\documentclass[11pt]{article}

\renewcommand{\labelenumi}{(\alph{enumi})}
\renewcommand{\labelenumii}{\arabic{enumii}}

\usepackage{amsmath}
\usepackage{amsfonts}
\DeclareMathSymbol{\C}{\mathalpha}{AMSb}{"43}
\DeclareMathSymbol{\Q}{\mathalpha}{AMSb}{"51}
\DeclareMathSymbol{\R}{\mathalpha}{AMSb}{"52}
\DeclareMathSymbol{\Z}{\mathalpha}{AMSb}{"5A}

\def\choice#1{\left\{ \begin{array}{ll} #1 \end{array} \right.}
\def\({\left(}
\def\){\right)}
\def\[{\left[}
\def\]{\right]}

\def\x{\times}
\def\B{{\cal B}}
\def\Cn{{\cal C}_n}



\title{18.901 Problem Set 4}
\author{Chris Laas}
\date{May 5, 2000}

\begin{document}
\maketitle




\section*{Problem 1}

\subsection*{32.4}

This proof follows the lines of Theorem 32.1 very closely; only the
first paragraph changes, to this:

Let $X$ be a regular Lindel\"of space.  Let $A$ and $B$ be disjoint
closed subsets of $X$.  Each point $x$ of $A$ has a neighborhood $U$
not intersection $B$.  Using regularity, choose a neighborhood $V$ of
$x$ whose closure lies in $U$.  By choosing such a neighborhood for
each $x$ in $A$, we constructa covering of $A$ by open sets whose
closures do not intersect $B$.  Since $X$ is Lindel\"of, there exists
a countable subcovering, which we denote by $\{U_n\}$.

The rest of the proof is identical to pages 200-201; we have then
proven that all regular Lindel\"of spaces are normal.


\subsection*{34.8}

Let $X$ be a regular Lindel\"of space that is locally metrizable.  For
every point $x$ of $X$ there is a neighborhood $U$ of $x$ which is
metrizable in the subspace topology.  Since $X$ is regular, there is a
neighborhood $V$ of $x$ such that $\bar{V} \subset U$.  Since closed
subspaces of Lindel\"of spaces are Lindel\"of, $\bar{V}$ is
Lindel\"of, and since subspaces of metric spaces are metrizable,
$\bar{V}$ is metrizable.  Thus, by the result of problem 30.5b,
$\bar{V}$ has a countable basis; thus, $V$ has a countable basis,
since $V$ is a subspace of $\bar{V}$.

Thus, for every point $x$ of $X$, we can construct a neighborhood $V$
of $x$ with a countable basis.  The collection of all these
neighborhoods is an open covering of $X$, and since $X$ is Lindel\"of,
there exists a countable subcovering of $X$, $\{V_n\}$.  Thus, since
$X = \bigcup V_n$, $X$ is a countable union of open subspaces, each of
which has a countable basis; thus, $X$ has a countable basis.

Since $X$ is regular and has a countable basis, $X$ is metrizable, by
the Urysohn metrization theorem.







\section*{Problem 2: 33.4}

Let $X$ be normal and let $A$ be a subset of $X$.

\subsection*{Existence of $f$ implies $A$ is closed $G_\delta$}

If there exists a continuous function $f: X \to [0,1]$ such that $f(x)
= 0$ for $x \in A$, and $f(x) > 0$ for $x \notin A$, then
$A = f^{-1}(\{0\})$.  Since $f$ is continuous and $\{0\}$ is closed in
$[0,1]$, $A$ is closed in $X$.  Let
$U_n = f^{-1}\(\[0,\frac{1}{n}\)\)$ for all $n \in \Z_+$; since
$\[0,\frac{1}{n}\)$ is open in $[0,1]$, $U_n$ is open in $X$.  If
$x \in A$, then $f(x) = 0 \in \[0,\frac{1}{n}\)$ for all $n$; thus,
$x \in \bigcap U_n$; and thus, $A \subset \bigcap U_n$.  Conversely,
if $x \subset \bigcap U_n$, then $f(x) \in \[0,\frac{1}{n}\)$ for all
$n$, and thus $f(x) = 0$.  Thus, $x \in A$; therefore, $\bigcap U_n = A$.
Thus, $A = \bigcap U_n$, i.e. $A$ is the intersection of a countable
collection of open sets of $X$, and thus $A$ is a closed $G_\delta$
set in $X$.

