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\title{18.901 Problem Set 3}
\author{Chris Laas}
\date{April 14, 2000}

\begin{document}
\maketitle




\section*{Problem 1: 26.8}

\subsection*{$f$ continuous implies $G_f$ closed}

First let us assume $f$ is continuous, and prove that
$X \times Y - G_f$ is open.

Take any point $(x,y) \in X \x Y - G_f$.  Then $y \neq f(x)$.  Thus,
there exist disjoint neighborhoods $U$ of $y$ and $V$ of $f(x)$, since
$Y$ is Hausdorff.

Thus, $f^{-1}(V)$ is open, and hence $B = f^{-1}(V) \x U$ is a basis
element of $X \x Y$.

Since $f(x) \in V$ and $y \in U$, $(x,y) \in B$.

Now, assume there exists some point $(x',f(x')) \in G_f$ such that
$(x',f(x')) \in B$.  Then, $x' \in f^{-1}(V)$, and thus $f(x') \in V$.
But we also have $f(x') \in U$, so $f(x') \in U \cap V$.  But this is
a contradiction, since U and V are disjoint.  Thus, $B$ does not
intersect $G_f$.

Thus, for any point $(x,y) \in X \x Y - G_f$, there exists a basis
element $B$ such that $(x,y) \in B \subset X \x Y - G_f$.  Thus,
$X \x Y - G_f$ is open.  Thus, $G_f$ is closed.


\subsection*{$G_f$ closed implies $f$ continuous}

Now let us assume that $G_f$ is closed, and prove that $f$ is continuous.

Take any $x_0 \in X$ and any neighborhood $V$ of $f(x_0)$.

Since $V$ is open, $Y-V$ is closed; hence $X \x (Y-V)$ is closed.
Thus, $G_f \cap (X \x (Y-V))$ is closed.

By the result of problem 7, since $Y$ is compact, the projection
$\pi_1: X \x Y \to X$ is a closed map.  Thus, since
$G_f \cap (X \x (Y-V))$ is closed,
$A = \pi_1(G_f \cap (X \x (Y-V)))$ is closed.  $A$ is the set of all
$x \in X$ such that $f(x) \notin V$.

Let $U = X - A$.
Then, since $U$ is the complement of a closed set, $U$ is open in $X$.
Since $f(x_0) \in V$, $x_0 \in U$.  And, since $U$ is the set of all
$x \in X$ such that $f(x) \in V$, $f(U) \subset V$.

Hence, for any $x_0 \in X$ and any neighborhood $V$ of $f(x_0)$, there
is a neighborhood $U$ of $x_0$ such that $f(U) \subset V$.
Thus, $f$ is continuous.





\section*{Problem 2:  26.9}

Let $A$ and $B$ by subspaces of $X$ and $Y$, respectively; let $N$ be
an open set in $X \x Y$ containing $A \x B$.  Also let $A$ and $B$ be
compact.

\subsection*{Lemma:
For any $x \in A$, there exist open sets $U_x \subset X$ and
$V_x \subset Y$ such that $x \x B \subset U_x \x V_x$
and $U_x \x V_x \subset N$.
}

Since $x \x B \subset N$ and $N$ is open, there exists an open
covering of $x \x B$ by basis elements $U \x V$ lying in $N$.
And since $x \x B$ is compact (by homeomorphism to the compact space
$B$), there is a finite subcovering $\{ U_i \x V_i \}$.

Now, let us define $U_x = U_1 \cap \cdots \cap U_n$ and
$V_x = V_1 \cup \cdots \cup V_n$; $U_x$ and $V_x$ are clearly open.

Then, clearly $x \in U_x$ and $B \subset V_x$, and so
$x \x B \subset U_x \x V_x$.

Now, to show $U_x \x V_x \subset N$, take any $(x',y') \in U_x \x V_x$.
Clearly $y' \in V_i$ for some $V_i$, and since $x' \in U_j$ for all
$U_j$, $(x',y') \in U_i \x V_i$ for some $U_i \x V_i$.  Since all
$U_i \x V_i$ were defined to be subsets of $N$, $(x',y') \in N$.
Thus, $U_x \x V_x \subset N$.

