Finished open and closed set section
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chapters/sections/open_closed_sets.tex
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chapters/sections/open_closed_sets.tex
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\documentclass[../../script.tex] {subfiles}
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%! TEX root = ../../script.tex
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\begin{document}
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\section{Open and Closed Sets}
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\begin{defi}[Inner points and Boundary points]
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Let $\metric$ be a metric space, $A \subset X$ and $x \in X$.
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\begin{enumerate}[(i)]
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\item $x$ is said to be an inner point of $A$, if \[ \exists \epsilon > 0: ~~ \oball(x) \subset A \]
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\item $x$ is said to be a boundary point of $A$ if
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\[
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\forall \epsilon > 0: ~~\underbrace{\oball(x) \cap A \ne \emptyset}_{\substack{\oball(x) \text{ contains} \\ \text{points from } A}}
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\wedge \underbrace{\oball(x)
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\cap (X \setminus A) \ne \emptyset}_{\substack{\oball(x) \text{ contains points}\\ \text{from outside of } A}}
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\]
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\item The set
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\[
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\set[x \text{ is inner point of } A]{x \in X}
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\]
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is called the interior of $A$, and is denoted as $\interior{A}$.
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\item The set
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\[
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\set[x \text{ is boundary point of} A]{x \in X}
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\]
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is called the boundary of $A$, and is denoted as $\partial A$.
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\item $A \cup \boundary{A}$ is said to be the closure of $A$, and is denoted as $\closure{A}$.
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\end{enumerate}
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\begin{center}
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\begin{tikzpicture}
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\draw [->, thick] (-6, -3.5) -- (-6, 3);
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\draw [->, thick] (-6, -3.5) -- (2, -3.5);
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\draw [thick, fill=lightgray] plot[smooth, tension=.7] coordinates {(-4,2.5) (-3,3) (-2,2.8) (-0.8,2.5) (-0.5,1.5) (0.5,0) (0,-2)(-1.5,-2.5) (-4,-2) (-3.5,-0.5) (-5,1) (-4,2.5)};
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\node[below right] at (-6, 3) (X) {$X$};
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\node[below right] at (-4.5, 2) (A) {$A$};
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\draw[fill] (-2, -1) circle [radius=1pt];
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\node[below right] at (-2, -1) (C) {$x$};
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\draw[dashed] (-2, -1) circle [radius=20pt];
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\node[below] at (-2, -1.75) (C) {$\oball(x)$};
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\draw[fill] (-0.63, 2) circle [radius=1pt];
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\node[right] at (-0.63, 2) (C) {$x$};
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\end{tikzpicture}
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\end{center}
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\end{defi}
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\begin{eg}
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Consider $X = \realn^2$. Then
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\begin{align*}
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A &= \set[0 \le y < 1]{(x, y) \in \realn} \\
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\mathring{A} &= \set[0 \le y < 1]{(x, y) \in \realn^2} \\
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\boundary{A} &= \set[y = 1 \vee y = 0]{(x, y) \in \realn^2} \\
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\closure{A} &= \set[0 \le y \le 1]{(x, y) \in \realn^2}
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\end{align*}
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\end{eg}
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\begin{rem}
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\begin{enumerate}[(i)]
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\item $\interior{A} \subset A$
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\item Boundary points of $A$ can be elements of $A$ or not.
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\item $A \subset \interior{A} \cup \boundary{A}$, $~~\interior{A} \cap \boundary{A} = \varnothing$
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\item $\boundary{A} = \boundary{X \setminus A}$
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\end{enumerate}
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\end{rem}
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\begin{thm}
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Let $\metric$ be a metric space, $A \subset X$ and $x$ an interior point or boundary point of $A$. Then
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\[
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\exists \anyseqdef{A}: ~~x_n \conv{} x
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\]
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\end{thm}
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\begin{proof}
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If $x \in A$ then this is trivial, so let $x \notin A$. Then
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\begin{equation}
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\forall n \in \natn ~\exists x_n \in \left(\oball[\frac{1}{n}](x) \cap A \ne \varnothing\right)
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\end{equation}
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We need to show that $\seq{x}$ converges to $x$.