\subsection*{Lemma}



\subsection*{$A$ is closed $G_\delta$ implies existence of $f$ }

Conversely, let us assume that $A$ is a closed $G_\delta$ set in $X$.
Then $A = \bigcap U_n$ for a countable collection $\{U_n\}$ of open
sets of $X$.  Then, for each $n$, $A$ and $X-U_n$ are disjoint closed
subsets of $X$.  Since $X$ is normal, there exists a continuous map
$f_n: X \to \[0,\frac{1}{n}\]$ such that $f_n(x) = 0$ for every $x$ in
$A$, and $f_n(x) = \frac{1}{n}$ for every $x$ in $X - U_n$, by the
Urysohn lemma.

Then, we can construct the function $f: X \to [0,1]$ by
$$ f(x) = \sup \{ f_n(x) \mbox{\ such that\ } n \in \Z_+ \} $$

It is trivial to show that $f(x) = 0$ for all $x \in A$:  if
$x \in A$, then $f_n(x) = 0$ for all $n$, and thus
$f(x) = \sup \{0\} = 0$.  Likewise, to show that $f(x) > 0$ for all
$x \notin A$: if $x \notin A$, then $x \notin U_n$ for some $n$; thus,
$x \in X - U_n$, and thus $f(x) \geq \frac{1}{n}$.

The difficult part is to prove that $f$ is continuous.
I'm not sure how to do this.

%  To this end,
%we will take any $x_0 \in X$ and any interval $(c,d)$ such that
%$c < f(x_0) < d$, and show that there exists an neighborhood $U$ of
%$x_0$ such that $f(U) \subset (c,d)$.  It suffices to consider the
%case $\frac{1}{n+1} < c < d < \frac{1}{n-1}$, since we can
%always shrink the interval to such a one without loss of generality.
%
%We will choose a real $e$ such that $f(x_0) < e < d$, and then let
%$U = f^{-1}((c,e))$.  Then, by the definition of $f$,
%$y \in U$ if and only if there exists $i$ such that $f_i(y) > c$ and,
%for all $j$, $f_j(y) < d$.  




\section*{Problem 3}

\subsection*{Lemma:  Let $X$ be a compact Hausdorff space.  $X$ is
metrizable if and only if $X$ has a countable basis.}

Since $X$ is compact Hausdorff, it is normal and Lindel\"of.  If $X$
has a countable basis, then $X$ is metrizable, by the Urysohn
metrization theorem.  If $X$ is metrizable, then, since $X$ is
Lindel\"of, $X$ has a countable basis, by the result of problem 30.5b.

\subsection*{Lemma:  If $X$ is locally compact Hausdorff, $X$ is regular.}

Take any point $x$ of $X$ and a neighborhood $U$ of $x$.  By theorem
29.2, there is a neighborhood $V$ of $x$ such that $\bar{V}$ is
compact and $\bar{V} \subset U$.  Thus, by lemma 31.1, $X$ is regular.

\subsection*{34.4a}

Since $X$ is locally compact Hausdorff, $X$ is regular.  Thus, since
$X$ is regular and has a countable basis, $X$ is metrizable, by the
Urysohn metrization theorem.

\subsection*{34.4b and 34.5a}

I couldn't prove these, nor find counterexamples.  Lots of scribbled
notes and facts though...

\subsection*{34.5b}

Since $Y$ is compact, $Y$ is Lindel\"of.  Thus, if $Y$ is metrizable,
$Y$ has a countable basis, by the results of problem 30.5b.  Thus $X$
has a countable basis, as a subspace of $Y$.




\end{document}

