4.5 Interior, Closure, Boundary
Definition 4.5.1 (Interior). Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
$A \in \cn(x)$.
There exists $U \in \fB$ with $U \subset A$.
There exists $U \in \cn(x)$ with $U \subset A$.
The set of all points satisfying the above is the interior $A^{o}$ of $A$, which is the largest open set contained in $A$.
Proof. $(1) \Rightarrow (2)$: Since $\fB$ is a fundamental system of neighbourhoods, there exists $U \in \fB$ with $U \subset A$.
$(2) \Rightarrow (3)$: $\fB \subset \cn(x)$.
$(3) \Rightarrow (1)$: By (F1) of $\cn(x)$, $A \in \cn(x)$.
Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by Lemma 4.4.3. By (3), $U \subset A^{o}$, so $A^{o}$ contains every open subset of $A$. On the other hand, for any $x \in A^{o}$, (2) implies that there exists $U \in \cn^{o}(x)$ with $U \subset A$. In which case, $U \subset A^{o}$, and $A^{o} \in \cn(x)$. Therefore $A^{o}$ itself is open by Lemma 4.4.3.$\square$
Definition 4.5.2 (Closure). Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
For every $B \supset A$ closed, $x \in B$.
For every $U \in \cn(x)$, $U \cap A \ne \emptyset$.
For every $U \in \fB$, $U \cap A \ne \emptyset$.
There exists a filter $\fF \subset 2^{A}$ that converges to $\fF$.
The set $\ol{A}$ of all points satisfying the above is the closure of $A$ in $X$. By (1), it is the smallest closed set containing $A$.
Proof. $\neg (2) \Rightarrow \neg (1)$: Let $U \in \cn(x)$ such that $U \cap A = \emptyset$ and $V \in \cn^{o}(x)$ with $V \subset U$. Since $x \in V$ by (V1), $V^{c} \supset A$ is closed with $x \not\in V^{c}$.
$(2) \Rightarrow (3)$: $\cn(x) \supset \fB$.
$(3) \Rightarrow (4)$: Let $\fF = \bracs{V \cap A| V \in \cn(x)}$. For any $V \cap A \in \fF$, there exists $W \in \fB$ such that $W \subset V$. Since $W \cap A \ne \emptyset$, $V \cap A \ne \emptyset$ as well. Thus $\emptyset \not\in \fF$, and $\fF$ is a filter in $A$, and $\fF$ converges to $x$ by definition.
$\neg (1) \Rightarrow \neg (4)$: Let $B \supset A$ such that $x\not\in B$, then $B^{c} \in \cn^{o}(x)$ with $B^{c} \cap A = \emptyset$. Thus there exists no filter on $A$ containing $B^{c} \cap A$, and no filter on $A$ converging to $x$.$\square$
Proposition 4.5.3. Let $X, Y$ be topological spaces, $A \subset X$, and $f: X \to Y$ be continuous, then $f(\ol{A}) \subset \ol{f(A)}$.
Proof. Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.$\square$
Proposition 4.5.4. Let $X$ be a topological space and $\seqf{E_j}\subset 2^{X}$, then $\bigcup_{j = 1}^{n} \ol{E_j}= \ol{\bigcup_{j = 1}^n E_j}$.
Proof. Since $\bigcup_{j = 1}^{n} \ol{E_j}$ is closed, $\bigcup_{j = 1}^{n} \ol{E_j}\supset \ol{\bigcup_{j = 1}^n E_j}$. On the other hand, $\ol{\bigcup_{j = 1}^n E_j}\supset E_{j}$ for each $1 \le j \le n$. Thus $\ol{\bigcup_{j = 1}^n E_j}\supset \ol{E_j}$ for each $1 \le j \le n$, and $\ol{\bigcup_{j = 1}^n E_j}\supset \bigcup_{j = 1}^{n}\ol{E_j}$.$\square$
Definition 4.5.5 (Dense). Let $X$ be a topological space and $A \subset X$, then the following are equivalent:
$\ol{A}= X$.
For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.
For every $U \subset X$ open, $\overline{A \cap U}\supset U$.
If the above holds, then $A$ is a dense subset of $X$.
Proof. $(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of Definition 4.5.2, $U \cap A \ne \emptyset$.
$(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of Definition 4.5.2, $x \in \overline{A \cap U}$.
$(3) \Rightarrow (1)$: $X$ is open.$\square$
Proposition 4.5.6. Let $\seqi{X}$ be topological spaces and $\seqi{A}$ such that each $A_{i} \subset X_{i}$ is dense, then $\prod_{i \in I}A_{i}$ is dense in $\prod_{i \in I}X_{i}$.
Proof. Let $x \in \prod_{i \in I}X_{i}$ and $U \in \cn(x)$, then there exists $J \subset I$ finite and $\seqf{U_j}$ such that $U_{j} \in \cn(\pi_{j}(x))$ for each $j \in J$, and $x \in \bigcap_{j \in J}\pi_{j}^{-1}(U_{j}) \subset U$. By density of each $A_{j}$, there exists $y \in X$ such that $y_{j} \in U_{j}$ for each $j \in J$. Thus $y \in U$, and $A$ is dense.$\square$
Proposition 4.5.7. Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^{n} X_{j}$ is also separable.
Proof. Let $\seq{A_j}$ such that $A_{j} \subset X_{j}$ is countable and dense for each $1 \le j \le n$. By Proposition 4.5.6, $\prod_{j = 1}^{n} A_{j}$ is countable and dense in $\prod_{j = 1}^{n} X_{j}$.$\square$
Lemma 4.5.8. Let $X$ be a topological space and $A \subset X$, then $(A^{o})^{c} = \overline{A^c}$.
Proof. Let $x \in X$. By (3) of Definition 4.5.1, $x \in (A^{o})^{c}$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^{o})^{c}$ if and only if $U \cap A^{c} \ne \emptyset$ for all $U \in \cn(x)$. By (2) of Definition 4.5.2, this is equivalent to $x \in \ol{A^c}$.$\square$
Definition 4.5.9 (Boundary). Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
For every $V \in \cn(x)$, $V \cap A \ne \emptyset$ and $V \cap A^{c} \ne \emptyset$.
For every $V \in \fB$, $V \cap A \ne \emptyset$ and $V \cap A^{c} \ne \emptyset$.
$x \in \overline{A}\setminus A^{o}$.
$x \in \overline{A}\cap \overline{A^c}$.
The set $\partial A$ of all points satisfying the above is the boundary of $A$.
Proof. $(1) \Rightarrow (2)$: $\cn(x) \supset \fB$.
$(2) \Rightarrow (3)$: By (2) of Definition 4.5.2, $x \in \overline{A}$. Since $V \cap A^{c} \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of Definition 4.5.1, $x \not\in A^{o}$.
$(3) \Rightarrow (4)$: By Lemma 4.5.8.
$(4) \Rightarrow (1)$: By (2) of Definition 4.5.2.$\square$