\documentclass[11pt]{book} \usepackage{amsmath, amssymb,latexsym, amsthm, amsfonts, amscd} \usepackage{verbatim} \usepackage{epsfig} \usepackage{graphics} \usepackage{makeidx} \title{Topology\protect\\ \small{M367K}\protect\\ %Fall 2003 } \author{Michael Starbird\protect\\ \&\protect\\ Cynthia Verjovsky Marcotte\protect\\ } \date{Fall 2004} \theoremstyle{remark} \newtheorem{thm}{Theorem}[chapter] \newtheorem{THM*}{Theorem} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \newtheorem{scol}[thm]{Scholium} \newtheorem{ax}{Axiom} \newtheorem{pr}[thm]{Problem} \newtheorem{question}[thm]{Question} \newtheorem{exercise}[thm]{Exercise} \newtheorem{example}{Example} \newtheorem{examples}[example]{Examples} \theoremstyle{remark} \newtheorem*{dfn}{Definition} \newtheorem*{dfns}{Definitions} \newtheorem*{note}{Note} \newtheorem*{hint}{Hint} \newtheorem*{rem}{Remark} \newtheorem*{notation}{Notation} \newtheorem*{Zorn}{Zorn's Lemma} \newtheorem*{AxChoice}{Axiom of Choice} \newtheorem*{WellOrPrin}{Well-ordering Principle} \newtheorem*{Continuum}{Continuum Hypothesis} \newtheorem*{LebegueNumThm}{Lebesgue Number Theorem} \numberwithin{equation}{section} \newcommand{\nibf}{\noindent \textbf} \newcommand{\niit}{\noindent \textit} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\D}{\mathbb{D}} \newcommand{\C}{\mathbb{C}} \newcommand{\PP}{\mathbb{R}\mathrm{P}} \newcommand{\Sph}{\mathbb{S}} \newcommand{\T}{\mathbb{T}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\K}{\mathbb{K}} \newcommand{\B}{\mathbb{B}} \newcommand{\A}{\mathbb{A}} \newcommand{\N}{\mathbb{N}} \newcommand{\wt}{\widetilde} \newcommand{\bs}{\backslash} \newcommand{\beq}{\begin{equation}} \newcommand{\eeq}{\end{equation}} \newcommand{\bea}{\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}} \newcommand{\bi}{\begin{itemize}} \newcommand{\ei}{\end{itemize}} \newcommand{\be}{\begin{enumerate}} \newcommand{\ee}{\end{enumerate}} %\renewcommand{\baselinestretch}{2} \renewcommand{\arraystretch}{1} \newcommand{\no}{\noindent} \newcommand{\ol}{\overline} \newcommand{\ul}{\underline} \newcommand{\al}{\alpha} \def\dim{\mathop{\rm dim}\nolimits} \def\image{\mathop{\rm Im}\nolimits} \def\interior{\mathop{\rm Int}\nolimits} \def\kernel{\mathop{\rm Ker}\nolimits} \def\bd{\mathop{\rm Bd}\nolimits} \def\ext{\mathop{\rm Ext}\nolimits} \def\num{\mathop{\#}\limits} \def\cl{\mathop{\rm Cl}\nolimits} \def\lub{\mathop{\rm lub}\nolimits} \newcommand{\relation}{\mathcal R} \newcommand{\cantor}{X_{C}} \newcommand{\varep}{\varepsilon} \newcommand{\np}{\vfill\eject} \newcommand{\vup}{\vspace{-.08in}} \theoremstyle{remark} \newtheorem{normlemma}[thm]{Normality Lemma} \newtheorem{HBthm}[thm]{Heine-Borel Theorem} \newtheorem{ASthm}[thm]{Alexander Sub-basis Theorem} \newtheorem{urysohn}[thm]{Urysohn Lemma} \newtheorem{Tietze}[thm]{Tietze Extension Theorem} \newtheorem{Tych}[thm]{Tychonoff Theorem} \newtheorem{mthm}[thm]{Metatheorem} \makeindex \begin{document} \maketitle \tableofcontents \chapter{Cardinality and the Axiom of Choice} \label{cardinality} At the end of the nineteenth century, mathematicians embarked on a program whose aim was to axiomatize all of mathematics. That is, the goal was to emulate the format of Euclidean geometry in the sense of explicitly stating a collection of definitions and unproved axioms and then proving all mathematical theorems from those definitions and axioms. The foundation on which this program rested was the concept of a set. Axioms for set theory were proposed and then the goal was to cast known mathematical theorems in set theoretic terms. So the challenge for mathematicians was to take familiar objects, such as the real line, and familiar concepts, such as continuity and convergence, and recast them in terms of sets. From this effort arose the concept of a topological space and the field of topology. We begin our exploration of set-theoretic topology by starting with perhaps the most basic mathematical idea---counting---and finding a way to generalize that notion to apply reasonably to infinite sets. We begin by exploring some of the most familiar infinite sets. \begin{dfn}[number sets] Throughout these notes we will use the following notation: \be \item $\N=$% \index{$\N$} the set of natural numbers (\emph{i.e.}, the positive integers). \item $\Z=$% \index{$\Z$} the set of all integers. \item $\Q=$% \index{$\Q$} the set of rational numbers. \item $\R=$% \index{$\R$} the set of real numbers. \ee \end{dfn} The most basic counting question is deciding how to tell when two sets have the same size. Finite sets have the same size if they have the same number of elements. But `the same number of elements' does not seem too useful a phrase when dealing with infinite sets, since there is no number that describes the size of an infinite set. Instead, we notice that if two finite sets have the same number of elements, then we can put the elements of one set in one-to-one correspondence with the elements of the other set. This idea of pairing up the elements of one set with the elements of the other is what we need to generalize the concept of size to infinite sets. The idea of a 1-to-1 function between two sets is the fundamental idea on which the exploration of the size or \emph{cardinality} of infinite sets rests. So here then is the basic definition of two sets having the same cardinality. \begin{dfn}[cardinality]% \index{cardinality} Two sets, $A$ and $B$, have the \emph{same cardinality\/} if and only if there exists a one-to-one, onto function $f: A\to B$. The cardinality of a set $A$ is denoted $|A|$. \end{dfn} \begin{dfn}[finite set, infinite set] A set $X$ is \emph{finite\/}% \index{finite set}\index{set!finite}\index{cardinality!finite} if and only if it is empty or there is a $1$--$1$, onto function $f:X\to \{ 1,2,\ldots, n\}$ where $n$ is an element of $\N$. A set that is not finite is \emph{infinite\/}. \end{dfn} The cardinality of a finite set is simply the number of elements in that set: 0, 1, 2, 3, .... A basic fact about the natural numbers $\N$, which you should feel free to use in your proofs, is that every non-empty set of natural numbers has a least element. \begin{thm} The even positive integers have the same cardinality as the natural numbers. \end{thm} \begin{thm} $|\N| = |\Z|$. \end{thm} \begin{thm} Every subset of $\N$ is either finite or has the same cardinality as $\N$. \end{thm} \begin{dfn}[countable set] \index{countable set}\index{cardinality!countable}% \index{set!countable} A set that has the same cardinality as a subset of $\N$ is \emph{countable}. \end{dfn} So, a countable set is either finite or has the same cardinality as $\N$. The next theorem shows that the set of natural numbers is in some sense the smallest infinite set. \begin{thm} Every infinite set has a countably infinite subset. \end{thm} \begin{thm} A set is infinite if and only if there is a one-to-one function from the set into a proper subset of itself. \end{thm} \begin{thm} $\Q$ is