Closure and interior

Closure

1. Closure

   Let $X$ be a space and $D\subseteq X$. The closure of $D$ is the intersection of all closed sets that contain $D$. It is denoted by $\overline{D}$

2. Examples
   • If $X$ has indiscrete topology, then $\overline{A}=X$ if and only if $A\ne\emptyset$.
   • If $X$ has discrete topology, then for all $A\subseteq X$ we have $\overline{A}=A$.
   • If $X$ has cofinite topology, then $\overline{A}=A$ if and only if $A$ is finite.

3. Lemma

   $D$ is closed if and only if $D=\overline{D}$.

Characterization of closure

1. Properties of closure

   The closure of $D\subseteq X$ satisfies:

   • $\overline{D}$ is closed.
   • $D\subseteq \overline{D}$.
   • If $A\subseteq X$ is closed and $D\subseteq A$, then $\overline{D}\subseteq A$.

2. Characterization of closure

   The first three properties in the last theorem characterize $\overline{D}$, in the sense that if $R$ is a closed set that contains $D$, and that is contained in any closed set that contains $D$, then we must have $R=\overline{D}$.

Closure and subspaces

1. Note

   If $Y$ is a subspace of $X$ and $D\subseteq Y$, the closure of $D$ in $Y$ and the closure of $D$ in $X$ might be different. The notation $\overline{D}$ will always denote the closure in $X$.

2. Closure and subspaces

   Let $Y$ be a subspace of $X$ and $D\subseteq Y$. Then the closure of $D$ in $Y$ equals $\overline{D}\cap Y$.

3. Proof

   We prove that $\overline{D}\cap Y$ satisfies the properties that characterize the closure of $D$ in $Y$.

Other characterizations of closure

1. Intersecting sets

   We will say that the set $A$ intersects the set $B$ if $A\cap B\ne\emptyset$

2. Neighborhood

   If $x\in X$, a neigborhood of $x$ is an open set $U$ such that $x\in U$.

3. Points in closure

   Let $X$ be a space, $A\subseteq X$ and $x\in X$. Then

   • $x\in \overline{A}$ if and only if every neighborhood of $x$ intersects $A$.
   • If the topology of $X$ is given by the basis $\mathcal{B}$, then $x\in \overline{A}$ if and only if every $B\in \mathcal{B}$ that contains $x$ intersects $A$.

Limit points

1. Limit point

   Let $X$ be a space, $A\subseteq X$ and $x\in X$. We say that $x$ is a limit point of $A$ if every neighborhood of $x$ intersects $A$ in a point different from $x$. We denote the set of all limit points of $A$ by $A'$.

2. Closure and limit points

   Let $X$ be a space and $A\subseteq X$. Then $\overline{A}=A\cup A'$.

3. Proof
   • Let $x\in \overline{A}$. If $x\in A$, then $x$ is in the set on the right side. If $x\not\in A$, let $U$ be a neighborhood of $x$. Then $U$ intersects $A$ in a point different from $x$.
   • From the definition it follows immediately that $A'\subseteq \overline{A}$, and also $A\subseteq \overline{A}$ always. $\Box$

4. Remark

   It follows that $A$ is closed if and only if $A'\subseteq A$.

Interior

1. Interior

   Let $X$ be a space and $D\subseteq X$. The interior of $D$ is the union of all open sets contained in $D$. It is denoted by $D^{\circ}$.

2. Properties of interior

   The interior of $D\subseteq X$ satisfies:

   • $D^{\circ}$ is open.
   • $D^{\circ}\subseteq D$.
   • If $U\subseteq X$ is open and $U\subseteq D$, then $U\subseteq D^{\circ}$.
   • $U$ is open if and only if $U=U^{\circ}$.

Links

• Closure (topology) - Wikipedia, the free encyclopedia
• Interior (topology) - Wikipedia, the free encyclopedia
• Limit point - Wikipedia, the free encyclopedia

Exercises

• True or false? For all $A,B\subseteq X$, $\overline{A\cap B}=\overline{A}\cap \overline{B}$.
• True or false? For all $A,B\subseteq X$, $\overline{A\cup B}=\overline{A}\cup \overline{B}$.
• True or false? For all $A,B\subseteq X$, $(A\cap B)^{\circ}=A^{\circ}\cap B^{\circ}$.
• True or false? For all $A,B\subseteq X$, $(A\cup B)^{\circ}=A^{\circ}\cup B^{\circ}$.
• We define the boundary $\partial A$ as $\partial A= \overline{A}\cap \overline{X-A}$. Prove that $\overline{A}=A\cup\partial A$, that $A$ is closed if and only if $\partial A\subseteq A$, and that $A$ is open if and only if $\partial A\cap A=\emptyset$.
• We define the exterior $\mathrm{ext}(A)$ as $\mathrm{ext}(A)=(X-A)^{\circ}$. Prove that for all $A\subseteq X$, $X=A^{\circ}\cup \partial A\cup \mathrm{ext}(A)$, and this union is disjoint.
• Let $X$ be a totally ordered set with the order topology. Show that $\overline{(a,b)}\subseteq [a,b]$. Is equality always true?
• Show that $A'$ is closed for any $A\subseteq X$.