This completes the proof of the lemma.

\subsection*{Theorem:
There exist open sets $U$ and $V$ of $X$ and $Y$, respectively, such
that $A \x B \subset U \x V \subset N$.
}

Now, using the lemma, we can construct the collection of open sets
of the form $U_x \x V_x$ defined by the lemma for all $x \in A$.
Since, for each $U_x \x V_x$, $x \x B \subset U_x \x V_x$, this is an
open covering of $A \x B$.

\def\Ux#1{U_{x_#1}}
\def\Vx#1{V_{x_#1}}

Since $A \x B$, as the product of two compact spaces, is compact,
there is a finite subcovering $\{ \Ux{i} \x \Vx{i} \}$.

Let us define $U = \Ux{1} \cup \cdots \cup \Ux{n}$ and
$V = \Vx{1} \cap \cdots \cap \Vx{n}$.  Clearly $U$ and $V$ are open in
$X$ and $Y$ respectively.  Since $\{ \Ux{i} \x \Vx{i} \}$ covers
$A \x B$, it's clear that $A \subset U$; and, since $B \subset \Vx{i}$
for all $\Vx{i}$, $B \subset V$.  Thus, $A \x B \subset U \x V$.

Lastly, to show that $U \x V \subset N$, take any $(x',y') \in U \x V$.
Then $x'$ is in some $\Ux{i}$, and since $y' \in V$, $y' \in \Vx{i}$.
Thus $(x',y')$ is in $\Ux{i} \x \Vx{i}$, and since all $\Ux{i} \x \Vx{i}$
are subset of $N$, $(x',y') \in N$.  Thus, $U \x V \subset N$.

Therefore, $U$ and $V$ are open in $X$ and $Y$, respectively, and
$A \x B \subset U \x V \subset N$.  This completes the proof.







\section*{Problem 3:  28.4}

Let $X$ be a $T_1$ space.

\subsection*{Limit point compactness implies countable compactness}

Assume $X$ is limit point compact.

For purposes of contradiction, assume $X$ is not countably compact.
Then there exists a countable open covering $\{U_n\}$ such that no
finite subcollection covers $X$.

Then (using the axiom of choice) we may choose
$x_n \notin U_1 \cup \cdots \cup U_n$ for each $n \in \Z_+$; such an
$x_n$ exists because $U_1, \dots, U_n$ does not cover $X$.

Then, since $X$ is limit point compact, and $\{x_n\}$ is an infinite
subset of $X$, $\{x_n\}$ has a limit point in $X$; call it $x$.

Now, since $\{U_n\}$ is an open covering of $X$, $x$ is in some $U_i$.
But since $X$ is a $T_1$ space, $x$ is a limit point of $\{x_n\}$, and
$U_i$ is a neighborhood of $x$, $U_i$ contains infinitely many $x_n$.
However, this is a contradiction, since by the construction of
$\{x_n\}$, only finitely many (up to $i$) $x_n$ can be elements of any
$U_i$.

Thus, if $X$ is $T_1$ and limit point compact, then $X$ is countably
compact.

\subsection*{Countable compactness implies limit point compactness}

Assume $X$ is countably compact.

For purposes of contradiction, assume $X$ is not limit point compact.
Then there exists an infinite subset $A \subset X$ with no limit
point.  Since $A$ is infinite, it contains a countable subset
$\{x_n\}$, which also has no limit point.

Since none of the $\{x_n\}$ are limits points of $\{x_n\}$, and $X$ is
a $T_1$ space, for each of the $x_n$ there exists a neighborhood $U_n$
which contains only finitely many of the $\{x_n\}$.

Also, since $\{x_n\}$ has no limit points, it is closed; thus
$U_0 = X - \{x_n\}$ is open.