countable. \end{thm} \begin{thm} The union of two countable sets is countable. \end{thm} \begin{thm} The union of countably many countable sets is countable. \end{thm} \begin{thm} The set of all finite subsets of a countable set is countable. \end{thm} \begin{exercise} Suppose a submarine is moving in a straight line at a constant speed in the plane such that at each hour, the submarine is at a lattice point. Suppose at each hour you can explode one depth charge at a lattice point that will kill the submarine if it is there. You do not know where the submarine is nor do you know where or when it started. Prove that you can explode depth charges in such a way that you will be guaranteed to eventually kill the submarine. \end{exercise} \begin{dfn}[power set] For any set $A$, $2^A$ (or ${\mathcal P}(A)$) denotes the set of all subsets of $A$. (The empty set, denoted $\emptyset$, is a subset of any set.) $2^A$ is called the \emph{power set\/}% \index{power set} of $A$. \end{dfn} \begin{exercise} Suppose $A=\{a,b,c\}$, then write down $2^A$ (or ${\mathcal P}(A)$), the power set of $A$. \end{exercise} \begin{thm} For any set $A$, there is a $1$--$1$ function $f$ from $A$ into $2^A$. \end{thm} \begin{thm} For a set $A$, let $P$ be the set of all functions from $A$ to the two point set $\{0,1\}$. Then $|P| = | 2^A|$. \end{thm} \begin{thm} There is a $1$--$1$ correspondence between $2^{\N}$ and infinite sequences of $0$'s and $1$'s. \label{0-1fn} \end{thm} \begin{thm}[Cantor] \label{power} There is no function from a set $A$ onto $2^A$. \end{thm} Note that Cantor's Theorem implies that $2^{\N}$ is not a countable set. A set that is not countable is called uncountable. So $2^{\N}$ is an uncountable set. In fact, Cantor's theorem implies that there are infinitely many different infinite cardinal numbers: \begin{cor} There are infinitely many different infinite cardinalities. \end{cor} \begin{thm} There is a $1$--$1$, onto function $f: [0,1]\to [0,1)$. \end{thm} \begin{thm}[Schroeder-Bernstein] \index{Schroeder-Bernstein theorem} If $A$ and $B$ are sets such that there exist one-to-one functions $f$ from $A$ into $B$ and $g$ from $B$ into $A$, then $|A| =|B|$. \end{thm} \begin{hint} We need to produce a $1$--$1$, onto function ${h:A\to B}$. A useful picture is to depict $A$ and $B$ as parallel, equal length vertical lines and show $f$ as a shrinking $A$ into three-quarters of $B$ and $g$ shrinking $B$ into three-quarters of $A$ by drawing slanted lines between the top of $A$ to the three-quarters point on $B$ and vice versa, thinking of the bottom points going to one another under $f$ and $g$. When defining $h$, for each point $x\in A$, either $h(x)=f(x)$ or $h(x)=g^{-1}(x)$. For some points $x$ in $A$, you can not use $g^{-1}$, you must use $f$. Shade that interval, and shade its image under $f$. Now look at $g$ of that interval, which in the picture is an interval in $A$. Could you use $g^{-1}$ on those points? Why not? Continue the process and describe those points on which you must use $f$ in your definition of $h$, and on which points you must use $g^{-1}$. \end{hint} \begin{thm} $|\R| = |(0,1)| = |[0,1]|$. \end{thm} \begin{thm} There is a $1$--$1$ function from $\R\to 2^{\N}$. \end{thm} \begin{thm} $|\R| = |2^{\N}|$. \end{thm} \section{*Zorn's Lemma, Axiom of Choice, Well-Ordering Principle} Three important statements in foundational mathematics are Zorn's Lemma, the Axiom of Choice, and the Well-Ordering Principle. These three statements are equivalent. They are accepted as fundamental axioms and used freely in most standard mathematics. We will use them in this course. In this section we give the relevant definitions and then state Zorn's Lemma, the Axiom of Choice, and the Well-Ordering Principle. \begin{dfn}[partially ordered set, poset] A set $X$ is \emph{partially ordered\/}% \index{partially ordered set}\index{set!partially ordered}\index{poset} by the relation $\relation$ if and only if, for any elements $x$, $y$, and $z$ in $X$, \begin{enumerate} \item $x \relation x$, \vup \item if $x\relation y$ and $y\relation z$ then $x\relation z$, \vup \item if $x\relation y$ and $y\relation x$ then $x=y$. \end{enumerate} A partially ordered set is sometimes called a {\it poset}. \end{dfn} \begin{example} For any set $X$, the power set of $X$, $2^X$ is a poset under the relation $\subseteq$. \end{example} \begin{dfn}[least element]\index{least element} Let $X$ be a poset with relation $\relation$. Then an element $a$ in $X$ is a {\it least element\/} if and only if for any $x\in X$, $a \relation x$. \end{dfn} \begin{dfn}[maximal element]\index{maximal element} Let $X$ be a poset with relation $\relation$. Then an element $m$ in $X$ is a {\it maximal element\/} if and only if for any $x$ in $X$, $m\relation x$ implies that $m=x$. \end{dfn} \begin{example} Recall we saw above that the power set $2^A$ is partially ordered by set inclusion. The set $A$ is a maximal element, and, in fact, the only maximal element in this ordering. \end{example} \begin{dfn}[totally ordered set] A poset is {\it totally ordered\/}% \index{totally ordered set}\index{set!totally ordered} if and only if it is partially ordered and every two elements are comparable (that is, for all $x$, $y$, either $x \relation y$ or $y \relation x$). \end{dfn} In general, we will use $\leq$ rather than $\relation$ when talking about the relation in a totally ordered set. \begin{dfn}[well-ordered set]\index{well-ordered set}\index{set!well-ordered} A set is {\it well-ordered\/} if and only if it is totally ordered and every non-empty subset has a least element. \end{dfn} The natural numbers are well-ordered. \begin{exercise} Show that the ordinary ordering on the reals is not a well-ordering. \end{exercise} \begin{Zorn}\index{Zorn's lemma} Let $X$ be a partially ordered set in which each totally ordered subset has an upper bound in $X$. Then $X$ has a maximal element. \end{Zorn} \begin{AxChoice}\index{Axiom of Choice} Let $\{ A_\alpha\}_{\alpha\in\lambda}$ be a collection of non-empty sets. Then there is a function $f: \lambda\to \bigcup_{\alpha\in\lambda} A_\alpha$ such that for each $\alpha$ in $\lambda$, $f(\alpha)$ is an element of $A_\alpha$. \end{AxChoice} \begin{WellOrPrin}\index{Well-ordering principle} Every set can be well-ordered. That is, every set is in $1$--$1$ correspondence with a well-ordered set. \end{WellOrPrin} \section{*Ordinal numbers} In common English an ordinal number refers to the numerical position of an object: first, second, third, and so on. We will use the arabic numerals to denote the ordinal numbers with which we are most familiar: $0,1,2,3,\ldots$. We can define ordinals in a manner that allows us to produce an ordered set of ordinals that includes infinite ordinals. We start with the empty set, $\emptyset$. This set corresponds to~$0$. The next ordinal, corresponding to~1, is the set containing the empty set, $\{\emptyset\}$. The next ordinal, corresponding to~2, is the set of its