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\centerline{\bf More on second countable  spaces}
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\bigskip
This  is an addendum to Section VI.1 in which we describe how one can
find examples of spaces that are separable but not Lindel\"of,
Lindel\"of but not separable, and  both separable and Lindel\"of
but not second countable.  By the  results of Section VI.1 it follows 
that such spaces examples of the first two types are neither second 
countable nor homeomorphic to metric spaces, and examples of the third
type are likewise not homeomorphic to metric spaces.
\medskip
{\it Separability does not imply the Lindel\"of property. }
In one of the exercises there is an example of a separable Hausdorff
space $X$ which has a closed subspace $A$ that is both uncountable and
discrete.  We claim this space cannot be Lindel\"of.  If it were,
then the closed subspace $A$ would also be Lindel\"of by another of
the exercises.  Since the open covering of $A$ by one point subsets
does not have a countable subcovering, it follows that  $A$ and hence
$X$ cannot be Lindel\"of.\bull
\medskip
{\it The Lindel\"of property does not imply separability. }  Since
compact spaces are Lindel\"of, it suffices to find a compact space
that is not separable.  We can do this using Tychonoff's Theorem.
Specifically, let $X = \{0,1\}{}^A$ be an uncountable product of 
spaces homeomorphic
to $\{0,1\}$ with the discrete topology, and let $A$ be the indexing 
set for the family of spaces that are factors of  $X$.  It will
be convenient to assume that $|A| > 2^{\aleph_0}$.  Then $X$ is 
compact (by Tychonoff's Theorem) and Hausdorff (since this is true
for all products of Hausdorff spaces).  We claim that $X$ is not
separable.

The first thing to note is that if $Y$ is a separable space then
the cardinality of the set of continuous real valued functions
on $Y$ is precisely $2^{\aleph_0}$; the constant functions are
a subset with this cardinality, and we may show there are at most
$2^{\aleph_0}$ such functions as follows:  If $D$ is a countable
dense subset of $Y$, then a continuous real valued function is
completely determined by its restriction to $D$, and thus an
upper bound for the cardinality of the set of continuous real 
valued functions on $Y$ is given by the cardinality of the set
of all set-theoretic real valued functions on $D$.  But the
latter cardinality is 
$$\left(2^{\aleph_0}\right)^{\aleph_0}~~=~~
2^{\aleph_0}\,\cdot \,2^{\aleph_0} ~~=~~ 2^{\aleph_0}~.$$
Therefore, {\it if we can show that the cardinality of $X = 
\{0,1\}{}^A$ is greater than $2^{\aleph_0}$ it will follow 
that $X$ is not separable}.

Given $\alpha\in A$ let $f_\alpha$ be given by projection
onto the $\alpha$ factor of $\{0,1\}{}^A$ followed by the
inclusion of $\{0,1\}$ in ${\bf R}$.  Then $\alpha\neq
\beta$ implies $f_\alpha\neq f_\beta$ because their 
zero sets are different, and therefore we have a subset of
continuous real valued functions on $\{0,1\}{}^A$ that is in
1--1 correspondence with $A$.  Since $|A| > 2^{\aleph_0}$ it
follows that $X$ cannot have a countable dense subset.

One can also prove that $X$ is not separable if $|A| =
2^{\aleph_0}$, but the argument is significantly
more complicated.\bull
\medskip
{\it A separable Hausdorff space that also satisfies the Lindel\"of
property is not necessarily second countable. }
Every countable space is automatically separable and Lindel\"of
(the space is its own countable dense subset, and given an open
covering one can extract a countable subcovering indexed by the space
itself --- for each $x$ let $U_x$ be an open subset in the covering
that contains $x$).  We also know that second countable implies
first countable.  Therefore it will suffice to construct a countable
Hausdorff space that is not first countable.

Define an equivalence relation on the space ${\bf Q}^2$ of points
in ${\bf R}^2$ with rational coordinates such that the equivalence
classes are all one point sets $\{(x,y)\}$, where $x$ and $y$ are 
rational with $y\neq 0$, together with the rational points on the
$x$-axis.  This space $QPP$ is sometimes known as the {\it rational 
pinched plane}, and by construction it is countable.  We need to show 
that this space is Hausdorff but not first countable.

To verify that the space is Hausdorff, suppose we have two distinct
points in $QPP$.  It follows immediately that quotient space 
projection defines a homeomorphism from ${\bf Q}\times\left(
{\bf Q}- \{0\}\right)$ to the open subset $QPP-\{L\}$,
where $L$ is the equivalence class given by the $x$-axis.  
Therefore the open subset $QPP-\{L\}$ is Hausdorff, and there
are disjoint open neighborhoods for each pair of distinct
points in $QPP-\{L\}$.  It remains to consider the case where
one distinct point is $L$ and the other comes from $(x,y)$ where
$y\neq 0$.  By the definition of the quotient topology, these two
points have disjoint open neighborhoods if and only if there are
disjoint open subsets of ${\bf Q}^2$ containing $(x,y)$ and 
${\bf Q}\times \{0\}$.  But this is true because ${\bf Q}^2$ is
metrizable and hence ${\bf T}_{\bf 3}$.  

Finally we need to show that $QPP$ is not first countable; in 
fact, we claim that there is no sequence of open sets $\{\,U_n\,\}$
in $QPP$ such that each one contains $L$ and every open
subset $W$ containing $L$ contains some $U_n$.  Let $\pi$
be the quotient space projection onto $QPP$; by the definition of
the quotient topology the open neighborhoods of $L$ in $QPP$
are precisely the sets of the form $\pi(V)$ where $V$ is an open
subset of ${\bf Q}^2$ containing the $x$-axis.  Therefore it will
suffice to show {\it there is an open subset $W_0\subset {\bf Q}^2$
containing the $x$-axis such that $\pi(W_0)$ does not contain
{\bf any} of the open sets $U_n$}.  Since $U_n$ is the image of
$\pi^{-1}(U_n)$ under $\pi$, it will suffice to show that one
can find an open subset $W_0$ such that $W_0$ does not contain
any of the open sets $\pi_{-1}(U_n)\subset {\bf Q}^2$.  Note
that each set $\pi^{-1}(U_n)$ is an open subset containing the
rational points of the $x$-axis.

Given a positive integer $n$ there is an $\varepsilon_n>0$
such that $N_{\varepsilon_n}(\,(n,0)\,)$ is contained in
$\pi^{-1}(U_n)$.  If we define $W_0$ to be the open set
$$\left(\bigcup_n~(-n - 1,n+1)\times (-\varepsilon_n/2,
\,+\varepsilon_n/2)\,\right)~~\bigcap~~{\bf Q}^2$$
then $W_0$ contains the rational points of the $x$-axis
but for each $n$ we know that $W$ does not contain the
open segment $\{n\}\times (-\varepsilon_n,\,+\varepsilon_n)$, 
which is contained in $\pi^{-1} (U_n)$; it might be useful 
to draw a picture as an aid to understanding this.  Therefore $W_0$
cannot contain any of the open sets $\pi^{-1} (U_n)$, and as
noted above this implies that $QPP$ is not first (or
second) countable.\bull
\bye

