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diff --git a/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.tex b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.tex new file mode 100644 index 00000000000..406ce622ad2 --- /dev/null +++ b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.tex @@ -0,0 +1,246 @@ +%% topology-hw-1.tex +%% Copyright 2021 Rebecca B. Turner. +% +% This work may be distributed and/or modified under the +% conditions of the LaTeX Project Public License, either version 1.3 +% of this license or (at your option) any later version. +% The latest version of this license is in +% http://www.latex-project.org/lppl.txt +% and version 1.3 or later is part of all distributions of LaTeX +% version 2005/12/01 or later. +% +% This work has the LPPL maintenance status `maintained'. +% +% The Current Maintainer of this work is Rebecca B. Turner. +% +% This work consists of the files: +% README.md +% rbt-mathnotes.tex +% rbt-mathnotes.sty +% rbt-mathnotes.cls +% rbt-mathnotes-util.sty +% rbt-mathnotes-messages.sty +% rbt-mathnotes-hw.cls +% rbt-mathnotes-formula-sheet.cls +% examples/cheat-sheet.tex +% examples/multivar.tex +% examples/topology-hw-1.tex +% and the derived files: +% rbt-mathnotes.pdf +% examples/cheat-sheet.pdf +% examples/multivar.pdf +% examples/topology-hw-1.pdf + +\documentclass[twocolumn, noxcolor, maketitle]{rbt-mathnotes-hw} +\mathnotes{ + instructor = Prof.~Ruth Charney , + name = Rebecca Turner , + email = rebeccaturner@brandeis.edu , + course = \textsc{math} 104a (Intro to Topology) , + institution = Brandeis University , + semester = Spring 2020 , +} +\title{Homework 1} +\date{2020-01-18} +\def\T{\mathcal{T}} +\def\basis{\mathcal{B}} +\usepackage[ + letterpaper, + margin = 1in, +]{geometry} +\raggedbottom +\begin{document} +\maketitle + +\section{Topological Spaces} +\subsection{Open Sets and the Definition of a Topology} +\begin{problem}[1.7] + Define a topology on $\R$ (by listing the open sets within it) that contains + the open sets $(0,2)$ and $(1,3)$ that contains as few open sets as possible. +\end{problem} +$\T = \{ \emptyset, (0,2), (1,3), (1,2), (0,3), \R \}$. + +\subsection{Basis for a Topology} +\begin{problem}[1.10] + Show that $\mathcal{B} = \{[a,b) \subset \R : a < b\}$ is a basis for a + topology on $\R$. +\end{problem} +\begin{enumerate} + \item \textbf{$\emptyset \in \T$, $\R \in \T$.} $\emptyset \in \T$ (by the + definition of the completion of a basis to a topology). + + Next, we show $\R \in \T$. For all $n \in \Z_{\ge 0}$, $[n-1, n) \in \basis$ + and $[-n + 1, -n) \in \basis$. We know that if $b_1, b_2 \in \basis$, $b_1 + \union b_2 \in \T$, so these short intervals can be gathered together (``unionized'') + to produce $\R$: + \[ \bigunion_{n=1}^{\infty} \left( [n-1, n) \union [-n + 1, -n] \right) = \R, \] + so $\R \in \T$. + + \item \textbf{$\T$ contains all finite intersections of elements of $\T$.} + Suppose we have two intervals $[a, b)$ and $[c, d)$. Then, we define + \begin{alignat*}{1} + a' &= \max(a, c) \\ + b' &= \min(b, d). + \end{alignat*} + If $a' > b'$, the intersection $[a, b) \inter [c, d) = \emptyset$, which is + in $\T$. Otherwise, the intersection is $[a', b')$, which is an element of + $\basis$. All elements of the basis are in $\T$, so the intersection of two + elements is in the topology. + + Thankfully, the intersection is itself always a basis element, so we can + use the same process to show that finite intersections are in $\T$ by + induction. + + \item \textbf{Unions of elements of $\T$ are in $\T$.} By the definition of + the completion of a basis to a topology, this is true (all unions of basis + elements are included in $\T$). +\end{enumerate} + +\begin{problem}[1.12] + % See example 1.9 for defn of \R_l + Determine which of the following are open sets in $\R_l$. In each case, prove + your assertion. + \[ A = [4,5) + \quad B = \{3\} + \quad C = [1,2] + \quad D = (7,8) \] +\end{problem} +\begin{enumerate} + \item $A$ is open in $\R_l$; $[4,5) \in \basis$. + \item $B$ is not an open set in $\R_l$; there is no $[a, b) \subset \R$ where + both $b > a$ and $|[a, b)| = 1$ (because $[0,1) \cong \R$, i.e.~all + intervals contain infinitely many points). + + (Where $\cong$ means ``is isomorphic to.'') + \item $C$ is not open in $\R_l$ because the upper bound of an open set in + $\R_l$ is never inclusive. There is no set of intervals $[a_1, b_1), \dots$ + where the union or intersection of the intervals has an inclusive upper bound. + \item $D$ is open because we can take + \[ D = \lim_{n\to\infty} \left[7 + \frac{1}{n}, 8\right), \] + where $[7 + 1/n, 8) \in \basis$ for any $n \in \R$ with $n \ne 0$. +\end{enumerate} + +\pagebreak +\begin{problem}[1.15] + An arithmetic progression in $\Z$ is a set + \[ A_{a,b} = \{\dots, a - 2b, a - b, a, a + b, a + 2b, \dots\} \] + with $a,b \in \Z$ and $b \ne 0$. Prove that the collection of arithmetic + progressions + \[ \mathcal{A} = \{ A_{a,b} : a, b \in \Z \text{ and } b \ne 0 \} \] + is a basis for a topology on $\Z$. The resulting topology is called the + arithmetic progression topology on $\Z$. +\end{problem} + +\begin{proof} + \def\Ar#1#2{A_{#1,#2}} + \def\Ars{\mathcal{A}} + Let us describe the \emph{minimal form} of an arithmetic progression $\Ar ab$ + to be the progression $\Ar{a'}{b'} = \Ar ab$ with $a', b' > 0$ and the + smallest possible $a'$; in particular, that $a' < b'$. + + We can obtain the minimal form of the progression like so: + \begin{alignat*}{1} + a' &= a \bmod b \\ + b' &= |b|, \\ + \Ar{a'}{b'} &= \Ar ab. + \end{alignat*} + + \begin{remark} + Two arithmetic progressions have the same elements if their minimal forms are + the same; this give an equivalence relation on $\Ars$. + \end{remark} + + Now, suppose we have two arithmetic progressions $\Ar ab$ and $\Ar cd$. We + assume that the progressions are in minimal form without loss of generality. We + also assume that $b \le d$ (by swapping $(a,b)$ with $(c,d)$ if necessary), + again without loss of generality. + + If $b \mid d$ and $a = c$, we have $\Ar ab \subset \Ar cd$. In particular, $\Ar + ab \inter \Ar cd = \Ar cd$. + + If $b \mid d$ and $a \ne c$, we have $\Ar ab \inter \Ar cd = \emptyset$. + + If $b \nmid d$, we have a different progression. An intersection is generated by + an index $(n_1, n_2)$, where + \begin{alignat*}{1} + a + b n_1 &= c + d n_2. \\ + \intertext{We can then solve for $n_1$:} + t(n) &= c - a + dn \\ + n_1 &= \frac{t(n_2)}{b}. \\ + \intertext{Next, we have an infinite \emph{set} of possibilities for $n_2$:} + n_2 &\in \left\{n \in \Z : t(n) \mid b \right\}. + \intertext{Sorting the possible values of $n_2$ by absolute value, let us call + the smallest two values $i_1$ and $i_2$. Then, the difference between + adjacent elements in the intersection progression $\Ar ab \inter \Ar cd$ + is $i_2 - i_1$. + \endgraf + Let} + a' &= a + bi_1 \\ + b' &= i_2 - i_1 \\ + \Ar ab \inter \Ar cd &= \Ar{a'}{b'}. + \end{alignat*} + This isn't super rigorous, admittedly (we're missing some inductive reasoning + about the integers to prove that there are an infinite set of valid values of + $n_2$, in particular), but I have some fairly convincing Haskell code. And the + missing steps are mostly boilerplate, and it's late at night already\dots + + In all cases, the intersection of two arithmetic progressions is either empty or + another arithmetic progression (i.e.~either the empty set or another basis + element), so the same argument given above for $\R_l$ holds (namely that we can + extend this to all finite intersections of elements of $\Ars$ inductively). + + Therefore, finite intersections are in the basis. Unions are in the completion + of the basis (again by definition). The special element $\emptyset$ is in the + completion (by definition), and $\Z = \Ar01$, so $\Z \in \basis$. Therefore, + $\Ars$ forms the basis of a topology on $\Z$. +\end{proof} + +\pagebreak +\subsection{Closed Sets} +\begin{problem}[1.27(a)] + The infinite comb $C$ is the subset of the plane illustrated in Figure~1.17 + and defined by + \begin{multline*} + C = \{(x,0) : 0 \le x \le 1\} \;\union \\ + \bigg\{ \left( \frac{1}{2^n}, y \right) : n = 0,1,2, \dots \\ + \text{ and } 0 \le y \le 1 \bigg\}. + \end{multline*} + Prove that $C$ is not closed in the standard topology on $\R^2$. +\end{problem} +\begin{proof} + Suppose $C$ is closed in the standard topology on $\R^2$. Then, its complement + $C^c = \R^2 \setminus C$ must be an open set. + + The point $(0, 1)$ is not in $C$, so $(0, 1) \in C^c$. Every open ball in $\R^2$ + containing $(0, 1)$ also contains a smaller open ball centered about $(0, 1)$. + (For example, the open ball about $(-1, 1)$ of radius $1.1$ contains the open + ball centered about $(0, 1)$ of radius $0.1$.) + + However, every open ball centered about $(0, 1)$ contains infinitely many + points of $C$; if the ball has radius $r$, all the comb's ``tines'' at $x = + 1/2^n$ for $n > - \log_2 r$ intersect with the ball. + + Therefore, every open ball containing $(0, 1)$ also contains points in $C$. As + a result, $C^c$ is not open, which contradicts our assumption. Therefore, $C$ + is not closed. +\end{proof} + +\begin{problem}[1.32] + Prove that intervals of the form $[a, b)$ are closed in the lower limit + topology on $\R$. +\end{problem} +\begin{proof} + Take some interval $[a, b)$. Its complement is given by $(-\infty, a) \union + [b, \infty)$. Given that + \begin{alignat*}{1} + (-\infty, a) &= \bigunion_{n=1}^\infty [a-n, a) \\ + [b, \infty) &= \bigunion_{n=1}^\infty [b, b+n), + \end{alignat*} + the complement of $[a, b)$ is the union of a number of lower-limit intervals + in $\R$, i.e.~the basis elements. The basis elements and its unions are open + sets, so the complement of $[a, b)$ is an open set. Then, by the definition of + a closed set, $[a, b)$ is closed in $\R_l$. +\end{proof} + + +\end{document} |