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src/topology/main/interiorclosureboundary.tex

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+\section{Interior, Closure, Boundary}
+\label{section:icb}
+
+\begin{definition}[Interior]
+\label{definition: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:
+    \begin{enumerate}
+        \item $A \in \cn(x)$.
+        \item There exists $U \in \fB$ with $U \subset A$.
+        \item There exists $U \in \cn(x)$ with $U \subset A$.
+    \end{enumerate}
+    The set of all points satisfying the above is the \textbf{interior} $A^o$ of $A$, which is the largest open set contained in $A$.
+\end{definition}
+\begin{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 \ref{lemma:openneighbourhood}. 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 \ref{lemma:openneighbourhood}.
+\end{proof}
+
+\begin{definition}[Closure]
+\label{definition: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:
+    \begin{enumerate}
+        \item For every $B \supset A$ closed, $x \in B$.
+        \item For every $U \in \cn(x)$, $U \cap A \ne \emptyset$.
+        \item For every $U \in \fB$, $U \cap A \ne \emptyset$.
+        \item There exists a filter $\fF \subset 2^A$ that converges to $\fF$.
+    \end{enumerate}
+
+    The set $\ol{A}$ of all points satisfying the above is the \textbf{closure} of $A$ in $X$. By (1), it is the smallest closed set containing $A$.
+\end{definition}
+\begin{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 $x$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:closure-of-image}
+    Let $X, Y$ be topological spaces, $A \subset X$, and $f: X \to Y$ be continuous, then $f(\ol{A}) \subset \ol{f(A)}$.
+\end{proposition}
+\begin{proof}
+    Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.
+\end{proof}
+
+
+\begin{definition}[Dense]
+\label{definition:dense}
+    Let $X$ be a topologicial space and $A \subset X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\ol{A} = X$.
+        \item For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.
+        \item For every $U \subset X$ open, $\overline{A \cap U} \supset U$.
+    \end{enumerate}
+    If the above holds, then $A$ is a \textbf{dense} subset of $X$.
+\end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \ref{definition:closure}, $U \cap A \ne \emptyset$.
+
+    $(2) \Rightarrow (3)$: 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 \ref{definition:closure}, $x \in \overline{A \cap U}$.
+
+    $(3) \Rightarrow (1)$: $X$ is open.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:dense-product}
+    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 $\prod_{i \in I}X_i$.
+\end{proposition}
+\begin{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.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:closurecomplement}
+    Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.
+\end{lemma}
+\begin{proof}
+    Let $x \in X$. By (3) of \ref{definition:interior}, $x \in (A^o)^c$ 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 \ref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
+\end{proof}
+
+
+\begin{definition}[Boundary]
+\label{definition: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:
+    \begin{enumerate}
+        \item For every $V \in \cn(x)$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
+        \item For every $V \in \fB$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
+        \item $x \in \overline{A} \setminus A^o$.
+        \item $x \in \overline{A} \cap \overline{A^c}$.
+    \end{enumerate}
+
+    The set $\partial A$ of all points satisfying the above is the \textbf{boundary} of $A$.
+\end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: $\cn(x) \supset \fB$.
+
+    $(2) \Rightarrow (3)$: By (2) of \ref{definition:closure}, $x \in \overline{A}$. 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 \ref{definition:interior}, $x \not\in A^o$.
+
+    $(3) \Rightarrow (4)$: By \ref{lemma:closurecomplement}.
+
+    $(4) \Rightarrow (1)$: By (2) of \ref{definition:closure}.
+\end{proof}