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\begin{equation}
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\forall \epsilon > 0 ~\epsilon N \in \natn: ~~\frac{1}{N} < \epsilon
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\end{equation}
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For $n \ge N$ we have
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\begin{equation}
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\frac{1}{n} \le \frac{1}{N} < \epsilon
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\end{equation}
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and thus
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\begin{equation}
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d(x_n, x) < \frac{1}{n} < \epsilon
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\end{equation}
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\end{proof}
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\begin{defi}[Open and Closed sets]
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Let $\metric$ be a metric space. $A \subset X$ is said to be
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\begin{enumerate}[(i)]
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\item open, if every point in $A$ is an interior point
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\item closed, if $A$ contains all its boundary point
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\item neighbourhood of $x \in A$, if $x$ is an interiot point of $A$
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\end{enumerate}
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\end{defi}
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\begin{thm}
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Let $\metric$ be a metric space and $A \subset X$.
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\[
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A \text{ open} \iff X \setminus A \text{ closed}
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\]
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\end{thm}
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\begin{proof}
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\begin{subequations}
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\begin{align}
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A \text{ open} &\iff \forall x \in A: ~~x \in \interior{A} \\
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&\iff \forall x \in A: ~~x \in \boundary{A} \\
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&\iff X \setminus A \text{ contains all boundary point of } A \\
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&\iff X \setminus A \text{ contains all boundary points of } X \setminus A \\
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&\iff X \setminus A \text{ closed}
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\end{align}
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\end{subequations}
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\end{proof}
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\begin{rem}
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That doesn't mean $A$ has to be either open and closed.
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\end{rem}
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\begin{eg}
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Let $\metric$ be a metric space, $x \in X$ and $r > 0$. Then
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\begin{align*}
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\Oball(x) &= \set[d(x, y) < r]{y \in X} \text{ is open} \\
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\Cball(x) &= \set[d(x, y) < r]{y \in X} \text{ is closed}
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\end{align*}
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\end{eg}
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\begin{rem}
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Consider the special case $a, b \in \realn$ with $a < b$
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\begin{align*}
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(a, b) &= \oball[\frac{b - a}{2}]\left(\frac{a + b}{2}\right) \text{ open} \\
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[a, b] &= \cball[\frac{b - a}{2}]\left(\frac{a + b}{2}\right) \text{ closed}
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\end{align*}
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\end{rem}
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\begin{thm}
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Let $\metric$ be a metric space and $A \subset X$.
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\[
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A \text{ closed} \iff \forall \anyseqdef{A} \text{ convergent}: ~~\limn{x_n} \in A
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\]
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\end{thm}
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\begin{proof}
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Assume $A$ is closed. Let $\anyseqdef{A}$ be convergent to $x$. then
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\begin{equation}
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\forall \epsilon > 0 ~\exists N \in \natn: ~~x_n \in \oball(x) ~~\forall n \ge N
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\end{equation}
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This means that every $\epsilon$-ball around $x$ contains at least one point from $A$.
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I.e. $x$ is always a point (or a boundary point) of $A$. From $A$ closed follows $x \in A$.
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Now assume $x \in \boundary{A}$. Then
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\begin{equation}
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\exists \anyseqdef{A}: ~~\seq{x} \conv{} x
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\end{equation}
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According to the prerequisites, $x \in A$.
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\end{proof}
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\begin{thm}
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Let $\metric$ be a metric space, and $\tau$ the set of all open subsets. Then
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\begin{enumerate}[(i)]
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\item $\varnothing \in \tau$, $~~X \in \tau$
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\item The union of any number of sets from $\tau$ is an open set
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\[
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\bigcup_{t \in \tau} t \in \tau
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\]
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\item The intersection of finitely many sets from $\tau$ is an open set
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\[
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\bigcap_{t \in \tau} t \in \tau
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\]
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\end{enumerate}
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\end{thm}
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\begin{proof}
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\reader
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\end{proof}
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\begin{rem}
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\begin{enumerate}[(i)]
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\item $\tau$ is said to be the topology induced by $d$
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\item \begin{itemize}
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\item $\varnothing$, $X$ are also closed
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\item The intersection of any number of closed sets is closed
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\item The union of finitely many closed sets is closed
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\end{itemize}
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\item Infinitely many intersections of open sets are not open in general.
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\end{enumerate}
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\end{rem}
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\begin{thm}
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Let $\metric$ be a metric space and $A \subset X$. Then
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\[
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\interior{A} \text{ open} \implies \boundary{A}, \closure{A} \text{ closed}
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\]
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\end{thm}
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\begin{proof}
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Let $\interior{A}$ be open and $x \in \interior{A} \subset A$. This means
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\begin{equation}
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\exists \epsilon > 0: ~~\oball(x) \subset A
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\end{equation}
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We have to show that $\oball(x) \subset \interior{A}$. Let $y \in \oball(x)$.