predecessors, namely $\{\emptyset,\{\emptyset\}\}$. The next ordinal, corresponding to~3, is the set of its predecessors, $\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$. Notice that each finite ordinal number $k$ described in this way is a set containing $k$ elements. Continuing in this fashion, we can define each subsequent ordinal as the set of its predecessors. For example, the first infinite ordinal, called $\omega_0$, is the set of all the finite ordinals, namely, the set $\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}$, $\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\},\ldots\}$. The next ordinal is called $\omega_0+1$, then $\omega_0+2$, $\omega_0+3,\ldots$, then $2\omega_0,2\omega_0+1,\ldots$; $\ldots k\omega_0, k\omega_0+1,k\omega_0+2,\ldots$, etc. Note that every ordinal number has an immediate successor; however, not every ordinal has an immediate predecessor. For example, $\omega_0$ has no immediate predecessor. Note also that each ordinal is a set and, consequently, has a cardinality. The ordinal $\omega_0$ is the first infinite ordinal and has the same cardinality as $\N$, denoted by $\aleph_0$, aleph nought. Each of the ordinal numbers ${\omega_0+1}$, ${\omega_0+2}$, ${\omega_0+3}$, and so on has countable cardinality, $\aleph_0$. The first uncountable ordinal is called $\omega_1$. It is the set of all the countable ordinals. Every ordinal preceding it, which is the same as in it, is countable. The cardinality of $\omega_1$ is less than or equal to the cardinality of $2^{\N}$ (or $\R$). However, the question of whether the cardinality of $\omega_1$ is equal to the cardinality of $\R$ is the content of the Continuum Hypothesis. The Continuum Hypothesis asserts that these two cardinalities are in fact equal; however, it has been proved that the Continuum Hypothesis is independent of the standard Axioms of Set Theory (the Zermelo-Fraenkel axioms). That is, the Continuum Hypothesis can be neither proved nor disproved. \begin{Continuum}\index{Continuum Hypothesis} The real numbers have the same cardinality as $\omega_1$, the first uncountable ordinal. \end{Continuum} Ordinal numbers are well-ordered by $\subseteq$, because the intersection of any set of ordinals is the smallest ordinal in the set, so every non-empty subset has a smallest element. \begin{thm} Let $\{\alpha_i\}_{i\in\omega_0}$ be a countable set of ordinal numbers where each $\alpha_i <\omega_1$. Then there is an ordinal $\beta$ such that $\alpha_i<\beta$ for each $i$ and $\beta<\omega_1$. \end{thm} \chapter{General Topology} In this chapter, we will start with the real number line and investigate some of its properties. We will then define a topological space as an abstraction of features of the real line. The topological ideas of limit point, convergence, open and closed sets, and continuity are all the result of capturing essential characteristics that we find in the real numbers. \section{The Real Number Line} We will not present an axiomatic definition of the real numbers. Instead, we will rely on our understanding of the real number line as the set of all decimal numbers ordered in their familiar way. %We will assume a basic property of the real numbers namely: %\begin{dfn}[least upper bound axiom] %In the real number line $\R$ , every bounded set has a least upper bound. %\end{dfn} Let us first review the concepts necessary to define convergence of sequences and continuity of functions on the real number line. \begin{dfn}[open interval] In the real number line $\R$ define an \emph{open interval}\index{open interval} $(a,b)$ as the set $\{x\in\R|a0$ the \emph{open interval centered at $x$ of radius $\varep$}, $B(x,\varep)$ is the open interval $(x-\varep,x+\varep)$. \end{dfn} \begin{dfn}[open set in $\R$] In $\R$ a set $U$ is \emph{open} \index{open set!$\R$} if and only if for every point $x\in U$ there is an $\varep_x>0$ such that $(x-\varep_x,x+\varep_x)\subseteq U$. \end{dfn} \begin{thm} The empty set is open and $\R$ is open. \end{thm} \begin{thm} If $U_1$ and $U_2$ are open sets, then $U_1\cap U_2$ is open. In fact, the intersection of finitely many open sets is open. \end{thm} \begin{thm} The union of any collection of open sets in $\R$ is open. \end{thm} \begin{thm} If $U$ is open, then $U$ is the union of open intervals. \end{thm} Let us recall the definition of convergence of a sequence in $\R$: \begin{dfn}[convergent sequence] We say a sequence $\{x_i\}_{i\in\N}\subseteq\R$ \emph{converges} to $x$, or that $x$ is the \emph{limit of the sequence}, written as $x_i\to x$ if and only if for every $\varep>0$ there is an $N\in \N$ such that $|x-x_i|<\varep$ for all $i>N$. \end{dfn} We are in the process of recasting ideas from the real line in set-theoretic terms. So let's rephrase the definition of convergence \emph{without} using distance, but, instead, in terms of open intervals. The definition above says that any open interval of radius $\varep$ centered at $x$ contains \emph{all but finitely many} of the elements of the sequence. Instead of restricting ourselves to open intervals centered at $x$, we can consider \emph{any} open set containing $x$ and define convergence as follows: \begin{dfn}[convergent sequence] \index{convergence of a sequence!in $\R$}% \index{sequence!convergence!in $\R$} If $\{x_i\}_{i\in\N}\subseteq\R$, then we say that $x$ \emph{converges} to $x$, written as $x_i\to x$ if and only if for any open set containing $x$, there is an $N\in \N$ such that $x_i\in U$ for all $i>N$. \end{dfn} If instead of studying sequences we wish to look at sets in $\R$ in general, we would like to define what we mean by saying that a point is ``close'' to (other) points in that set. One way is to abstract the concept of distance, but we don't actually need to have a distance, we can also use the (more general) concept of an open set. Let us see how it would play out in $\R$. We could say that $x$ is ``close'' to a set $A$ if there is a sequence of elements of $A$ that converges to $x$. If $x$ is in $A$, however, we could always cheat and pick the constant sequence $\{x_i\}$ where $x_i=x$ for all (or almost all) $i$. This definition would not include the idea that $x$ has nearby points from A. To avoid this problem we could say that $x$ is ``close'' to a set $A$ if there is a sequence $\{x_i\}$ of elements of $A-\{x\}$ that converges to $x$. But a more general definition would be to dispense with sequences altogether, and use the condition that the intersection of any open set containing $x$ with $A-\{x\}$ is never empty. So we thus come to the following definition: \begin{dfn}[limit point in $\R$] \index{limit point!in $\R$} Let $A$ be a subset of $\R$ and $x$ be a point in $\R$. Then $x$ is a \emph{limit point\/} of $A$ if and only if for every open set $U$ containing $x$ $(U-\{x\})\cap A)$ is not empty. \end{dfn} In other words, a limit point of a set is one that cannot be isolated from the rest of the set with an open set. \begin{dfn}[closed