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Defn  $A$ is {\it dense} in $X$ is $\overline{A} = X$.  
$X$ is {\it separable} is separable if it has a countable dense subset.

Ex:
 

Thm 30.3b:  2nd countable implies separable.

Proof:  Let $\{B_n ~|~ n \in {\bf N} \}$ be a countable basis for $X$.  Let $A = $

\v

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 
 $X \subset \cup_{i=1}^\infty U_{a_i}$.


Thm 30.3a:  2nd countable implies Lindelof.

Proof:  Let $\{B_n ~|~ n \in {\bf N} \}$ be a countable basis for $X$.  Suppose $X \subset \cup_{a \in A} U_a$, $U_a$ open 

\v

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

Separable:

1st countable:  

Not 2nd countable.

Lindelof.

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



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

Hence the product of Lindelof spaces need not be Lindelof.


A subspace a of Lindelof space need not be Lindelof. 

Ex:  The ordered square $= [0, 1] \times [0, 1]$ with the order topology 
using the dictionary order is Lindelof, but  the subspace $[0, 1] \times 
(0, 1)$ is not.
\end
\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