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Since $\oball(x)$ is open
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\begin{equation}
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\exists \delta > 0: ~~\oball[\delta](y) \subset \oball(x) \subset A
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\end{equation}
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This means that $y \in \oball(x)$ is interior point $A$. I.e. $\subset(x) \subset \interior{A}$, and thus $x$ is interior point of $\interior{A}$.
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Let $B = X \setminus A$. Then $\boundary{A} = \boundary{B}$
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\begin{equation}
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X = A \cup B = \interior{A} \cup \boundary{A} \cup \interior{B} \cup \boundary{B} = \interior{A} \cup \boundary{A} \cup \interior{B}
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\end{equation}
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Then
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\begin{subequations}
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\begin{align}
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A \text{ and } B \text{ are disjoint } &\implies \interior{A}, \interior{B} \text{ disjoint} \\
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&\implies \boundary{A} \text{ disjoint to } \interior{A}, \interior{B}
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\end{align}
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\end{subequations}
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This results in
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\begin{equation}
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\boundary{A} = X \setminus (\underbrace{\interior{A} \cup \interior{B}}_{\text{open}}) \implies \boundary{A} \text{ closed}
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\end{equation}
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and
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\begin{equation}
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\closure{A} = A \cup \boundary{A} = \interior{A} \cup \boundary{A} = X \setminus \interior{B} \text{ closed}
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\end{equation}
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\end{proof}
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\begin{thm}
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Let $\metric$ be a metric space and $A \subset X$
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\begin{align*}
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\bigcup_{\substack{O \text{ open} \\ O \subset A}} O = \interior{A} && \text{and} && \bigcap_{\substack{C \text{ closed} \\ A \subset C}} C = \closure{A}
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\end{align*}
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\end{thm}
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\begin{proof}
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Let $\interior{A}$ is open and $\interior{A} \subset A$
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\begin{equation}
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\implies \bigcup_{O \subset A \text{ open}} \supset \interior{A}
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\end{equation}
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Now let $O \subset A$ be open and $x \in O$, i.e.
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\begin{equation}
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\exists \epsilon > 0: ~~\oball(x) \subset O \subset A \implies x \in \interior{A}
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\end{equation}
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This implies that $O \subset \interior{A}$. Since this holds for all open $O \subset A$, this statement is proven.
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The other statement follows from the complement.
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\end{proof}
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\begin{thm}
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Let $\metric$ be a complete space and $A \subset X$ be closed. Then $(A, d_A)$ is complete.
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\end{thm}
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\begin{proof}
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\reader
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\end{proof}
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\begin{rem}
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Topological terms (open, closed, continuous, compact) don't just depend on $A$, but also on $X$.
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\end{rem}
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\begin{defi}
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Let $\metric$ be a metric space and $x \in X$.
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\begin{enumerate}[(i)]
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\item $x$ is said to be an isolated point if $\exists \epsilon > 0$ such that $\oball(x) = \set{x}$.
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\item $x$ is said to be a limit point if it's not an isolated point.
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\end{enumerate}
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\end{defi}
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\begin{defi}[Punctured neighbourhood, Punctured ball]
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$\dot{U} \subset X$ is said to be a punctured neighbourhood, if there is a neighbourhood $U$ of $x$ with $\dot{U} = U \setminus \set{x}$
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A punctured ball is $\dot{\oball}(x) = \oball \setminus \set{x}$.
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\end{defi}
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\begin{defi}[Limit of mappings]
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Let $(X, d_X), (Y, d_Y)$ and $x$ a limit point of $X$. Let $\dot{U}$ be a punctured neighbourhood of $x$ and $f: \dot{U} \rightarrow Y$.
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Then $f$ converges to $y \in Y$ in $x$ ($y$ is said to be the limit of $f$ in $x$), if
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\[
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\forall \epsilon > 0 ~\exists \delta > 0: ~~f(\tilde{x}) \in \oball(y) ~~[d(f(\tilde{x}), y) < \epsilon]
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\]
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if $\tilde{x} \in \dot{\oball}(x)$ [$d(\tilde{x}, x) < \delta$]
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\end{defi}
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\begin{eg}
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Let $f, g: \realn^2 \setminus \set{0} \rightarrow \realn$.