set in $\R$] A set in $\R$ is \emph{closed} if and only if it contains all of its limit points. \end{dfn} \begin{thm} The intersection of any collection of closed sets in $\R$ is closed. \end{thm} \begin{thm} The union of two closed sets in $\R$ is closed. In fact, the union of finitely many closed sets in $\R$ is closed. \end{thm} Closed sets and open sets are related by the following theorem. \begin{thm} A set in $\R$ is open if and only if its complement is closed. \end{thm} Let us now review the definition of continuity that you probably first encountered in calculus---the $\varep$-$\delta$ definition. \begin{dfn}[continuous function in $\R$] A function $f: D\subseteq \R\to \R$ is \emph{continuous at $x$} if and only if for every $\varep>0$ there is a $\delta>0$ such that if $z\in D$ and $|x-z|<\delta$ then $|f(x)-f(z)|<\varep$. We say $f$ is \emph{continuous} if it is continuous for every point $x$ in its domain $D$. \end{dfn} We wish to convert this definition into the language of open sets and intervals. It is saying that if a function is continuous in its domain, then if we pick an open interval $I=(f(x)-\varep,f(x)+\varep)$, then we can find an open interval $J=(x-\delta,x+\delta)$ that is mapped into $I$. So, in general, if we pick an open set $U\subseteq \R$ that contains a point $f(x)$, then we can find an open set $V\subseteq \R$ containing $x$ whose image is contained inside $U$. This set theoretic view of continuity allows us to re-word the concept of continuity in the language of open sets as follows: \begin{dfn}[continuous function in $\R$]% \index{continuous function!in $\R$}\index{function!continuous!in $\R$} A function $f: \R\to \R$ is a \emph{continuous function\/} if and only if for every open set $U$ in $\R$, $f^{-1}(U)$ is open in $\R$. \end{dfn} \begin{thm} The two definitions of continuity for real-valued functions on $\R$ are equivalent. \end{thm} \section{Open Sets and Topologies} We have now seen that several important concepts in analysis (convergence, limit points, closed sets, and continuity) can be defined using ideas about sets and their intersections and unions. In our familiar world of the real numbers, open sets were the central players in all these concepts. Now we would like to extend these concepts to spaces other than the familiar real numbers with their usual concept of open set. Our strategy is to abstract a more general concept of an open set from our experience with the real numbers. To that end, we isolate some of the conditions that were satisfied by the usual open sets of $\R$ and use those properties to define a \emph{topology} and a \emph{topological space}. \begin{dfn}[topology]\index{topology} Suppose $X$ is a set. Then $\mathcal T$ is a \emph{topology} for $X$ if and only if ${\mathcal T}$ is a collection of subsets of $X$ such that \begin{enumerate} \item $\emptyset \in {\mathcal T}$, \item $X\in {\mathcal T}$, \item if $U\in {\mathcal T}$ and $V\in {\mathcal T}$, then $U\cap V\in {\mathcal T}$, \item if $\{ U_\alpha\}_{\alpha\in\lambda}$ is any collection of sets each of which is in $\mathcal T$, then $\bigcup_{\alpha\in\lambda} U_\alpha \in {\mathcal T}$. \end{enumerate} A \emph{topological space\/}% \index{topological space} is a pair $(X,{\mathcal T})$ where $X$ is a set and $\mathcal T$ is a topology for $X$. If $(X,{\mathcal T})$ is a topological space, then $U \subseteq X$ is called an \emph{open set\/}% \index{open set}\index{set!open} in $(X,{\mathcal T})$ if and only if $U\in {\mathcal T}$. \end{dfn} \begin{thm} Let $\{ U_i\}_{i=1}^n $ be a \emph{finite} collection of open sets in a topological space $(X,{\mathcal T})$. Then $\bigcap_{i=1}^n U_i$ is open. \end{thm} Our first step toward understanding this abstract definition of a topological space is to confirm that the definition has captured relevant features of the prototype (that is, the real line) that spawned it. So our first example of a topological space will be the real number line where the collection of open sets in $\R$ that we talked about in the previous section is the standard topology on $\R$. \begin{example}[standard topology on $\R$] The \emph{standard topology} $\mathcal T_\textrm{std}$ for $\R$ is defined as follows: a subset $U$ of $\R$ belongs to $\mathcal T_\textrm{std}$ if and only if for each point $p$ of $U$ there is an open interval $(a_p,b_p)$ such that $p\in (a_p,b_p)\subset U$. \index{standard topology on $\R$}\index{topology!standard!$\R$} \end{example} Let us consider some other examples of topological spaces. Note that $(X,\mathcal T)$ and $(X,\mathcal T')$ are \emph{different\/} topological spaces if $\mathcal T\neq \mathcal T'$, even though the underlying set $X$ is the same. Keep in mind that open sets $U$ are \emph{elements} of the topology $\mathcal{T}$, and \emph{subsets} of the space $X$. Elements of $X$, on the other hand are what we call the points of the space $X$. \begin{example}[discrete topology] For a set $X$, let $2^X$ be the set of all subsets of $X$. Then ${\mathcal T} = 2^X$ is called the \emph{discrete topology\/}% \index{discrete topology}\index{topology!discrete} on $X$. The space $(X,2^X)$ is called a \emph{discrete topological space\/}. \end{example} Note the spelling: \emph{discrete} topology, not \emph{discreet} topology! \begin{example}[indiscrete topology] For a set $X$, ${\mathcal T} =\{ \emptyset, X\}$ is called the \emph{indiscrete topology\/}% \index{indiscrete topology}\index{topology!indiscrete} for $X$. So $(X,\{ \emptyset, X\})$ is an indiscrete topological space. \end{example} Notice that the discrete topology has the maximum possible collection of open sets that any topology can have while the indiscrete topology has the minimum possible collection of open sets. \begin{example}[finite complement or co-finite topology] For any set $X$, the \emph{finite complement (or co-finite) topology\/}% \index{finite complement topology}\index{co-finite topology}\index{topology!finite complement}% \index{topology!co-finite} for $X$ is described as follows: a subset $U$ of $X$ is open if and only if $U=\emptyset$ or $X-U$ is finite. \end{example} Recall that a countable set is one that is either finite or countably infinite. \begin{example}[countable complement topology] For any set $X$, the \emph{countable complement topology\/}% \index{countable complement topology}% \index{topology!countable complement} for $X$ is described as follows: a subset $U$ of $X$ is open if and only if $U=\emptyset$ or $X-U$ is countable. \end{example} \begin{exercise} Verify that all the examples given above are indeed topologies; in other words, that they satisfy all four conditions needed to be a topology. \end{exercise} \begin{exercise} \be \item Describe some of the open sets you get if $\R$ is endowed with the topologies described above (standard, discrete, indiscrete, co-finite, and countable complement). Specifically, identify sets that demonstrate the differences among these topologies, that is, find sets that are open in some topologies but not in others. \item For each of the topologies, determine if the interval $(0,1)\in\R$ is an open set in that topology. \ee \end{exercise} We can generalize the standard topology on $\R$ to the Euclidean spaces $\R^n$. Rather than using open intervals to generate open sets, we use \emph{open balls}: \begin{example} Let $\R^n$ be the set of all $n$-tuples of real numbers. \be \item The \emph{Euclidean distance}% \index{Euclidean distance} $d(x,y)$ between points $x= (x_1,x_2,\ldots,x_n)$ and $y=(y_1,y_2,\ldots,y_n)$ is given by the equation \[d(x,y) = \biggl( \sum_{i=1}^n (x_i-y_i)^2\biggr)^{1/2}\ .