sets $U_{A}$, $U_{B}$ containing $x$ with $U_{A} \subset A$ and $U_{B} \subset B$. Their intersection is open in $X$, contains $x$, and belongs to $A \cap B$, hence $x \in \text{int}(A \cap B)$. ---- #### ---- #### 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 > ```