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\begin{align*}
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f(x) := \norm{x}^2 && g(x) := \frac{1}{\norm{x}}
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\end{align*}
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Then $\limes{x}{0} f(x) = 0$, because for $\epsilon > 0$ and $\delta = \sqrt{\epsilon}$ we have
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\[
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d(\tilde{x}, 0) = \norm{\tilde{x} - 0} = \tilde{x} < \delta \implies d(f(\tilde{x}), 0) = \abs{\norm{\tilde{x}}^2 - 0} = \norm{\tilde{x}}^2 < \epsilon = \delta^2
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\]
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\end{eg}
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\begin{thm}
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\[
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f \text{ converges to } y \in Y \text{ in } x \iff \forall \anyseqdef{X}: ~~f(x_n) \conv{x_n \rightarrow x} y
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\]
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\end{thm}
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\begin{proof}
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Let $\anyseqdef{X}$ with $x_n \conv{} x$. Let $\epsilon > 0$, then
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\begin{equation}
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\exists \delta > 0: ~~f(\tilde{x}) \in \oball(y) \text{ if } \tilde{x} \in \oball[\delta](x)
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\end{equation}
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Furthermore
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\begin{equation}
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\exists N \in \natn: ~~x_n \in \oball[\delta](x) ~~\forall n \ge N
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\end{equation}
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Then
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\begin{equation}
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f(x_n) \in \oball(y) ~~\forall n \ge N
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\end{equation}
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To prove the other direction, assume $f$ doesn't converge to $y$ in $y$. This means
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\begin{equation}
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\exists \epsilon > 0: ~~\exists \tilde{x} \in \oball[\delta](x) \text{ but } f(\tilde{x}) \notin \oball(y) ~~\forall \delta > 0
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\end{equation}
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Therefore
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\begin{equation}
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\forall n \in \natn: ~~\exists x_n \in \oball[\frac{1}{n}](x)
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\end{equation}
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We know that $x_n \conv{} x$ since $d(x_n, x) < \frac{1}{n}$, but $f(x_n)$ doesn't converge to $y$ since $d(f(x_n), y) \ge \epsilon$.
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\end{proof}
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\begin{cor}
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Let $\metric$ be a metric space, $x \in X$ a limit point and $\dot{U}$ a punctured neighbourhood of $x$. Let $f, g: \dot{U} \rightarrow \field$ with
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\begin{align*}
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\limes{\tilde{x}}{x} f(\tilde{x}) = y_1 && \limes{\tilde{x}}{x} g(\tilde{x}) = y_2
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\end{align*}
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Then
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\begin{align*}
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\limes{\tilde{x}}{x} (f + g)(\tilde{x}) = y_1 + y_2 && \limes{\tilde{x}}{x} (f \cdot g)(\tilde{x}) = y_1 \cdot y_2 \\
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\limes{\tilde{x}}{x} \left(\frac{f}{g}\right)(\tilde{x}) = \frac{y_1}{y_2}
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\end{align*}
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\end{cor}
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\begin{hproof}
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Draw parallels back to number sequences
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\end{hproof}
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\end{document}
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@ -6,4 +6,5 @@
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\subfile{sections/metr_and_normed.tex}
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\subfile{sections/seq_ser_limits.tex}
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\subfile{sections/open_closed_sets.tex}
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\end{document}
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script.pdf
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script.pdf
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@ -59,7 +59,6 @@
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\newcommand{\series}[2][\infty]{\sum_{#2 = 1}^{#1}}
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\newcommand{\finite}{\text{ finite}}
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\newcommand{\conj}[1]{\overline{#1}}
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\newcommand{\closure}[1]{\overline{#1}}
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\newcommand{\must}[1][=]{\stackrel{!}{#1}}
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\newcommand{\metric}[1][X]{(#1, d)}
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@ -72,6 +71,10 @@
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\newcommand{\Oball}{\oball[r]}
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\newcommand{\Cball}{\cball[r]}
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\newcommand{\interior}[1]{\mathring{#1}}
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\newcommand{\boundary}[1]{\partial #1}
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\newcommand{\closure}[1]{\bar{#1}}
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\newcommand{\conv}[1]{\xrightarrow{\makebox[2em][c]{$\scriptstyle#1$}}}
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\newcommand{\convinf}{\conv{n \rightarrow \infty}}
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