131 lines
7.1 KiB
TeX
131 lines
7.1 KiB
TeX
\section{Interior, Closure, Boundary}
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\label{section:icb}
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\begin{definition}[Interior]
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\label{definition:interior}
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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:
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\begin{enumerate}
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\item $A \in \cn(x)$.
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\item There exists $U \in \fB$ with $U \subset A$.
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\item There exists $U \in \cn(x)$ with $U \subset A$.
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\end{enumerate}
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The set of all points satisfying the above is the \textbf{interior} $A^o$ of $A$, which is the largest open set contained in $A$.
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\end{definition}
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\begin{proof}
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$(1) \Rightarrow (2)$: Since $\fB$ is a fundamental system of neighbourhoods, there exists $U \in \fB$ with $U \subset A$.
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$(2) \Rightarrow (3)$: $\fB \subset \cn(x)$.
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$(3) \Rightarrow (1)$: By (F1) of $\cn(x)$, $A \in \cn(x)$.
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Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \autoref{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 \autoref{lemma:openneighbourhood}.
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\end{proof}
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\begin{definition}[Closure]
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\label{definition:closure}
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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:
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\begin{enumerate}
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\item For every $B \supset A$ closed, $x \in B$.
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\item For every $U \in \cn(x)$, $U \cap A \ne \emptyset$.
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\item For every $U \in \fB$, $U \cap A \ne \emptyset$.
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\item There exists a filter $\fF \subset 2^A$ that converges to $x$.
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\end{enumerate}
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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 $A$.
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\end{definition}
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\begin{proof}
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$\neg (2) \Rightarrow \neg (1)$: 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 $x \not\in V^c$.
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$(2) \Rightarrow (3)$: $\cn(x) \supset \fB$.
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$(3) \Rightarrow (4)$: 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.
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$\neg (1) \Rightarrow \neg (4)$: 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 $x$.
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\end{proof}
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\begin{proposition}
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\label{proposition:closure-of-image}
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Let $X, Y$ be topological spaces, $A \subset X$, and $f: X \to Y$ be continuous, then $f(\ol{A}) \subset \ol{f(A)}$.
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\end{proposition}
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\begin{proof}
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Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.
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\end{proof}
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\begin{proposition}
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\label{proposition:closure-finite-union}
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Let $X$ be a topological space and $\seqf{E_j} \subset 2^X$, then $\bigcup_{j = 1}^n \ol{E_j} = \ol{\bigcup_{j = 1}^n E_j}$.
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\end{proposition}
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\begin{proof}
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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 $\ol{\bigcup_{j = 1}^n E_j} \supset \bigcup_{j = 1}^n\ol{E_j}$.
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\end{proof}
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\begin{definition}[Dense]
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\label{definition:dense}
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Let $X$ be a topological space and $A \subset X$, then the following are equivalent:
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\begin{enumerate}
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\item $\ol{A} = X$.
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\item For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.
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\item For every $U \subset X$ open, $\overline{A \cap U} \supset U$.
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\end{enumerate}
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If the above holds, then $A$ is a \textbf{dense} subset of $X$.
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\end{definition}
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\begin{proof}
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$(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \autoref{definition:closure}, $U \cap A \ne \emptyset$.
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$(2) \Rightarrow (3)$: 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 \autoref{definition:closure}, $x \in \overline{A \cap U}$.
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$(3) \Rightarrow (1)$: $X$ is open.
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\end{proof}
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\begin{proposition}
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\label{proposition:dense-product}
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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 $\prod_{i \in I}X_i$.
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\end{proposition}
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\begin{proof}
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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.
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\end{proof}
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\begin{proposition}
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\label{proposition:separable-product}
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Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
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\end{proposition}
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\begin{proof}
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Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \autoref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
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\end{proof}
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\begin{lemma}
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\label{lemma:closurecomplement}
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Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.
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\end{lemma}
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\begin{proof}
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Let $x \in X$. By (3) of \autoref{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 \autoref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
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\end{proof}
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\begin{definition}[Boundary]
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\label{definition:boundary}
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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:
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\begin{enumerate}
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\item For every $V \in \cn(x)$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
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\item For every $V \in \fB$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
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\item $x \in \overline{A} \setminus A^o$.
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\item $x \in \overline{A} \cap \overline{A^c}$.
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\end{enumerate}
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The set $\partial A$ of all points satisfying the above is the \textbf{boundary} of $A$.
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\end{definition}
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\begin{proof}
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$(1) \Rightarrow (2)$: $\cn(x) \supset \fB$.
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$(2) \Rightarrow (3)$: By (2) of \autoref{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 \autoref{definition:interior}, $x \not\in A^o$.
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$(3) \Rightarrow (4)$: By \autoref{lemma:closurecomplement}.
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$(4) \Rightarrow (1)$: By (2) of \autoref{definition:closure}.
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\end{proof}
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