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\begin{document}
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\section*{IV.Normal Space}
\subsection* {1. Normal space }

\begin{defn}
A topological space $X$ is  \textbf{normal($T_4$)} if it is
Haussdorff and for
any two disjoint closed sets of $X$, there are separating open neighborhoods, i.e.,
for any disjoint closed subsets $C_1$ and $C_2$ of $X$, there
exist open sets $O_1$ and $O_2$ s.t. $C_1\subset O_1$ and $C_2\subset O_2$
and $O_1\cap O_2=\phi$.
\end{defn}

\textbf{Recall}    
$X$ is regular if there are separating open neighborhoods for a point and a disjoint
closed set.

\begin{ex}
A compact Hausdorff space is normal.
\end{ex}

\begin{proof}
A closed subset of compact space is compact. We can separate two
disjoint compact subsets of a Hausdorff space.
\end{proof}

\begin{ex}
A metric space $X$ is normal.
\end{ex}

\begin{proof}
For given disjoint closed subsets $C_1$ and $C_2$ in $X$, let\\
$O_1:=\{x\in X|d(x,C_1)<d(x,C_2)\}$ and
$O_2:=\{x\in X|d(x,C_1)>d(x,C_2)\}$.\\
Then clearly $O_1\cap O_2=\phi$ and $C_1\subset O_1, C_2\subset
O_2$. Now let's show that $O_1$ and $O_2$ are open.
Since $g(x)=d(x,C_1)-d(x,C_2)$ is a continuous function and
$O_1=g^{-1}(-\infty,0)$ and
$O_2=g^{-1}(0,\infty)$. Hence a metric space is normal.
\end{proof}

\begin{prop}
$X$ is normal. \\
$\Leftrightarrow \forall$ closed $A\subset X$ and open $U\supset A$,
there exists open $V$ s.t. $A\subset V\subset\overline{V}\subset U$.
\end{prop}
\begin{proof}
($\Rightarrow$)$U^c$ is closed.\\
$U^c$ and $A$ are two disjoint closed sets.\\
$\Rightarrow \exists$ disjoint open sets V and W s.t. $A\subset V$ and $U^c\subset W$.\\
$A\subset V\subset W^c$ and $W^c$ is closed\\
$\Rightarrow\overline{V}\subset W^c\\
\Rightarrow \overline{V}\cap W=\phi$.\\
Since $U^c\subset W$, $\overline{V}\cap U^c=\phi$ and then $\overline{V}\subset U$.\\
($\Leftarrow$)Let $C$ and $D$ be two disjoint closed sets.
Then $U:=D^c$ is an open set containing $C$. Hence there exists open $V$ s.t. $A\subset V\subset\overline{V}\subset U$. Also $V\cap \overline{V}^c=\phi$.
Then $V$ and $\overline{V}^c$ are two
seperating open neighborhoods of $C$ and $D$.
\end{proof}\\

Note that similar statement also holds for a regular space.\\

\textbf{Homework}. Prove the followings.\\
(a) A subspace of a regular space is regular.\\
(b) A product of regular spaces is regular.

\begin{rem}
A product of normal spaces need not be normal: It is not difficult to show that
$\re_l$ is normal (and hence regular). But $\re_l\times\re_l$(=Sorgenfrey plane) is not normal.
(See Munkres p198.) But notice that $\re_l\times\re_l$ is regular.\\
(regular + 2nd countable $\Rightarrow$ normal. See p200.)
\end{rem}


\begin{rem}
If $J$ is uncountable, the product space $\re^J$ is not normal. (A difficult exercise: p206)
\end{rem}

This example serves three purposes. Firstly, it shows that a regular
space $\re^J$ need not be normal. Secondly, it shows that a subspace
of a normal space need not be normal, for $\re^J\cong (0,1)^J\subset
[0,1]^J$. We can easily see that $[0,1]^J$ is normal since it is
compact by Tychonoff theorem and Hausdorff. (It is easy to show that a closed subset of normal space is normal.) Lastly, it shows that
a product of normal space need not be normal.
\end{document}