Hence, the collection consisting of $U_0$ and all $U_n$ is a countable
open covering of $X$, since all $x_n \in U_n$ and all
$x \notin \{x_n\}$ are in $U_0$.

Thus, since $X$ is countably compact, there is a finite subcovering of
$X$.  However, this is a contradiction: because $U_0$ contains none of
the $\{x_n\}$, and each $U_n$ contains only finitely many $\{x_n\}$,
any finite union of sets in the collection can contain only finitely
many $\{x_n\}$; thus, since $\{x_n\}$ is infinite, there must be some
$x_i \in X$ which is not in any set in the finite subcollection.  This
contradicts the existence of a finite subcovering.

Thus, if $X$ is $T_1$ and countably compact, then $X$ is limit point
compact.








\section*{Problem 4: 30.5}
\begin{enumerate}

\item%(a)

Let $X$ be a space with metric $d: X \x X \to \R$ and a countable
dense subset $A$.

Then let the collection $\B$ be the collection of all $d$-balls
$B\(a,\frac{1}{n}\)$ with $a \in A$ and $n \in \Z_+$.

Take any $d$-ball $B(x,\epsilon)$ for $x \in X$ and $\epsilon \in \R$.
Then, since $A$ is dense in $X$, there exists
$a \in A \cap B(x,\epsilon)$.  Thus, if we pick $n \in \Z_+$ such that
$\frac{1}{n} < \epsilon - d(x,a)$, the $d$-ball
$B\(a,\frac{1}{n}\) \in \B$ is a subset of $B(x,\epsilon)$.

Thus $\B$ is a basis of $X$.

Since $\B$ is a countable union of countable sets
(i.e. $\B = \bigcup_{a \in A} \{ B(a,\frac{1}{n}) \}$)
$\B$ is countable.

Thus $\B$ is a countable basis of $X$.


\item%(b)

Let $X$ be a Lindel\"of space with metric $d: X \x X \to \R$.

For every $n \in \Z_+$, let $\Cn$ be the collection of all
$d$-balls $B\(x,\frac{1}{n}\)$ for $x \in X$.  Then, since
$\Cn$ is an open covering of $X$, and $X$ is Lindel\"of,
$\Cn$ contains a countable subcovering $\B_n$.

Let $\B = \bigcup \B_n$.  Then, since it is a countable union of
countable collections, $\B$ is a countable collection.

Take any $d$-ball $B(x,\epsilon)$: let $n \in \Z_+$ be such that
$\frac{1}{n} < \frac{\epsilon}{2}$.  Then, since $\B_n$ covers $X$,
$x \in B\(y,\frac{1}{n}\) \in \B_n \subset \B$ for some $y$.  And,
since $d(x,y) < \frac{1}{n} < \frac{\epsilon}{2}$,
$B\(y,\frac{1}{n}\) \subset B(x,\epsilon)$.

Thus, $\B$ is a countable basis of $X$.



\end{enumerate}







\section*{Problem 5: 30.6}

(I heard a rumor this was due in addition to the other problems.  If
it's not, feel free to ignore this section.)

If $\R_\ell$ were metrizable, then, since $\R_\ell$ is Lindel\"of,
$\R_\ell$ would have a countable basis (by the result of the previous
problem).  But since we know $\R_\ell$ is not second countable, this
is a contradiction.  Thus, $\R_\ell$ is not metrizable.

Since $I_o^2$ is compact, it is Lindel\"of.  Thus, if it were
metrizable, it would have a countable basis $\B$.  Then choose, for
each $x \in [0,1]$, and element $B_x$ of $\B$ such that
$x \x 0 \in B_x \subset \{x\} \x \[0,\frac{1}{2}\)$.  If $x \neq y$,
then $B_x \neq B_y$, since $x \x 0 = \inf B_x$ and $y \x 0 = \inf B_y$.
Therefore, since $[0,1]$ is uncountable, $\B$ must be uncountable,
which is a contradiction.  Thus, $I_o^2$ is not metrizable.




\end{document}