\] \item The open ball of radius $\varep>0$ around point $p\in\R^n$ is the set $B(x,\varep) =\{ x\mid d(p,x)<\varep\}$% \index{open ball} \item A topology $\mathcal T$ for $\R^n$ is defined as follows: a subset $U$ of $\R^n$ belongs to $\mathcal T$ if and only if for each point $p$ of $U$ there is a $\varep_p>0$ such that $B(x,\varep_p)\subseteq U$. \ee This topology $\mathcal T$ is called the \emph{standard topology for\/} $\R^n$. \index{standard topology on $\R^n$}\index{topology!standard!$\R^n$} \end{example} \begin{exercise} Give an example of a topological space and a collection of open sets in that topological space to show that the \emph{infinite} intersection of open sets need not be open. \end{exercise} \section{Limit Points and Closed Sets} As in $\R_\textrm{std}$, we will define the concept of a limit point using open sets, and then define closed sets as those sets that contain all their limit points. \begin{dfn}[limit point] Let $(X,{\mathcal T})$ be a topological space, $A$ be a subset of $X$, and $p$ be a point in $X$. Then $p$ is a \emph{limit point\/}% \index{limit point} of $A$ if and only if for each open set $U$ containing $p$, ${(U-\{ p\})\cap A\neq \emptyset}$. Notice that $p$ may or may not belong to $A$. \end{dfn} In other words, $p$ is a limit point of $A$ if \emph{all\/} open sets containing $p$ intersect $A$ at \emph{some point other than} itself. Thus, the concept of a limit point gives us a way of capturing the idea of a point ``being arbitrarily close" to a set \emph{without} using the concept of distance. Instead we use the idea of open sets in a topology. \begin{dfn}[isolated point]\index{isolated point} Let $(X,{\mathcal T})$ be a topological space, $A$ be a subset of $X$, and $p$ be a point in $X$. If $p\in A$ but $p$ is not a limit point of $A$, then $p$ is an \emph{isolated point\/} of $A$. \end{dfn} If $p$ is an isolated point of $A$, then there is an open set $U$ such that $U\cap A=\{p\}$. \begin{thm} Suppose $p\not\in A$ in a topological space $(X,\mathcal T)$. Then $p$ is \emph{not} a limit point of $A$ if and only if there exists an open set $U$ with $p\in U$ and $U\cap A =\emptyset$. \end{thm} \begin{exercise} Give examples of a set $A$ in a topological space and \be \item a limit point of $A$ that is an element of $A$; \item a limit point of $A$ that is not an element of $A$; \item an isolated point of $A$; \item a point not in $A$ that is not a limit point of $A$. \ee \end{exercise} \begin{dfn}[closure of a set]% \index{closure}\index{set!closure of} Let $(X,{\mathcal T})$ be a topological space, and $A\subseteq X$. Then the \emph{closure} of $A$, denoted $\ol{A}$ or $\cl(A)$, is $A$ together with all of its limit points. \end{dfn} \begin{dfn}[closed set]% \index{closed set} Let $(X,{\mathcal T})$ be a topological space and $A\subseteq X$. $A$ is \emph{closed} if and only if $\cl(A)=A$, in other words, if $A$ contains all its limit points. \end{dfn} \begin{thm} For any topological space $(X,{\mathcal T})$ and $A\subseteq X$, $\ol{A}$ is closed, that is, for any set $A$ in a topological space, $\overline{\overline{A}} = \overline{A}$. \end{thm} A basic relationship between open sets and closed sets in a topological space is that they are complements of each other. \begin{thm} Let $(X,{\mathcal T})$ be a topological space. Then the set $A$ is closed if and only if $X - A$ is open. \end{thm} \begin{thm} Let $(X,{\mathcal T})$ be a topological space, and let $U$ be an open set and $A$ be a closed subset of $X$. Then the set $U - A$ is open and the set $A - U$ is closed. \end{thm} The properties of a topological space can be captured by focusing on closed sets instead of open sets. From that perspective, the four defining properties of a topological space are captured in the following theorem about closed sets. \begin{thm} Let $(X, \mathcal{T})$ be a topological space: \bi \item[$i$)] $\emptyset$ is closed. \item[$ii$)] $X$ is closed. \item[$iii$)]The union of finitely many closed sets is closed. \item[$iv$)]Let $\{ A_\alpha\}_{\alpha\in\lambda}$ be a collection of closed subsets in $(X,\mathcal{T})$. Then $\bigcap_{\alpha\in\lambda} A_\alpha$ is closed. \ei \end{thm} \begin{exercise} Give an example to show that the union of infinitely many closed sets in a topological space may be a set that is not closed. \end{exercise} \begin{exercise} Give examples of topological spaces and sets in them that: \be \item are closed, but not open; \item are open, but not closed; \item are both open and closed; \item are neither open nor closed. \ee \end{exercise} \begin{exercise} State whether each of the following sets are open, closed, both or neither. \begin{enumerate} \item In $\Z$ with the finite complement topology: $\{0, 1, 2\}$, $\{ \text{prime numbers} \}$, $\{ n \ : \ |n| \geq 10\}$. \item In $\R$ with the standard topology: $(0,1), (0,1], [0,1], \{0,1\}, \{ 1/n \ | \ n\in \N\}$. \item In $\R^2$ with the standard topology: $\{(x,y) \ | \ x^2+y^2 = 1\}$, $\{(x,y) \ | \ x^2+y^2 > 1\}$, $\{(x,y) \ | \ x^2+y^2 \ge 1\}$, \item Which sets are closed in a set $X$ with the discrete topology? indiscrete topology? \end{enumerate} \end{exercise} \begin{thm} For any set $A$ in a topological space $X$, the closure of $A$ equals the intersection of all closed sets containing $A$, that is, \[ \cl(A)=\bigcap_{A\subseteq C\textrm{, }C\in\mathcal{C}}C \] where $\mathcal{C}$ is the collection of all closed sets in $X$. \end{thm} Informally, we can say $\ol{A}$ is the ``smallest" closed set that contains $A$. \begin{exercise} Pick several different subsets of $\R$, and find their closure in: \be \item the discrete topology; \item the indiscrete topology; \item the finite complement topology; \item the standard topology. \ee \end{exercise} \begin{thm} Let $A$, $B$ be subsets of a topological space. Then \begin{enumerate} \item $A\subseteq B$ $\Rightarrow$ $\overline{A} \subseteq \overline{B}$; and \item $\overline{A\cup B} = \overline{A} \cup \overline{B}$. \end{enumerate} \end{thm} \begin{exercise} In $R^2$ with the standard topology, describe the limit points and closure of the following two sets: \be \item The topologist's sine curve:\index{topologist's sine curve} \[ S=\left\{\left(x,\sin\left({\frac{1}{x}}\right)\right)| x\in(0,1)\right\} \] \item The topologist's comb:\index{topologist's comb} \[ C=\{(x,0)|x\in[0,1]\}\cup \bigcup_{n=1}^{\infty}\left\{\left(\frac{1}{n},y\right)\left|\right. y\in[0,1]\right\} \] \ee \end{exercise} The following exercise is difficult. \begin{exercise} In the standard topology on $\R$, describe a non-empty subset $C$ of the closed unit interval $[0,1]$ that is closed, contains no non-empty open interval, and where no point of $C$ is an isolated point. \end{exercise} \section{Interior, Exterior, and Boundary} Just as we have the concept of the ``smallest" closed set containing $A$, we can considered what is the ``largest" open set \emph{contained in} $A$. \begin{dfn}[interior of a set] \index{interior of a set}\index{set!interior} The \emph{interior} of a set $A$ in a topological space $X$, denoted $\stackrel{\circ}{A}$ or $\interior(A)$, is defined as: \[ \interior(A)=\bigcup_{U\subseteq A\textrm{, }U\in\mathcal{T}}U. \] \end{dfn} \begin{exercise} Pick several different subsets of $\R$, and find their interior in: \be \item the discrete topology; \item the indiscrete topology; \item the finite complement topology; \item the standard topology. \ee \end{exercise} We sometimes say, `$X$ is a topological space.' When we say that, we mean that there is a topology $\mathcal T$ on $X$ that is implicit. \begin{thm} Let $A \subseteq X$ be a subset of topological space $X$. Then $\interior(A)$ is the collection of points $p$ such that there exists an open set $U$ with $p \in U \subset A$. \end{thm} \begin{dfn}[boundary]\index{boundary of a set}\index{set!boundary} The \emph{boundary} of $A$, $\bd(A)$ or $\partial A$, is $Cl(A)\cap Cl(X-A)$. \end{dfn} Finally, let us see how we would define the convergence of a sequence in a general topological space: \begin{dfn}[limit of a sequence]\label{top.conv}% \index{limit of a sequence}% \index{convergence of a sequence}\index{convergence!sequence} Given a sequence $\{ x_i\}$ in a topological space $X$, we say $x$ is a \emph{limit of the sequence}, written as $x_i \rightarrow x$, if and only if for every open set $U$ containing $x$, $U$ contains all but finitely many of the $ x_i$'s. We also say $x_i$ \emph{converges} to $x$. \end{dfn} \begin{thm} Let $A \subseteq X$ be a subset of the topological space $X$. If $\{ x_i \}_{i\in\N} \subset A$ and $x_i \rightarrow x$, then $x$ is in the closure of $A$. \end{thm} \begin{question} Is the converse of the preceding theorem true? That is, if $A \subseteq X$ where $X$ is a topological space and $x$ is a limit point of $A$, then does there exist $\{ x_i \}_{i\in\N} \subset A$ such that $x_i \rightarrow x$? \end{question} \begin{exercise} Find an example of a topological space and a convergent sequence in that space, where the limit of the sequence is not unique. \end{exercise} \begin{exercise} Consider sequences in $\R$ with the finite complement or co-finite topology. Which sequences converge? To what value(s) do they converge? \end{exercise} We will leave the study of continuity in general topological spaces for a later chapter. For now, we will focus on different ways of creating and describing topological spaces. \section{Bases}\label{bases} Because arbitrary unions of open sets are open, a topological space can have extremely complicated open sets. It is often convenient to describe a (simpler) subcollection of open sets that \emph{generate}---in a prescribed way---all open sets in a given topology. So instead of having to specifically describe all of the open sets in a topological space $(X,\mathcal T)$, we can more conveniently specify a subcollection, called a {\em basis} for the topology $\mathcal T$. Recall, for instance, that in order to define the open sets in the standard topology in $\R$ (respectively, $\R^n$) we used the concept of open intervals (respectively, open balls). We called a set $U$ an open set if we could find an open interval (respectively, open ball) contained in $U$ around every point in $U$. Thus, we could think of open sets as being made by taking arbitrary unions of these simpler open sets. \begin{dfn}[basis of a topology]\index{basis of a topology}\index{topology!basis} Let $\mathcal T$ be a topology on a set $X$ and let $\mathcal B\subseteq\mathcal T$. Then $\mathcal B$ is a \emph{basis\/} for the topology $\mathcal T$ if and only if every element of $\mathcal T$ is the union of elements in $\mathcal B$. If $B\in \mathcal{B}$, we say $B$ is a \emph{basis element}% \index{basis element} or \emph{basic open set}. \index{basic open set}\index{set!basic} Note that $B$ is an \emph{element} of the basis, but a \emph{subset} of the space. \end{dfn} \begin{thm} Let $(X,{\mathcal T})$ be a topological space and $\mathcal B$ be a collection of subsets of $X$. Then $\mathcal B$ is a basis for $\mathcal T$ if and only if \be \item ${\mathcal B}\subseteq {\mathcal T}$, \item $\emptyset \in {\mathcal B}$, \item for each set $U$ in $\mathcal T$ and point $p$ in $U$ there is a set $V$ in $\mathcal B$ such that $p\in V\subseteq U$. \ee \end{thm} \begin{thm} Let ${\mathcal B}_1 = \{ (a,b) \subseteq \R \mid a,\ b \in \Q\}$, then ${\mathcal B}_1$ is a basis for the \emph{standard topology} on $\R$. Let ${\mathcal B}_2 = \{ (a,b)\cup(c,d) \subseteq \R \mid a,\ b,\ c,\ d{\textrm{ are distinct irrational numbers}}\}$, then ${\mathcal B}_2$ is also a basis for the \emph{standard topology} on $\R$. \end{thm} Suppose you are given a set $X$ and a collection $\mathcal B$ of subsets of $X$. Under what circumstances is there a topology for which $\mathcal B$ is a basis? This question is answered in the following theorem. There is a subtle difference between the following theorem and the theorem two before this one. The former theorem started with a given topology and explored the question of when a collection of sets could form a basis for that particular topology. The following theorem explores the question of whether a given collection of sets could be a basis for \emph{some topology} on $X$. \begin{thm} \label{basis.thm} Suppose $X$ is a set and $\mathcal B$ is a collection of subsets of $X$. Then $\mathcal B$ is a basis for a topology for $X$ if and only if the following conditions hold. \be \item $\emptyset \in {\mathcal B}$, \item for each point $p$ in $X$ there is a set $U$ in $\mathcal B$ with $p\in U$, and \item if $U$ and $V$ are sets in $\mathcal B$ and $p$ is a point in $U\cap V$, there is a set $W$ in $\mathcal B$ such that $p\in W \subseteq (U\cap V)$. \ee \end{thm} Theorem~\ref{basis.thm} allows us to describe topological spaces by first specifying a set $X$ and then a collection $\mathcal{B}$ of subsets of $X$ satisfying the $3$ conditions listed in the theorem. Then the topology $\mathcal T$ with basis $\mathcal B$ is the collection of all possible unions of basis elements. \begin{example}[lower limit topology] We can define an alternative topology on $\R$, called the \emph{lower limit topology},% \index{lower limit topology}\index{topology!lower limit} generated by a basis consisting of all sets of the form $[a,b) = \{ x\in \R \mid a\le x< b\}$. Denote this space by $\R_{\mathrm{LL}}$.