\documentclass[10pt]{article}
\pagestyle{empty}
\usepackage{amsmath,amssymb,verbatim,graphicx}
\newlength{\up}\setlength{\up}{-\baselineskip}
\setlength{\textwidth}{6.5in} \setlength{\textheight}{9.5in}
\setlength{\topmargin}{-.75in} \setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\parindent10pt
\newcommand{\RR}{\mathbb{R}}
\newcommand{\va}{\bar{a}}
\newcommand{\vb}{\bar{b}}
\newcommand{\aut}{\mathrm{Aut}}
\newcommand{\inn}{\mathrm{Inn}}
\newcommand{\iso}{\mathrm{ISO}}
\newcommand{\ao}{\mathrm{AO}}
\newcommand{\oo}{\mathrm{O}}
\newcommand{\so}{\mathrm{SO}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\id}{\text{Id}}
\newcommand{\ggal}{\text{Gal}}
\begin{document}
\hfill Winter 2010
\begin{center} Ralph Hutchison's Topology Midterm Review\end{center}
~\\
{\bf \S 31 The Separation Axioms}
{\bf Definition.} Suppose that one-point sets are closed in $X$. Then $X$ is said to be
{\bf \em regular} if for each pair consisting of a point $x$ and a closed set $B$ disjoint from $x$,
there exist disjoint open sets containing $x$ and $B$, respectively. The space $X$ is said
to be {\em \bf normal} if for each pair $A$, $B$ of disjoint closed sets of $X$, there exist disjoint open sets containing $A$ and $B$ respectively.
{\bf Lemma 31.1}{\em Let $X$ be a topological space. Let one-point sets in $X$ be closed. (a) $X$
is regular if and only if given a point $x$ of $X$ and a neighborhood $U$ of $x$, there is a
neighborhood $V$ of $x$ such that $\bar V \subset U$. (b) $X$ is normal if and only if given a
closed set $A$ and an open set $U$ containing $A$, there is an open set $V$ containing $A$
such that $\bar V\subset U$.}
{\bf Theorem 31.2}{\em (a) A subspace of a Hausdorff space is Hausdorff. (b) A subspace of a regular
space is regular; a product of regular spaces is regular.}
{\bf \S 31 Example 1}{\em The space $\R_K$ is Hausdorff but not regular.}
{\bf \S 31 Example 2}{\em The space $\R_{\mathcal l}$ normal.}
{\bf \S 31 Example 3}{\em The Sorgenfrey plane $\R_{\mathcal l}^2$ is not normal.}
{\bf \S 32 Normal Spaces}
{\bf Theorem 32.1} {\em Every regular space with a countable basis is
normal.}
{\bf Theorem 32.2} {\em Every metrizable space is normal.}
{\bf Theorem 32.3} {\em Every compact Hausdorff space is normal.}
{\bf Theorem 32.4} {\em Every well-ordered set X is normal in the
order topology.}
{\bf \S 32 Example 1} {\em If J is uncountable,
the product space $\R^J$ is not normal}
{\bf \S 32 Example 2} {\em The product space $S_\Omega \times \bar S_\Omega$ is not normal.}
{\bf \S The Urysohn Lemma}
{\bf Theorem 33.1 (Urysohn lemma).} {\em Let $X$ be a normal space;
let $A$ and $B$ be disjoint closed subsets of $X$. Let $[a,b]$ be a
closed interval in the real line. Then there exists a continuous
map $f: X \rightarrow [a,b]$ such that $f(x)=a$ for every $x$ in
$A$, and $f(x)=b$ for every $x$ in $B$. }
{\bf \S 33 Definition.} if $A$ and $B$ are two subsets of the topological space $X$, and if
there is a continuous function $f: X \rightarrow [0,1]$ such that
$f(A) = \{0\}$ and $f(B) = \{1\}$, we say that $A$ and $B$
{\bf \em can be separated by a continuous function.}
{\bf \S 33 Defintion.} A space $X$ is {\em \bf completely regular} if one-point sets are closed
in $X$ and if for each point $x_0$ and each closed set $A$ not containing $x_0$, there is a
continuous function $f: X \rightarrow [0,1]$ such that $f(x_0)=1$ and $f(A) = \{0\}$.
{\bf Theorem 33.2} {\em A subspace of a completely regular space is completely regular. A product
of completely regular spaces is completely regular.}
{\bf \S 33 Example 1} The spaces $\R_{\mathcal l}^2$ and $S_\Omega \times \bar S_\Omega$ are completely regular but not normal.
\section{definitions recommended in class}
{\bf Definition.} Let $X$ and $Y$ be topological spaces. If $C$ is a compact subspace
of $X$ and $U$ is an open subset of $Y$, define $S(C,U) = \{f|f\in \mathcal C(X,Y) \mbox{ and } f(c)
\subset U\}.$ The sets $S(C,U)$ form a subbasis for a topology on $\mathcal C(X,Y)$ that is called
the {\em \bf compact-open topology}.
\end{document}