\magnification 2400
\parindent 0pt
\parskip 12pt
\hsize 7.2truein
\vsize 9.7 truein
\hoffset -0.45truein
\def\u{\vskip 0pt}
\def\w{\vskip -10pt}
\def\v{\vskip 0.5in}


Cor 29.3:  If $X$ is locally compact and if $A$ is closed in $X$, then $A$ is locally compact.

Cor 29.3:  If $X$ is locally compact Hausdorff and if $A$ is open in $X$, then $A$ is locally compact.

Cor 29.4:  $X$ is homeomorphic to an open subspace of a compact Hausdorff space if and only is $X$ is locally compact Hausdorff

\hrule

Thm 30.3b:  2nd countable implies separable.

Idea of proof:

Lemma:  If $X$ is 2nd countable, $A \subset X$, and the subspace topology on $A$ is the discrete
topology, then $A$ is countable.

Idea of proof:
\vfill

Thm 30.2a.  A subspace of a second countable space is second countable.  A countable product of second
countable spaces is second countable.
\eject

Thm 30.2b.  A subspace of a first countable space is first countable.  A countable product of first countable spaces is first countable.


Metric spaces are 1st countable.

$R^\omega$, product:

$R^\omega$, uniform:

$R^\omega$, box:



Defn:  $X$ is countably compact if $X \subset \cup_{n=1}^\infty U_n$, $U_n$ open implies there exists $n_1, ..., n_k$ such that 
 $X \subset \cup_{i=1}^k U_{n_i}$.


Defn:  $X$ is Lindelof if $X \subset \cup_{a \in A} U_a$, $U_a$ open implies there exists $a_i \in A$ such that 
 $X \subset \cup_{i=1}^\infty U_{a_i}$.

2nd countable does not imply countably compact.

Thm 30.3a:  2nd countable implies Lindelof.

\eject
${\cal R}_{\cal L}$ =  ${\cal R}$ with the lower limit topology is

Separable:

1st countable:

Not 2nd countable:

\vskip 0pt
Lindelof:

Step 1:  Every open covering has a countable subcover iff every open covering of basis elements 
has a countable subcover



\vfill
${\cal R}_{\cal L} \times {\cal R}_{\cal L}$ is not Lindelof.

Hence the product of Lindelof spaces need not be Lindelof.


HW:  A subspace a of Lindelof space need not be Lindelof. 

Hint:  The ordered square $= [0, 1] \times [0, 1]$ with the order topology using the dictionary order.

\eject

31 The separation axioms.
\u
Defn:  $X$ is $T_0$  if $x, y \in X$ $x \not= y$, then there exists an open set $U$ such that $x \in 
U$, $y \not\in U$ OR $y \in U$, $x \not\in U$.
\u
Ex:  $\{(a, \infty) ~|~ a \in {\cal R}\}$

Defn:  $X$ is $T_1$  if $x, y \in X$ $x \not= y$, then there exists open sets $U$, $V$ such that $x 
\in U$, $y \not\in U$ and $y \in V$, $x \not\in V$.
\u
Thm:  $X$ is $T_1$ iff points are closed in $X$.
\u
Ex:  finite complement topology

Defn:  $X$ is Hausdorff  if $x, y \in X$ $x \not= y$, then there exists disjoint open sets $U$, $V$ 
such that $x \in U$ and $y \in V$.
\u
Ex:  ${\cal R}_K$ where ${\cal R}_K$ has basis \hfil\break
$\{(a, b) ~|~ a, b \in {\cal R}, a < b \}
\cup \{(a, b) - K ~|~ a, b \in {\cal R}, a < b \}$ where $K = \{ {1 \over n} ~|~ n = 1, 2, 3, ...
\}$

Defn:  $X$ is regular if $X$ is $T_1$ and if  $x \in X$, $C$  closed in $X$ and $x \not\in C$, then there exists disjoint open sets $U$, $V$ such that $x \in U$ and $C \subset V$.
\u
Ex:  ${\cal R}_{\cal L} \times {\cal R}_{\cal L}$

Defn:  $X$ is normal if $X$ is $T_1$ and if  $C, D$ disjoint closed sets in $X$, then there exists disjoint open sets $U$, $V$ such that $C \subset U$ and $D \subset V$.

Ex: ${\cal R}_{\cal L}$

Lemma 31.1:  Let $X$ be $T_1$.
\w
a.)  $X$ is regular iff for all $x\in X$ and for all neighborhoods $U$ of $x$, there exists neighborhood $V$ of $x$ such that $\overline{V} \subset U$.
\w
b.)  $X$ is normal iff for all $C$ closed in $X$ and for all open set $U$ containing $C$, there exists an open set $V$ containing  $C$ such that $\overline{V} \subset U$.


Lemma 31.2:  A subspace of a $T_i$ space is $T_i$ for $i = 0, 1, 2, 3$, but not $i = 4$
\w
A product  of  $T_i$ spaces is $T_i$ for $i = 0, 1, 2, 3$, but not $i = 4$


32:  Normal Spaces:

Thm 32.1:  Every regular space with a countable basis is normal.

Thm 32.2:  Every metrizable space is normal.

Thm 32.3:  Every compact Hausdorff space is normal.

Thm 32.4:  If $X$ has the order topology then $X$ is normal.

\end

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