\index{$\R_{\mathrm{LL}}$} The real line with the lower limit topology is sometimes called the \emph{Sorgenfrey line}\index{Sorgenfrey line} or $\R^1$(bad). \end{example} \begin{thm}\label{std.in.lowerlim} Every open set in $\R$ with the standard topology is open in the lower limit topology, $\R_{\mathrm{LL}}$. \end{thm} \begin{exercise}\label{lowerlim.vs.std} Show that the lower limit topology and the standard topology are different topologies on $\R$. \end{exercise} \section{*Comparing Topologies} \begin{dfn}[finer topology]\index{topology!finer}\index{topology!coarser}% \index{finer topology}\index{coarser topology} Suppose a set $X$ is given $2$ topologies: $\mathcal{T}_1$ and $\mathcal{T}_2$. If $\mathcal{T}_1\subset\mathcal{T}_2$ we say that $\mathcal{T}_2$ is a \emph{finer\/} topology than $\mathcal{T}_1$ and that $\mathcal{T}_1$ is a \emph{coarser\/} topology than $\mathcal{T}_2$. If $\mathcal{T}_1\neq\mathcal{T}_2$ we would say \emph{strictly} finer or coarser. \end{dfn} It's hard to remember which is the finer and which is the coarser topology when $\mathcal{T}_1\subset\mathcal{T}_2$. A good way to remember which is which is that a comb with \emph{more} teeth per inch is \emph{finer} than one with fewer! In Exercises~\ref{std.in.lowerlim} and~\ref{lowerlim.vs.std} we showed that the lower limit topology on $\R$ is finer than the standard topology on $\R$. Note that it is possible that $\mathcal{T}_1\not\subseteq\mathcal{T}_2$ and $\mathcal{T}_2\not\subseteq\mathcal{T}_1$, in which case we say the topologies are \emph{not comparable}. \begin{exercise} Give an example of two topologies in $\R$ that are not comparable. \end{exercise} %\begin{dfn}[Cantor set]\index{Cantor set}\index{set!Cantor} %Let $X$ be the set of all countably infinite sequences of $0$'s and %1's, that is, for $x \in X$, $x = x_0x_1x_2\ldots$ where $x_i \in %\{0,1\}$ for all $i$. Let $W$ denote the set of all possible finite %words made from 0's and 1's. For $w \in W$, let $A_w \subseteq X$ be %the set of all infinite sequences which begin with the word $w$. %Then ${\mathcal B} = \{ A_w \ | \ w \in W\} \cup \{\emptyset\}$ is the basis for a %topology on $X$. We will call the set $X$ with this topology the %\emph{Cantor set}. %\end{dfn} %\begin{exercise} %\be %\item Show that ${\mathcal B}$ is in fact a basis for a topology. %\item Characterize each of the following subsets of the Cantor set %as open, closed, both or neither: %\be %\item $\{ x \ | \ x_0 =0 \}$ %\item $\{ x \ | \ x_p=0 \text{ for all prime } p\}$ %\item $\{ x \ | \ x_n = 1 \text{ for all } n > 10\}$. %\ee %\ee %\end{exercise} \section{Order Topology} \begin{dfn}[order topology]% \index{order topology}\index{topology!order} Let $X$ be a set totally ordered by $<$. Let $\mathcal B$ be the collection of all subsets of $X$ of one of the following three forms: \[\{ x\in X\ | \ x M$, $p_m\in U$. \end{dfn} \begin{thm} Suppose $x$ is a limit point of the set $A$ in a $1^{\textrm{st}}$ countable space $X$. Then there is a sequence of points in $A$ that converges to $x$. \end{thm} \begin{thm} Every uncountable set in a $2^{\textrm{nd}}$ countable space has a limit point. \end{thm} \newpage \section{Covering Properties} The next properties we will study are the ``covering" properties, so called because they involve collections of open sets that cover the space or a subset of the space. \begin{dfn}[cover, open cover]\index{open cover}\index{cover}\index{cover!open} Let $A$ be a subset of $X$ and let ${\cal C}= \{ C_\alpha\}_{\alpha\in\lambda}$ be a collection of subsets of $X$. Then $\cal C$ is a \emph{cover of\/} $A$ if and only if $A\subset \bigcup_{\alpha \in\lambda}C_\alpha$. $\cal C$ is an \emph{open cover\/} if and only if each $C_\alpha$ is open. \end{dfn} \begin{dfn}[compact]\index{compact}\index{topological space!compact} A space $X$ is \emph{compact\/} if and only if every open cover $\cal C$ of $X$ has a finite subcover $\cal C'$. That is, $\cal C'$ is a finite open cover of $X$ each of whose elements is a set in $\cal C$. \end{dfn} %\begin{dfn}[countably compact]\index{countably compact}% %\index{compact!countably}\index{topological space!countably compact} % A space $X$ is \emph{countably compact\/} if and only if every %countable open cover of $X$ has a finite subcover. %\end{dfn} %\begin{dfn}[Lindel\"of]\index{Lindel\"of}\index{topological space!Lindel\"of} %A space $X$ is \emph{Lindel\"of\/} if and only if every open %cover of $X$ has a countable subcover. %\end{dfn} %\begin{thm} %Every countably compact and Lindel\"of space %is compact. %\end{thm} %\begin{thm} %Every $2^{\textrm{nd}}$ countable space is Lindel\"of. %\end{thm} \begin{thm} Let $A$ be a closed subspace of a compact %(respectively, countably compact, Lindel\"of) space. Then $A$ is compact. %(respectively, countably compact, Lindel\"of). \end{thm} \begin{thm} Let $\cal B$ be a basis for a space $X$. Then $X$ is compact %(respectively, Lindel\"of) if and only if every cover of $X$ by basic open sets has a finite %(respectively, countable) subcover. \end{thm} \begin{thm} The closed subspace $[0,1]$ in the $\R_{\textrm{std}}$ topology is compact. \end{thm} \begin{thm} Let $A$ be a compact subspace of a Hausdorff space $X$. Then $A$ is closed. \end{thm} \begin{HBthm} Let $A$ be a subset of $\R^1$ with the standard topology. Then $A$ is compact if and only if $A$ is closed and bounded. \end{HBthm} %\begin{thm} %If $X$ is a Lindel\"of space, then every uncountable %subset of $X$ has a limit point. %\end{thm} %\begin{thm} %Let $X$ be a $T_1$ space. Then $X$ is countably %compact if and only if every infinite subset of $X$ has a limit point. %\end{thm} \begin{thm} A compact, Hausdorff space is normal. \end{thm} %\begin{thm} %A regular, Lindel\"of space is normal. %\end{thm} \section{Metric Spaces} The next category of topological spaces that we will consider are called metric spaces, so called because they rely on the idea of a distance between points. Metric spaces arise by considering the notion of the distance between two points in the familiar Euclidean spaces $\R^n$. The strategy is to look at that familiar idea of distance and cull from it central features, which then become the definition of a metric. \begin{dfn}[metric]\index{metric} A \emph{metric} on a set $M$ is a function $d:M\times M \to \R_+$, where $\R_+$ is the non-negative real numbers, such that for all $a,b,c \in M$: \be \item $d(a,b) \geq 0$; \item $d(a,b)=0$ if and only if $a=b$; \item $d(a,b) = d(b,a)$; and \item $d(a,b) + d(b,c) \geq d(a,c)$,. \ee The last property is called the triangle inequality. \end{dfn} A space with a given metric has a very natural topology. In fact, we have used the concept of distance in $\R^1$ and $\R^n$ already to define the standard topologies on these spaces. Let us give the general process. \begin{thm} Let $d$ be a metric on the set $X$. Then the collection of all open balls $B(p,\epsilon)=\{ y\in X | d(p,y)<\epsilon \}$ for every $p\in X$ and every $\epsilon>0$ forms a basis for a topology on X. The topology for which it is a basis is called the \emph{$d$-metric topology\/} for $X$. \end{thm} \begin{dfn}[metric space]\index{metric space} A topological space $(X, \cal T)$ is a \emph{ metric space} if and only if there is a metric $d$ on $X$ such that $\cal T$ is the $d$-metric topology. We will sometimes write a metric space as $(X,d)$ to stand for $X$ with the $d$-metric topology. \end{dfn} \begin{example} The following are metric spaces (prove this!):\\ a) $\R^1$ with Euclidean metric $d(x,y) = | x-y |$. \\ b) $\R^2$ with Euclidean metric $d((x_1,y_1),(x_2,y_2)) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} $. \\ c) $\R^2$ with the ``taxi-cab'' metric: $d((x_1,y_1),(x_2,y_2)) = | x_1-x_2 | + | y_1-y_2| $. \\ d) Any set M, with the metric: $d(a,a)=0$, and $d(a,b)=1$ if $a \neq b$.\\ e) $\Q$ with the metric: $d(\frac{a}{b}, \frac{m}{n}) = max(|a-m|, |b-n|)$. (All fractions reduced.)\\ %%hawaiian earing and bouquet of circles? \end{example} The same topology on a set may be generated by more than one metric. For example, show that the taxi-cab metric on $\R^2$ generates the standard topology on $\R^2$ just as the standard metric on $\R^2$ does. \begin{thm} If $X$ is a metric space and $Y \subset X$, then $Y$ is a metric space. \end{thm} \begin{thm} If $(X,\cal T)$ is a metric space, then there is a metric $d$ that generates $\cal T$ such that for each $x, y \in X$, $d(x,y) < 1$. \end{thm} %\begin{example} %The following aren't metric spaces:\\ %a) $\R$ with metric $d(x,y) = | xy | $\\ %b) $\N$ with metric %\[ d(a,b) = %\left\{ %\begin{array}{ll} %\frac{|a-b|}{2}, & \mbox{if $a,b$ both odd}\\ %|a-b|, & \mbox{otherwise} %\end{array} \right. . \] %\end{example} \begin{thm} If $M$ is a metric space, then $M$ is Hausdorff, regular, and normal. \end{thm} \begin{thm} A separable metric space is second countable. \end{thm} %\begin{thm} %\label{count.prop.metric} %In a metric space $X$, the following are %equivalent: %\begin{enumerate} %\item $X$ is $2^{\textrm{nd}}$ countable, %\item $X$ is separable, %\item $X$ is Lindel\"of, %\item every uncountable set in $X$ has a limit point. %\end{enumerate} %\end{thm} %\begin{thm} %If a metric space is countably compact, it is compact. %\end{thm} \begin{LebegueNumThm} Let $\{ U_\alpha\}_{\alpha\in\lambda}$ be an open cover of a compact set $A$ in a metric space $X$. Then there exists a $\delta >0$ such that for every point $p$ in $A$, $B(p,\delta) \subseteq U_\alpha$ for some $\alpha$. \end{LebegueNumThm} A $\delta$ satisfying the theorem above is called a \emph{Lebesgue number\/}. Now you can finish the exercise that was assigned at the beginning of the chapter. \begin{exercise} Make a grid with all our examples of topologies down the side and all separation, countability, and covering properties across the top. Fill in squares indicating which examples have what properties. \end{exercise} %\section{*Further Countability Properties} %\begin{dfn}[Souslin property]\index{Souslin property}\index{topological space!Souslin property} %A space $X$ has the \emph{Souslin property\/} if and only if %$X$ does \emph{not\/} contain an uncountable %collection of disjoint open sets. %\end{dfn} %\begin{thm} %A separable space has the Souslin property. %\end{thm} %Now that you know the definition of the Souslin property, you %extend Theorem~\ref{count.prop.metric}: %\begin{thm} %In a metric space $X$, the following are %equivalent: %\begin{enumerate} %\item $X$ is $2^{\textrm{nd}}$ countable, %\item $X$ is separable, %\item $X$ is Lindel\"of, %\item every uncountable set in $X$ has a limit point, %\item $X$ has the Souslin property, %\end{enumerate} %\end{thm} %\section{*Further Covering Properties} %The following uses the concept of a subbasis, as described in (optional) %Section~\ref{subbasis}: %\begin{ASthm} %Let $\mathcal S$ be a subbasis for a space $X$. Then $X$ is compact if and %only if every subbasic open cover has a finite subcover. (A subbasic %open cover is a cover of $X$ each element of which is a set in the %subbasis.) %\end{ASthm} %\begin{dfn}[locally finite]\index{locally finite} %A collection ${\cal B} = \{ B_\alpha\}_{\alpha\in\lambda}$ %of subsets of a space $X$ is \emph{locally finite\/} if and only if %for each point %$p$ in $X$ there is an open set $U$ containing $p$ such that $U$ %intersects only finitely many elements of $\cal B$. %\end{dfn} %\begin{thm} %Let ${\mathcal B}= \{ B_\alpha\}_{\alpha\in\lambda}$ %be a locally finite collection of subsets of a space $X$. Let $C$ be %a subset of $\lambda$. Then $\cl (\bigcup_{\alpha\in C} B_\alpha) = %\bigcup_{\alpha\in C} \overline{B}_\alpha$. %\end{thm} %\begin{example} %Let ${\cal B} = \{ [n,n+1] \subseteq \R\mid n$ is an %integer$\}$. Then $\cal B$ is a locally finite collection in $\R_\textrm{std}$. %\end{example} %\begin{dfn}[refinement of a cover]\index{refinement of a cover}\index{cover!refinement} %Let ${\cal B} = \{ B_\alpha\}_{\alpha\in\lambda}$ %be a cover of $X$. Then ${\cal C} = \{ C_\beta\}_{\beta\in\mu}$ is a %\emph{refinement\/} of $\cal B$ if and only if (i) $\cal C$ is a cover of $X$ %and (ii) for each $\beta\in\mu$ there is an $\alpha\in\lambda$ such that %$C_\beta \subseteq B_\alpha$. The collection $\cal C$ is an %\emph{open refinement\/} if and only if each $C_\beta$ is an open set. %\end{dfn} %\begin{dfn}[paracompact]\index{paracompact}\index{topological space!paracompact} % A space $X$ is \emph{paracompact\/} if and only if %every open cover %of $X$ has a locally finite open refinement and $X$ is Hausdorff. %\end{dfn} %\begin{thm} %Every compact, Hausdorff space is paracompact. %\end{thm} %\begin{thm} %Let $A$ be a closed subspace of a paracompact space. Then $A$ %is paracompact. %\end{thm} %\begin{thm} % A paracompact space is normal. %\end{thm} %\begin{thm} %A regular, $T_1$, Lindel\"of space is paracompact. %\end{thm} %\begin{thm} %A metric space is paracompact. %\end{thm} %\section{*Properties on the ordinals} %\begin{thm} $\omega_1$ is countably compact but not compact. %\end{thm} %\begin{thm} $\omega_1+1$ is compact. %\end{thm} %\nibf{Acknowledgements.} %\vspace*{2ex} %\medskip %\bibliographystyle{amsplain} %\begin{thebibliography}{99} %\bibitem{BS} %C.~Bankwitz and H.G.~Schumann, \emph{$\ddot{U}$ber Viergeflechte}, Abh. Math. % Sem. Univ. Hamburg \textbf{10} (1934), 263--284. %\bibitem{Ber} %J.~Berge, \emph{Some knots with surgeries yielding lens spaces}, %unpublished manuscript. %\end{thebibliography} %\newpage \printindex \noindent Michael Starbird\\ {\texttt{Department of Mathematics\\ RLM 8.100\\ The University of Texas at Austin\\ Austin, TX 78712}} \noindent{\texttt{starbird@math.utexas.edu}} \vspace*{5ex} \noindent Cynthia Verjovsky Marcotte\\ {\texttt{Department of Mathematics\\ St.\ Edward's University\\ 3001 South Congress Ave.\\ Austin, TX 78704}} \noindent{\texttt{cynthm@admin.stedwards.edu}} \end{document}